Lilongwe University of Agriculture and

Natural Resources

Bachelor of Science Education

PHY 32102
Oscillation and Waves II

Module Writer
Mfumu Kasakala

Reviewer

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 Copy right
All rights are reserved. No part of this module may be reproduced, stored in retrieval
system or transmitted in any form or by any means, electronic or mechanical, including
photocopying, recording or otherwise without copyright clearance from Malawi University
of Science and Technology.

Lilongwe University of Agriculture and Natural Resources (LUANAR)

Open, Distance and e-Learning
P.O. Box 219
Bunda
Lilongwe
Email: odl@luanar.ac.mw

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 Acknowledgements
Management of the Lilongwe University of Agriculture and Natural Resources
(LUANAR) for its commitment in the establishment of the centre of Open and Distance
Learning (ODL) and also for the continuous support of the activities in the directorate,
including the provision of funding for the development of this module.
All instructional designers for their dedication in development and review of the
modules.

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 Table of Contents
Catalog

Unit 1: Thin lenses and mirrors (plane and spherical mirrors) ..............................................4
1.1 Lenses ....................................................................................................................... 5
1.2 Mirrors ................................................................................................................... 12
Unit 2: Optical Instruments ................................................................................................. 24
2.1 The Eye ...................................................................................................................25
2.2 Camera ....................................................................................................................29
2.3 Simple Magnifier .................................................................................................... 30
2.4 Compound Microscope ...........................................................................................35
1.5 Telescope ................................................................................................................ 39
Unit 3: Simple Harmonic Motion ........................................................................................ 41
Key Terms .................................................................................................................... 41
3.1 Simple Harmonic Oscillator ................................................................................... 42
3.2 Simple Harmonic Oscillator Systems .....................................................................43
3.3 Damped Oscillations ...............................................................................................50
3.4 Free and Forced Oscillations .................................................................................. 54
Unit 4: Sound Waves ........................................................................................................... 61
4.1 Characteristics of Sound ......................................................................................... 62
4.2 Doppler’s Effect ......................................................................................................67
4.3 Waves on a String ...................................................................................................72
4.5 Resonance in air columns ....................................................................................... 78
Unit 5: Physical Optics ........................................................................................................ 81
5.1 Conditions for Interference .....................................................................................82
5.2 Young’s double slit experiment ..............................................................................83
5.3 Thin Film interference ............................................................................................ 89

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 5.4 Newton Rings ......................................................................................................... 93
5.5 Wedge film interference ......................................................................................... 94
5.6 Single slit diffraction .............................................................................................. 97
5.7 Fraunhofer vs. Fresnel diffraction Patterns .......................................................... 101
5.8 Resolving Power ...................................................................................................102
Unit 6: Polarization of Light ..............................................................................................108
6.1 Transverse nature of waves. ................................................................................. 109
6.2 Definition of Polarization ..................................................................................... 110
6.3 Types of Polarization ............................................................................................111
Glossary .............................................................................................................................. 119
References ...........................................................................................................................120
Module Test ........................................................................................................................ 121

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 List of Tables
Table 2.1: Parts of the eye and their functions ..................................................................... 26
Table 2.2: parts of the camera and their functions. ...............................................................30

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 List of Figures
Figure 1.1: convex and concave lenses ...................................................................................5
Figure 1.2: effect of convex and concave lenses on parallel beam of light ............................5
Figure 1.3: representation of lenses on Ray diagrams ............................................................6
Figure 1.4: representing Principal axis on Ray Diagrams ......................................................6
Figure 1.5: representing focal point on Ray diagrams ........................................................... 6
Figure 1.6: Object placed beyond focal point on a Ray diagram ........................................... 7
Figure 1.7: ray passing through optical centre not deflected ..................................................7
Figure 1.8: ray parallel to P.A. deflected by lens ................................................................... 7
Figure 1.9: Image formed where two rays meet ..................................................................... 8
Figure 1.10: image formation on concave lens .......................................................................8
Figure 1.11: compound lens system ..................................................................................... 11
figure 1.12: reflection of light rays on a plane surface ......................................................... 13
figure 1.13: image formation on a plane mirror ................................................................... 13
Figure 1.14: types of spherical mirrors .................................................................................14
Figure 1.15: concave lens ..................................................................................................... 14
figure 1.16: ray 1 ...................................................................................................................15
Figure 1.17: ray 2 ..................................................................................................................15
Figure 1.18: ray .....................................................................................................................16
Figure 1.19: object placed beyond C .................................................................................... 16
Figure 1.20: object placed between F and the convex lens .................................................. 17
figure 1.21: image formation by convex lens ....................................................................... 17
Figure 1.22: Ray 1 for convex lens .......................................................................................18
Figure 1.23: Ray 2 for convex lens .......................................................................................18
Figure 1.24: Ray 3 for convex lens .......................................................................................18
Figure 1.25 : Image formed by convex lens ......................................................................... 19

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Figure 1.26: geometric analysis of a concave lens ............................................................... 19
Figure 2.1: structure of the eye ............................................................................................. 25
Figure 2.2: image formation in the eye for object at a distance ........................................... 27
Figure 2.3: image formation for object near the eye ............................................................ 27
Figure 2.4: correcting shortsightedness using concave lens .................................................28
Figure 2.5: correcting long-sightedness using convex lens .................................................. 28
Figure 2.6: parts of a camera ................................................................................................ 29
figure 2.7: magnifying power of simple magnifier at N.A. ................................................. 31
Figure 2.8: magnifying power of simple magnifier when not at N.A. ................................. 33
Figure 2.7: Microsope ...........................................................................................................35
figure 2.8 .............................................................................................................................. 36
figure 2.9: magnifying power of compound microscope at N.A. .........................................36
Figure 3.1: Forces on a pendulum ........................................................................................ 44
Figure 3.2: Amplitude of a wave .......................................................................................... 46
Figure 3.3: circular vs SHM setup ........................................................................................48
Figure 3.4: Geometric representation of circular vs SHM ................................................... 48
figure 3.5: Trigonometric representations of oscillations .....................................................51
Figure 3.6: Spring oscillating system ................................................................................... 51
Figure 3.7: amplitude as a function of the driving frequency ...............................................57
Figure 4.1: sound detection by moths at separate points ...................................................... 65
Figure 4.2: Moving observer ................................................................................................ 68
Figure 4.3: Moving source ....................................................................................................69
Figure 4.4 ..............................................................................................................................72
Figure 4.5: Harmonics .......................................................................................................... 73
Figure 4.6: vibrating air columns ..........................................................................................75
Figure 4.7: Harmonics in columns closed on one end ..........................................................75
Figure 4.8:Figure 4.7: Harmonics in columns open on both ends ........................................77
Figure 4.9: demonstrating Natural frequency ......................................................................78

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Figure 5.1: constructive and destructive interference ...........................................................83
Figure 5.2: double slit setup ..................................................................................................84
Figure 5.3: formation of dark and bright fringes .................................................................. 85
Figure 5.3: locating exactly where bright fringes appear ..................................................... 86
Figure 5.4: Phase change due to reflection ...........................................................................88
Figure 5.5: rays passing through a thin film ......................................................................... 90
Figure 5.8: formation of newton rings ..................................................................................93
Figure 5.9: light rays passing through a wedge .................................................................... 95
Figure 5.10: interference pattern generated by a wedge .......................................................96
Figure 5.11: interference pattern due to a single slit ............................................................ 97
Figure 5.12: analysis of the slit .............................................................................................98
Figure 5.13: intensity variation in a single slit ................................................................... 100
Figure 5.14: Frensel vs Fraunhofer diffraction ...................................................................102
Figure 5.15: wave illustration of resolving power ..............................................................103
Figure 5.16: graphical illustration of resolving power ....................................................... 104
Figure 5.17: Rayleighs criterion on car headlights .............................................................105
Figure 6.1: generating a Transverse wave .......................................................................... 109
Figure 6.3: Polarization of Electromagnetic waves ............................................................110
Figure 6.4: Representing polarization with the aid of arrows ............................................ 111
Figure 6.5: unpolarized light passing through a polarizer .................................................. 112
Figure 6.6: Polarization by Reflection ................................................................................114
6.7: colour variations due to polarization of sunlight by air molecules ..............................116

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 Module Overview
This module aims to equip learners with the ability to analyze waves and anything that
moves in an oscillatory manner. It begins with an analysis of light and how it is affected by
lenses and mirrors through refraction and reflection respectively. Furthermore, the module
expands on the analysis of how lenses and mirrors can be used in various optical
instruments. This module analyzes light through both geometrical optics and Physical
optics perspectives. The module also dedicates a section sound waves to help learners have
a better understanding of how we perceive sound and how issues of sound quality can be
modeled into mathematical models. The module also dedicates a section to the careful
analysis of simple harmonic oscillators and how their motions can be predicated through
mathematical modelling.

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 Visual Icons
Text in this module is formatted to help you find information quickly and easily. The
visual icons in the left margin will help you locate information as described below.
Icon What the Icon Depicts
Introduction: This is an advance organizer that tells what you will learn
from a unit of study.

Objectives: These define the type of knowledge, skills and attitudes you
should be able to display after going through the lessons in the unit.
Key terms: These are words or phrases which will help you understand
lessons in the unit.
Lessons: This is content you must read and understand to achieve the
stated unit objectives.

Activity: This tells you the tasks you should perform to facilitate your
learning from the unit.
Answers to unit activities: This gives you the suggested answers to the
unit activities. Ensure that you begin by performing the activities on your
own before comparing your answers with the suggested answers.
Further readings: Shows you a suggested list of resources you will need
to study for your deeper understanding of lessons in the unit.

Summary: Reminds you of what you have learnt in a unit.

Unit Test: Examines your understanding of lessons in the unit.

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 Assessment
Students will be required to attend laboratory sessions, continuous assessment exams, end
of semester exams and projects, if need be, laboratory sessions, continuous assessments
and projects, will contribute 40% of the final grade while end of semester exam will
contribute 60%.

Please note: attendance is of paramount importance is this course.

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 Unit 1: Thin lenses and mirrors (plane
and spherical mirrors)
1.0 Introduction
Human beings perceive the natural world through their senses. At the centre of these
senses is the sense of sight. However, for humans to be able to see, it is hugely dependent
on the availability and quality of light. However, light being a wave is affected as it moves
from one medium to another (reflection, refraction e.t.c.). This unit is dedicated to studying
the role of light in producing images of the natural world and how various methods of
manipulating light (through lens and mirrors) affect the way we see our surroundings.

Learning Outcomes
By the end of this Unit, you should be able to:
 State the types of lenses and mirrors
 describe the basic properties of lenses
 describe the process of Image formation by lenses and mirrors
 construct ray diagrams for lenses and mirrors

Key Terms
Ensure that you understand the key terms or phrases used in this unit as listed below:
 Reflect
 Refract
 Divergent
 Convergent
 Virtual Image

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1.1 Lenses

A lens is a transparent material (e.g. glass) bound by one or two curved surfaces used for
concentrating or dispersing light rays. Lenses form images due to refraction of light rays as
they move from one medium to another (air to glass and back to air). There are two types
of lenses, namely; Convex (convergent) and (divergent) concave lenses. A convex lens is
thicker at the centre than the edges. On the other hand, Concave lenses are thicker at the
edges than the centre.

Figure 1.1: convex and concave lenses

Parallel light rays incident on a lens are bent (refracted) as they pass through the lens.
Parallel rays incident on a convex lens refract as they pass through the lens and converge
at a point after the lens. An image is therefore formed at the point which the light rays
converge.

Figure 1.2: effect of convex and concave lenses on parallel beam of light
Conversely, parallel rays incident on a concave lens diverge as they pass through the lens.
The diverged light rays when traced backwards seem to converge at a point before the lens.

1.1.2 Ray diagrams for lenses

Image formation can be illustrated by considering rays (represented by arrows) and tracing
the various paths they take from the object to form a corresponding image. For simplicity
sake, only a few rays are considered when constructing ray diagram.

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1.1.2.1 General Rules for Drawing Ray diagrams
I. In ray diagrams, a thin lens is represented by a straight line at which all the refraction
is considered to occur.

Figure 1.3: representation of lenses on Ray diagrams

II. A line called the principal axis is drawn perpendicular to the plane of the lens, passing
through the optical centre.

Figure 1.4: representing Principal axis on Ray Diagrams

III. The focal points are labeled on either side of the lens with respect to the focal length
of the lens

Figure 1.5: representing focal point on Ray diagrams

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IV. An object on one side of the lens is represented by an arrow:

Figure 1.6: Object placed beyond focal point on a Ray diagram

V. A ray is drawn from the object which passes through optical centre of the lens and is
not deflected

Figure 1.7: ray passing through optical centre not deflected

VI. A second ray is drawn from the same point and is drawn parallel to the principle axis.
This ray bends and passes through the focal point after passing through the lens.

Figure 1.8: ray parallel to P.A. deflected by lens

VII. The point where the two rays cross each other is where the image is formed

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 Figure 1.9: Image formed where two rays meet

Conversely, a ray diagram of an image formed by a concave lens would look like;

Figure 1.10: image formation on concave lens

Since the rays diverge after passing through the lens, the image is located by tracing back
the diverged rays to see where they meet before the lens. The image formed is upright.

NOTE: When ray diagrams are drawn to scale (preferably on a graph paper), they help in
locating the position of the image as well as its size.

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 In each case, state how far the image is formed from the lens as well as whether the
image is magnified or diminished.

1.1.3 The Lens Formula

The lens formula is an expression that gives the relationship between the Image distance
(v), Object distance (u) and the focal length (f).
1 1 1
�
=�+� (1.1)

where the image and object distances are measured from the optical centre of the lens to
the image and object respectively. On the other hand, the focal length (f) is the distance
from the optical centre of the lens to the the focal point (F).

1.1.4 Lateral Magnification

Lateral magnification refers to the ratio of the image size to the object size. It helps us to
know the exact number of times the image is larger or smaller than the object. It can be
calculated by considering the ratio of the image distance to the object distance. Thus:
����� ���� ����� ��������
Magnification, m = or m =
����� ���� ������ ��������
Using only the symbols, the magnification formula can be written;
v
m=
u
(1.2)

Now consider the lens formula as shown in Equation 1.1

1 1 1
= +
� � �
if we multiply the lens formula by image distance (v), we get;
� � �
= +
� � �

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 �
�
= �+1 (1.3)

Making magnification (m) the subject of the formula in Equation 1.3, we get at alternative
way to calculate magnification;
�
� = −1
�

Similarly, multiplying equation 1.1 (lens formula) by the object distance (u), we get
� � �
= +
� � �
� 1
�
=1+
�
(1.4)

Therefore magnification can be calculated using either of the equations (1.2 or 1.3 or 1.4).

An object is placed 12 cm in front of a diverging lens with a focal length of 7.9

cm. Find
(a) the image distance
(b) the magnification.

Solutions:
(a) We can make use of Equation 1.1
1 1 1
= +
� � �
Since we are dealing with a diverging (concave) lens, f =− 7.9 cm
1 1 1
= −
� ( − 7.9) 12
� =− �. � ��

(b) To find the magnification, we’ll make use of Equation 1.2

v 4.8 cm
m= = = �. ��
u 12 cm

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1.1.5 Compound Lens System
In practice, various equipment may use more than one lens to achieve its objective. A
compound Microscope is an example of an apparatus that makes use of two lenses to
achieve a required magnification. In systems that have more than one lens, there is need for
deriving a new expression for the calculation of the effective focal length of the combined
lenses.
Consider a system of two convex lenses combined as shown below:

Figure 1.11: compound lens system

In the absence of lens B, ray OP would pass through point I1 which would produce a real
image of lens A.
If u is the object distance and v1 is the image distance, then using the lens formula, it
follows that:
1 1 1
= + 1
�1 � �
With lens B in position, I1 acts as a virtual object for this lens forming an image at I. This
means that for lens B, the object distance is -v1 and the image distance is v. Thus using the
lens formula again it follows that:
1 1 1
= 1 +
�2 −� �
Combining the two equations through addition, we get:
1 1 1 1 1 1
�1
＋� = � + �1 − �1 + �
2

Hence

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 1 1 1 1
+ = +
�1 �2 � �

But since I is the image of O by refraction through both lenses, then using the initial lens
1 1 1
formula � = � + � where F is the combined focal length for both lenses.
1 1 1
Thus, �1
+� =�
2

Therefore, the expression for the effective focal length in a system of combined lens is:
1 1 1
�
=� +� (1.5)
1 2

NOTE: This formula applies for any two thin lenses that are in contact (whether convex or
concave). However, when the formula is used, the sign convention must be applied.

Since the image formed by the first lens acts as an object for the second lens, it is treated as
a virtual object for the second lens. The same procedure can be extended to the system of
more than two lenses.
Since the magnification due to the second lens is performed on an already magnified image
due to the first lens, the effective magnification of the image due to the combination of the
lenses is the product of the individual magnifications of each lens. Thus a combination of n
lenses will have the magnification:

m = m1× m1× m1×mn (1.6)

1.2 Mirrors

A mirror is simply a reflective surface. Mirrors are commonly made through coating one
side of a thin glass with highly reflective material (e.g. aluminium or silver). A thin layer
of lead is also added on top of the silver/aluminium layer. Mirrors come in different shapes
i.e. plane mirrors and spherical mirrors.

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1.2.1 Plane Mirrors
A plane mirror is simply a flat reflective surface. On a plane mirror, image formation is the
reflection of light when it is incident on the surface of the mirror.

figure 1.12: reflection of light rays on a plane surface

A minimum of two incident rays from a point object to the mirror are required in order to
locate the position of the image using a plain mirror. When produced backwards, the
reflect rays from a mirror appear to meet at a point and that is where the image is formed.

figure 1.13: image formation on a plane mirror

The image formed by the mirror is a virtual image as it cannot be projected onto a screen.
In addition, the distance from the object to the mirror (OD) and the distance from image to
mirror (ID) are always equal (i.e. OD =ID).

1.2.2 Spherical mirrors

Spherical mirrors get their name from the fact that they have the same shape as a section of
a sphere. Spherical mirrors exist in two forms, i.e. concave and convex mirrors.

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 Figure 1.14: types of spherical mirrors

concave mirror are mirrors that are curved inward (fig.1.4a) whereas convex mirrors are
curved inwards (fig. 1.4b).

1.2.3 image formation by concave mirrors

Concave lenses can be distinguished by how curved the surface of the mirror is. The term
radius of curvature is used to describe how curved a spherical mirror is. Consider
concave mirror of optical centre V,centre of curvature C and radius of curvature R :

Figure 1.15: concave lens

Now consider a point source of light placed at point � in (Fig. 1.15b), where O is any point
on the principal axis to the left of �. Two diverging light rays that originate at � are shown.
After reflecting from the mirror, these rays converge and cross at the image point �. They
then continue to diverge from � as if an object were there. If the object O is placed very
very far away from the lens, then its image is formed at the focal point of the lens (i.e. � ≈

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� for object placed very far away). The image formed in this case is a real image (since
light actually passes through the image point).

Ray diagrams can also be constructed to locate an image formed using a concave lens. To
draw a ray diagram, you must know the position of the object and the locations of the
mirror’s focal point (F) and center of curvature (C). The following steps may be followed:

i. Ray 1 is drawn from the top of the object parallel to the principal axis and

is reflected through the focal point F.

figure 1.16: ray 1

Ray 2 is drawn from the top of the object through the focal point and is reflected parallel
to the principal axis.

Figure 1.17: ray 2

ii. Ray 3 is drawn from the top of the object through the center of curvature C and is
reflected back on itself.

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 Figure 1.18: ray
Notice that the image is formed wherever any of the two rays meet. However the third ray
may simply as a check for the construction. However, for different object distances (from
the concave lens), images of various characteristics will be produced. For object placed
beyond centre of curvature (C), the image formed is real, diminished and inverted as
illustrated in figure 1.19.

Figure 1.19: object placed beyond C

However, when the object is placed in between the focal point (F) and the lens, the image
formed is virtual, magnified and upright as illustrate in fig..

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 Figure 1.20: object placed between F and the convex lens

1.2.4 image formation by convex mirrors

A convex mirror reflects light from the outer convex surface. It is sometimes called a
diverging mirror because the rays from any point on an object diverge after reflection as
though they were coming from some point behind the mirror.

figure 1.21: image formation by convex lens

The image in Fig 1.21 is virtual because the reflected rays only appear to originate at the
image point as indicated by the dashed lines. Furthermore, the image is always upright and
smaller than the object.

Ray diagrams can be constructed to determine the position where the image is formed by a
convex mirror using the following steps:

i. Ray 1 is drawn from the top of the object parallel to the principal axis and is

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reflected away from the focal point F.

Figure 1.22: Ray 1 for convex lens

ii. Ray 2 is drawn from the top of the object toward the focal point on the back

side of the mirror and is reflected parallel to the principal axis.

Figure 1.23: Ray 2 for convex lens

iii. Ray 3 is drawn from the top of the object toward the center of curvature C on

the back side of the mirror and is reflected back on itself.

Figure 1.24: Ray 3 for convex lens

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In a convex mirror, the image of an object is always virtual, upright, and reduced in size as
shown in Figure 36.13c. In this case, as the object distance decreases, the virtual image
increases in size and moves away from the focal point toward the mirror as the object
approaches the mirror.

Figure 1.25 : Image formed by convex mirror

1.2.5 The mirror equation

Consider an object at a distance p from a concave mirror of radius of curvature C and
radius of curvature R. we can use fig 1.26 below to calculate the image distance q. By
convection, the distances R,p and q are measured from the centre of curvature V.

Figure 1.26: geometric analysis of a concave mirror

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 ℎ
Using the large red right angled triangle in fig. 1.26, we see that tan � = �. Similarly, using

−ℎ'
the yellow right angled triangle, we see that tan � = �
. In this case, the negative is

introduced because the image is inverted. We can now calculate the magnification as:
h' q
M= h
=− p (1.7)

Taking into account the small green right angled triangle, we’ll see that:
−ℎ'
tan � =
�−�
(1.8)

Similarly, the observing the smaller red right angled triangle, we see that
ℎ
tan � = �−� (1.9)

It therefore, it follows that:

h' �−�'
h
=− �−� (1.10)

we can now compare equations 1.10 and 1.7 and we get:

�−�' q
=p (1.11)
�−�

Which can then be simplified as:

1 1 2
�
+ =
� �
(1.12)

If the object is very far from the mirror (that is, if p is so much greater than R that p can be
said to approach infinity), then 1/� ≈ 0, and Equation 1.12 shows that � ≈ �/2. Earlier
we also discussed that when the object is placed very far away from the concave mirror,
the image is formed at the focal point (F). Therefore, when the object is at infinity:

R
f= 2
(1.13)

Because the focal length is a parameter particular to a given mirror, it can be used to
compare one mirror with another. Combining Equations 1.12 and 1.13, the mirror
equation can be expressed in terms of the focal length:

1 1 1
�
+�=� (1.14)

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1. A spherical mirror has a focal length of 110.0 cm.

(a) Locate and describe the image for an object distance of 25.0 cm.

(b) Locate and describe the image for an object distance of 10.0 cm.

(c) Locate and describe the image for an object distance of 5.00 cm

Solutions:

(a) We’ll make use of Equation 1.14

1 1 1
+ =
� � �
1 1 1
= − = ��. � ��
� 10.0�� 25.0��
To adequately describe the image, we’ll also need to calculate its magnification
using Equation 1.7, hence

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 q 16.7 cm
M =− = = − �. ���
p 25.0 cm
The absolute value of M is less than unity, so the image is smaller than the
object, and the negative sign for M tells us that the image is inverted. Because q
is positive, the image is located on the front side of the mirror and is real.

(b) � = ∞
This result means that rays originating from an object positioned at the focal
point of a mirror are reflected so that the image is formed at an infinite distance
from the mirror; that is, the rays travel parallel to one another after reflection.
Such is the situation in a flashlight or an automobile headlight, where the bulb
filament is placed at the focal point of a reflector, producing a parallel beam of
light.

(c) � = �� �� ��� � =+ �. ��

The negative image distance tells us that the image is virtual and formed on the
side of the lens from which the light is incident, the front side. The image is
enlarged, and the positive sign for M tells us that the image is upright.

Summary
Lenses and mirrors are both apparatus that can be used to form images. Lenses form
images due to refraction of light while mirrors on the other had use the reflecting property
of light to form images.
Ray diagrams can be constructed (to scale) to help predict the characteristics of the images
to be formed by either a mirror or lens. In addition, mathematical calculations can also be
used to predict the characteristics of images formed by the mirrors and lenses (thus the
mirror equation and lens formula).

22
Further Reading
Spherical aberration and how it affects the process of image formation in cameras.

Unit Test
1. A thin lens has a focal length of 25.0 cm. Locate and describe the image when the object
is placed in front of the lens at a distance of:

(a) 26.0 cm

(b) 24.0 cm

2. An object is placed 20cm from a converging lens of focal length 15 cm.

Find the nature, position and magnification of the image.

3. An object is placed 50.0 cm from a concave spherical mirror with focal length of
magnitude 20.0 cm.

(a) Find the location of the image.

(b) What is the magnification of the image?

(c) Is the image real or virtual?

(d) Is the image upright or inverted?

Answers to Unit Activities

1. (a) 650 cm
(b) -600 cm
2. v=60 cm
3. (a) 33 cm in front of the mirror
(b) M= -0.666
(c) the image is real
(d) image is inverted

23
 Unit 2: Optical Instruments
2.0 Introduction
In the preceding unit (unit 1), we thoroughly discussed how lenses are capable of refracting
rays of light which subsequently results in image formation. This unit goes a step further to
help us understand how the same process of image formation is carried out by the human
eye. In addition to the human eye, this unit will aid us to carefully study a variety of optical
instruments designed to either correct or extend our abilities to see the natural world.

Learning Outcomes
By the end of this Unit, you should be able to:
 Describe the working of the human eye
 Describe the working of the Camera
 Describe the working of a simple magnifier
 Describe the working of a compound Microscope
 Describe the working of a Telescope

 Solve the magnifying power of various apparatus

 Explain the idea of resolving power

Key Terms
Ensure that you understand the key terms or phrases used in this unit as listed below:
 Accommodation
 Diverging
 Converging
 Virtual image
 Real image
 Normal adjustment

24
2.1 The Eye

The eye is a sensory organ that reacts to visible light to allow us to see things in our
surroundings. It consists of a nearly spherical ball with a bulge in front. It uses a convex
lens system to form small, inverted and real images of an object in front of it.

Figure 2.1: structure of the eye

Part Function
Cornea It is made out of a fairly dense, jelly like material which provides

protection for the eye. Its also acts as the first point where light
rays start to refract.

Aqueous Humour This is a watery liquid that helps to keep the cornea in a rounded
shape, similar to that of a lens.

Iris This controls the amount of light entering the eye. It opens and
closes to limit the amount of light that enters the eye.

Pupil The gap created by the opening and closing of the iris.

Lens A convex lens that controls the bending of light rays by change
of its shape.

Ciliary muscles These control the thickness of the lens during focusing. By

25
 contracting or squeezing the lens, they make it thicker and

vice versa.

Retina This is the light sensitive part of the eye and it is where images

are formed. It contains millions of tiny cells which are sensitive

to light. The cells send signals along the optic nerve to the brain.

Vitreous Humour This is a jerry-like substance that helps the eye to

keep its round shape. It is very close in optical density to the lens

material.

Yellow Spot This is a small area on the retina where the sharpest image,

that is, the finest detail can be seen.

Optic Nerve This is the nerve that transmits images received by the retina

to the brain for interpretation. The part of the eye where the optic

nerve joins the retina is called the blind spot because no images

can be observed at at this point.

Table 2.1: Parts of the eye and their functions

2.1.1 Accommodation of the Eye

Accommodation refers to the eye’s ability to see near and distant objects. The eye is
capable of focusing objects at different distances by automatic adjustment of the thickness
of the eye lens.

When focusing on distant objects, the ciliary muscles are relaxed which causes the lens to
be thinner which increases the focal length of the lens.

26
 Figure 2.2: image formation in the eye for object at a distance
Conversely, when focusing on near objects, the ciliary muscles contract which cause the
lens to be thicker which decreases its focal length. Thus:

Figure 2.3: image formation for object near the eye

2.1.2 Defects of vision and their corrections
The eye can accommodate a range of objects from far to near distances. As people get
older, the power of accommodation of the eye decreases. However, sight defects are not
limited to inability of the lens to adjust as they may also arise due to the eye ball being too
long or short. The most common defects of vision are short-sightedness (myopia) and long-
sightedness (hypermetropia).

2.1.2.1 Short-sightedness
It is a condition where one can clearly see nearby objects but cannot focus on distant
objects. This defect arises from the eyeball being too long or due to having too much
diffraction at the cornea. Either case results in the image of distant objects being formed in
front of the retina.

27
 (a) (b)
Figure 2.4: correcting shortsightedness using concave lens
This defect can be corrected by wearing a concave lens. The rays of light from the distant
object are diverged so that they appear to come from a point near so that they can now
easily be focused by the eye. This gives the illusion of having a distant object seemingly
nearer.

2.1.2.2 Long-sightedness

This refers to a condition whereby the individual can see distant objects clearly but cannot
focus on nearby objects. This condition is caused by either the eyeball being too short or
defectiveness of the cornea, such that when objects are close, the image is formed behind
the retina.

(a) (b)

Figure 2.5: correcting long-sightedness using convex lens

This defect can be corrected by wearing a convex lens. Rays from a nearby object are
converged such the rays leaving the lens are parallel which gives the illusion that object is
farther than it actually is and therefore can easily be focused by the eye.

Activity 2a:
1. It has been established that the lens in the eye is a convex lens. Explain why image
seeing would be deemed impossible if the lens of the eye was rather concave in nature.

28
2.2 Camera

A camera is a device that is used to take photographs. It is a device whose ability to

construct images is modeled on the human eye.

Figure 2.6: parts of a camera

A camera consists of a convex lens and a light sensitive film enclosed in a dark light-tight
box. The lens focuses rays to form real, inverted and diminished images on the film.
Unlike the eye, the lens on the camera lacks the ability to accommodate. However, the lens
on the camera is movable therefore its position can be physically adjusted for it to focus on
images at various distances to cover for its inability to accommodate.
Part Function
Aperture The space that is created due to the closing and opening of the
diaphragm
Lens focuses light from the object onto a light sensitive film. It is
moved to and fro so that a sharp image is formed on the film.

Diaphragm A set of sliding plates between the lens and the film. It

controls the diameter of a hole through which light passes.

Film This is where the image is formed. It is kept in darkness until the

shutter is opened. It is coated with light sensitive chemicals

29
 which are changed by the different shades and colors in the

image. When the film is processed, these changes are fixed and

the developed film is used to print the photograph.

Shutter Ensures that the camera in normally closed by controlling the

exposure time of the film. It opens and closes quickly to let a
small amount of light into the camera.

Table 2.2: parts of the camera and their functions.

Activity 2b:
1. Identify which parts of the eye have similar functions to the following:
(a) Film
(b) Aperture
(c) Diaphragm

2. The eye has no part which functions similarly to a shutter. Why do you think that is
the case?

2.3 Simple Magnifier

This sometimes referred to as a simple magnifying glass and used to view organisms or
parts of organisms that cannot easily be viewed by the naked eye. It consists of a single
convex lens which is used to produce a virtual, magnified and upright image of the object.
Recall that in unit 1, we were able to deduce that a convex lens only produces a virtual,
upright and magnified image when object is placed at a distance less than the focal length
of the lens. Therefore, when using a simple magnifier, the object is viewed at a distance
less than the focal length of the lens.

30
2.3.1 Magnifying Power of a simple magnifier
Magnifying power of an optical instrument is the ratio of the angle subtended by the image
at the eye (when the microscope is used) to the angle subtended by the object at the
unaided eye when the object is placed at the least distance of distinct vision.
Mathematically it is represented;

β
M=α (2.1)

Where � angle subtended by the image at the eye.

� angle subtended by the object at the unaided eye when the object is placed at the
least distance.

2.3.1.1 Magnification Power at normal adjustment

When a microscope is in normal use, the image is formed at the least distance of distinct
vision therefore least strain is exerted on the eye for getting it focused on the retina. This is
known as normal adjustment. Consider an object of height h placed at a given distance
from the lens as shown in figure 2.7 below;

figure 2.7: magnifying power of simple magnifier at N.A.

Let the β be the angle subtended by the image I to the lens. Therefore, from figure 2.7,

ℎ1
tan � =
�

31
Assuming that the eye is very close the lens such that � is very small, therefore tan � ≈ �,
hence;

ℎ1
β= �
(2.2)

Conversely, suppose that the object is viewed at the near point by the un-aided eye and that
it subtends an angle of α at the eye.

ℎ
tan � =
�

Again, for small �, tan � ≈ �, hence;

ℎ
α=
�
(2.3)

It follows that the magnifying power M is given by;

β
M=
α

h1
h1
M= D =
h h
D
h1
M=
h
(2.4)

h1
But we can also recall that h
also represent the linear magnification produced by a lens or

magnifying glass. Therefore using equation 1.3 (from unit 1), we get

�
�= −1= �
�
�
�= −1
�

Since the image is at the near point (least distance of district vision), the image distance v
is equal to – D, (negative for a virtual image).

−�
�=
�
−1 (2.5)

32
Equation 2.5 gives the maximum magnifying power of a simple magnifier (simple
microscope).

Note that in calculations, the value of the magnifying power is negative. The negative sign
can always be neglected since magnification cannot be negative. The object distance can
take any value in the range from the focal point to the point where it lies at the near point
and if the object is at the focal point, then the object distance is equal to the focal length
and the image is at infinity, and the microscope is not in normal adjustment.

2.3.1.2 Magnification Power of a microscope when it is not on normal

adjustment

When the microscope is not working on normal adjustment, them the image is formed at
infinity, hence the object distance is equal to the focal length.

Figure 2.8: magnifying power of simple magnifier when not at N.A.

In this case, angular magnification is still given by the formula;

β
M=
α

From figure 2.8;

ℎ
tan � =
�

ℎ1
tan � =
�

33
Again, assuming that the eye is very close the lens such that � is very small, therefore
tan � ≈ �, hence;

ℎ
β= � (2.6)

Conversely, using the same figure 2.8 while neglecting the use of the microscope (unaided
eye).

ℎ
tan � =
�

And for small �, tan � ≈ �, hence;

ℎ
α= � (2.7)

It therefore follows that angular magnification M,

h
D
M= f =
h f
D
D
�= f
(2.8)

This is the minimum magnifying power of the simple microscope. Note that, in this case,
D is positive since it is of a real image from the eye, and from the formula, angular
magnification is high for a lens of short local length.

34
2.4 Compound Microscope

We have already studied how a single lens can be used in

magnifying objects. However, some objects are too small
and require a higher magnifying power to been seen. In such
situations, two lenses are used. Using two lenses to attain a
higher magnification power forms the basis of the
compound microscope.
A compound microscope (figure 2.7) consists of two convex Figure 2.7: Microsope
lenses of short focal lengths referred to as the objective and
the eye piece.

The objective is nearest to the object and the eye piece is nearest to the eye of the observer.
The objective forms a real, magnified and inverted image. To achieve this, the object to be
viewed is placed at a distance just beyond the focal point of the objective. The real,
magnified and inverted image formed by the objective is formed at a point inside the eye
piece. In this case, the image from the objective then acts as an object for the eye piece
which then produces a virtual and magnified image. Therefore, the viewer looking through
the eye piece sees a magnified virtual image of a picture formed by the objective. Figure
2.8 below shows the schematic diagram of the image formation process in the compound
microscope.

35
 figure 2.8
An objective lens L1 forms a real magnified image I, of an object O just placed outside its
principal focus F0 . I, is formed just inside the principal focus Fe of the eye piece L2 , which
acts as a magnifying glass and produces a magnified, virtual image I2 of I1.

1.4.1 Magnifying Power of a Compound Microscope under

normal adjustment
As earlier discussed, when the eyes are relaxed, the image is at the near point and the
compound microscope is said to be in normal adjustment. We have already seen (figure 2.8)
that when a microscope is in normal use, the image I2 is formed at the least distance of
distinct vision, D from the eye. Thus v = D.

figure 2.9: magnifying power of compound microscope at N.A.

Suppose that the final image has a height h2 and is formed at a distance v from the eye
piece and that it subtends an angle β to the eye. The magnifying power will be again
calculated using Equation 2.1:

36
 β
M=
α

Where the β is the angle subtended by the image I to the eye piece and is considered to be
very small, such that tan β ≈ β. Careful analysis of figure 2.9 then shows us that:

h2
β≈ D
(2.9)

Similarly where the α represents the angle when viewed at the near point by the un-aided
eye and is considered to be very small, such that tan α ≈ α. Figure 2.9 then shows us that:

h
α≈
D

Therefore the magnifying power (M) will be calculated:

h2
h2
M= D
h =
h
(2.10)
D

We can now use a bit of mathematical manipulation to introduce the the height of the
��
image due to the objective (h1) by multiplying equation 2.10 by ��
, hence:

h2 ℎ
M= h
× ℎ1 (2.11)
1

Rearranging Equation 2.11, we get:

h2 ℎ1
M= ×
ℎ1 ℎ

�� ��
But we know that ��
is the linear magnification, me due to the eye piece and �
is the

linear magnification, m0 due to the objective, thus Equation 2.11 can be represented as:

M = me × m0 (2.12)

v
We have already seen that linear magnification is also given by m = f − 1 where v is the

image distance from the lens and f is the focal length.

37
 v0
It follows that linear magnification due to the objective lens, m0 = f
− 1, and that due to
ve
the eye piece, me = f
− 1. Therefore, we can alternatively represent equation 2.12 as:

ve v0
M=
f
−1
f
−1 (2.13)

1.4.2 Magnifying Power of a Compound Microscope when not

under normal adjustment

To have the microscope not function in normal adjustment, then the final image is formed
at infinity (i.e. v = ∞). Suppose that an object of height h is at a given position from the

objective lens, forming an image of height h1.

Therefore, the magnifying power will be again calculated using Equation 2.1:

β
M=
α

From figure 2.9 we can see that:

ℎ ℎ
tan � = �1 and tan � = �
�

For every v=-D very small, tan � ≈ � and tan � ≈ �, hence

ℎ1 ℎ
β= ��
and α = �

Substituting these into equation 2.1, we get:

h1
fe
M= h
D

h1 D h D
M= fe
× h = h1 × (2.14)
fe

h1 v0
But h
is the linear magnification of the objective which is also given by f
−1

Hence the magnifying power can be expressed as:

38
 D v0
M=
fe f
−1 (2.15)

Activity 1c:
A compound microscope has an eye piece of focal length 2.50cm and an
objective of focal length 1.60cm. If the distance between the objective and eye
piece is 22.1cm, calculate the magnifying power produced when the final image
is at infinity.

Solution:
If the final image is at infinity, the objective forms an image at the focal point of
the eye piece.

Let �� be the focal length of the eye piece and �0 of the objective.

�� = (separation) – (focal length of eye piece)

�� = (22.1 – 2.50) cm = 19.5 ��

Now we can make use of Equation 2.15

D v0
M= −1
fe f
But for a normal eye D = 25 cm, hence

1.5 Telescope

A telescope is an instrument that is used to view objects that are not necessarily small but
are hard to view with the naked eye simply because they are very far away. Distant objects
are difficult to see because light from them has spread out by the time it reaches the eyes,

39
and since our eyes are too small to gather much light. There are two main types of
Telescopes, namely; Refracting telescopes and Reflecting Telescopes.

Summary
Optical instruments help us improve the way we see our surroundings. The careful
understanding of how different types of lenses work in image formation is paramount to
coming up with various optical technologies by understanding their characteristics (focal
length, magnifying power, e.t.c.).
Different optical instruments are specialized to various functions including, aiding us to
see things that are too far to see with the naked eye (i.e. telescope) as well as those that are
simply too small to see (i.e. microscopes).

Further Reading
The various types of telescopes and how their magnifying power can be calculated.

Unit Test
1. Describe any two similarities between the Eye and the Camera.
2. Describe any two differences between the Eye and the Camera.
3. Describe any one similarity between the compound microscope and the Telescope in
terms of their functionality.
4. Describe any one difference between the compound microscope and the Telescope in
terms of their functionality.

40
 Unit 3: Simple Harmonic Motion
3.0 Introduction
This unit involves the careful study of systems that have periodic motion. This unit helps
us understand how phenomena that is periodic in nature can accurately be modeled with
respect to wave motion or rotational motion. The knowledge gained from such analysis of
oscillatory systems helps inform decisions in various engineering disciplines, e.g. When
designing bridges, when designing suspension systems of cars e.t.c.

Learning Outcomes
By the end of this Unit, you should be able to:
 Define a simple harmonic oscillator
 Describe examples of simple harmonic oscillating systems
 Describe free vibrations and forced vibrations
 Describe resonance and damped oscillations

Key Terms
Ensure that you understand the key terms or phrases used in this unit as listed below:
equilibrium
 Damped
 Resonance
 Oscillation
 Restoration
 Retardation

41
3.1 Simple Harmonic Oscillator

The motion which repeats itself after a regular interval of time is called periodic motion.
However, if a body in periodic motion moves along the same path to and fro about a
definite point (equilibrium position), then the body is said to be in oscillatory motion. All
objects that oscillate have one thing in common : each object is subjected to a restoring
force that increases with the increase in displacement from the equilibrium position.

A particle is said to execute Simple Harmonic Motion (S.H.M.) if the oscillatory motion of
the particle is in a straight line such that its displacement from the equilibrium position
varies sinusoidally with time. Since this is the simplest type of periodic motion (or
harmonic motion), it is called simple harmonic motion,

3.1.1 Characteristics of Simple Harmonic Motion

Displacement refers to the vector which shows the change in position from an object’s
equilibrium position. In oscillatory motion, displacement can be measure as length (in
metres) or as angular displacement (in radians).
Amplitude refers to the maximum displacement from the equilibrium position.
Amplitude is always taken to be positive and is measured in metres.
Wavelength refers to the distance between two points that are oscillating in phase on a
wave.
Period refers to the time it takes to complete one cycle. it is measured in seconds (s).
Frequency refers to the number of cycles completed per on second.

42
3.2 Simple Harmonic Oscillator Systems

We can identify systems that we use in our everyday lives that follow some kind of
periodic motion. Examples of such systems include but are not limited to; mass-spring
system, simple pendulum and particle moving in a circular manner. Each of these systems
can be observed to follow some kind of oscillatory motion and can be analyzed separately.

3.2.1. Mass-spring system

Consider a mass m connected to the end of a spring on a friction-less surface. When no
force is applied to the spring to neither stretch or compress it, it is said be it’s equilibrium
position. When at equilibrium position, mass m connected to the spring has not been
displaced therefore is considered to be at position x = 0. If mass m is displaced from this
equilibrium position, the spring will exert a force on the mass that is proportional to its
position as given by Hooke’s law:
�� =− �� (3.1)
This force is also referred to as the restoring force since it is always directed towards the
equilibrium position and therefore always opposite to the direction of displacement.
When the object is displaced from its equilibrium position, it undergoes an acceleration
due to the restoration force as given by Newton’s 2nd law of motion.
�� = ��� (3.2)
Therefore combining equations 3.1 and 3.2 we get,
−kx = max (3.3)
k
ax =− m
x (3.4)

hence, just as the force, the acceleration is also directly proportional to the position but
directed opposite to the direction of displacement of the block from equilibrium.
It is worth noting that an object is said to move with simple harmonic motion whenever its
acceleration is proportional to its position and is oppositely directed to the displacement
from equilibrium (i.e. follows the relationship in equation 3.4). Therefore equation a can
be summarized as:
ax ∝ x (3.5)

43
3.2.2. A Pendulum System
Consider a bob of mass m attached to a string of length L that is fixed at the upper end. The
motion happens in the vertical plane and is driven by the gravitational force.

Figure 3.1: Forces on a pendulum

The bob experiences two forces: Tension (T) and force due to gravity (mg). The tangential
component of the gravitational force mg sinθ always acts towards the equilibrium (θ=0).
Since the tangential component of the gravitational force acts towards the equilibrium, it
can also be considered as the restoring force, hence, using newton’s 2nd law
F = ma
F = − mgsinθ (3.6)
if the angle θ is very small, then sinθ ≈ θ, therefore the restoring force can be written as:
F = − mgθ (3.7)
d2 s
But we know that acceleration is the 2nd derivative of displacement a = dt2
, therefore
d2 s
F = ma → mgθ = m dt2 (3.8)
d2s
gθ= dt2 (3.9)

We also know that � = �� therefore:

d2 θ
gθ = L dt2 (3.10)
d2 θ g g
dt2
= L
θ→ a = L
θ (3.11)

44
Which also has the form � � θ. Needless to say that a pendulum also follows simple
harmonic motion since its acceleration is proportional to its position and is oppositely
directed to the displacement from equilibrium.

3.2.3. General Analysis Model For SHM

Consider the relationship between acceleration and displacement of the spring system
(equation 3.4). Since acceleration can also be represented as the 2nd derivative of
displacement, hence
k d2x k
a =− x → dt2
=− m
x
m

d2x k
dt2
=− m
x (3.12)
�
Substituting ω2 = � in equation above, we get;
d2 x
dt2
=− ω2 x (3.13)

In this case, equation 3.13 is a 2nd order differential equation with a solution of the form:
x(t) = A cos (ωt + ∅) (3.14)
where A = Amplitude
� = Angular frequency
∅ = Phase constant

To show that equation 3.14 is a solution that satisfies equation 3.12 :

��
��
= − ωA sin(ωt + ∅) (3.15)
�2 �
��2
= − ω2 A cos(ωt + ∅) = − ω2 x (3.16)

In this case, the Amplitude (A) represents the maximum displacement of the oscillation.

45
 Figure 3.2: Amplitude of a wave

The Angular frequency (�) on the other hand is measure of how rapidly the oscillations
are occurring. It has the units of radians per second and can be calculated using the
expressions:

For spring systems

k
ω=
m

g For Pendulum systems

ω=
L

The Phase constant (∅) is sometimes refereed to as the initial phase angle and is a
quantity that is determined by the position and velocity of the oscillation (or wave) at t=0s.
If the particle is at its maximum position x=A at t =0, then the phase constant is ∅=0,
hence our solution as defined in equation() simply reduces to:

x(t) = A cos (ωt) (3.17)

Note that the solution in equation 3.17 applies if and only if ∅=0.

If equation 3.14 is plotted on a graph, we notice that x(t) is a periodic function whose
value is the same whenever ωt increases by 2π radians. Now we can further investigate the
mathematical model for the simple harmonic oscillators by considering their period,
frequency, velocity and acceleration.

The Period T of the motion is the time interval required for the particle to go through one
full cycle of its motion. In this case a full cycle is covered when ωt increases by 2π
radians, hence

ωT= 2π

��
∴�= �
(3.18)

On the contrary, Frequency is simply the inverse of the Period, hence:

46
 1 �
f= = (3.19)
T ��

Rearranging equations 3.18 and 3.19, we get the alternative equivalent expressions for
angular frequency as:

2π
�= → � = 2πf (3.20)
�

Velocity of a particle undergoing simple harmonic motion can be calculated as a 1st

derivative of displacement (see equation 3.21). Conversely, the 2nd derivative of
displacement will also give us an expression for acceleration of a particle undergoing SHM
(see equation 3.23).

��
v= ��
= - ωA sin(ωt + ∅) (3.21)

�2 �
a= ��2
= - ω2 A cos(ωt + ∅) (3.22)

The maximum value for velocity is calculated when sin (ωt + ∅) = 1. Similarly, the
maximum acceleration for a particle undergoing SHM is calculated when cos(ωt + ∅) = 1.

���� = �A (3.23)

���� = �� A (3.24)

3.2.4. Particle moving around a circle

47
It is sometimes helpful to analyse simple harmonic motion in
terms of uniform circular motion. Consider a ball connected to
a turn-able rim of radius A as shown in figure 3.2. When the
rim is illuminated with a lamp from above, a shadow of the
ball is cast on the screen below. If the rim is rotated, the ball
rotates with it, consequently causing the shadow of the ball to
move back and forth in a simple harmonic motion fashion.
This arrangement shows the direct similarity between circular
motion and S.H.M.

Now consider figure 3.4 which a geometric diagram

Figure 3.3: circular vs
representing the motion in figure 3.3.
SHM setup

Figure 3.4: Geometric representation of circular vs SHM

The dotted circular path represents the motion of the ball as the rim rotates. This dotted
circular path is sometimes called the reference circle for the purposes of comparing
uniform circular motion to SHM. The x-axis on the other hand represents the various
positions of the shadow of the ball on the screen with respect to the rotation of the rim. Let
Point P on the circumference of the reference circle represent the starting position of the
ball’s rotation at time, with line OP making an angle ∅ with the x-axis at time t = 0s.

If the particle moves along the circle with constant angular speed ω, until line OP makes
an angle θ with the x-axis (figure 3.4(b)) at some time � > 0, then the angle between OP
and the x-axis is θ = ωt + ∅. Consequently, as the particle moves along the circle, the

48
projection of P on the x axis, labeled point Q, moves back and forth along the x axis
between the limits x =± A.

Notice that point P (on the circumference) and point Q (on the x-axis) always share the
same x-coordinate. Analysis of triangle OPQ shows that the x-coordinate

x(t) = cos (ωt + ∅) (3.25)

Equation 3.25 is just similar to Equation 3.14 which shows that the particle Q moves with
simple harmonic motion along the x-axis. Therefore, the motion of an object described by
the analysis model of a particle in simple harmonic motion along a straight line can be
represented by the projection of an object that can be modeled as a particle in uniform
circular motion along a diameter of a reference circle.

The geometric interpretation as seen in figure 3.4, shows that the period T for one complete
revolution on the reference circle equals the period of of motion of the SHM between x =
± A. Therefore the angular speed ω of point P around the reference circle equals the
angular speed of the SHM along the x-axis. You will recall that the relationship between
linear velocity (v) and angular velocity ( ω ) is v = rω. Therefore the magnitude of the
linear velocity of particle P can be regarded �� (where A is the radius of the reference
circle). From the geometry in figure 3.4(c) we see that the x-component of this velocity is
−ωA sin (ωt + ∅). On the other hand, the velocity of particle Q, by definition is given by
��
��
. In this case, differentiating Equation 3.25 with respect to time, w see that that the
��
velocity of particle Q is is the same as the x-component of the velocity of P (i.e. v(t)= ��

=−ωA sin (ωt + ∅)).

In similar fashion, the acceleration of point P on the reference circle is directed towards the
�2
centre O with magnitude �
= �2 �. From figure 3.4(d) we can deduce the x-component of

the acceleration of P as −ω2 A sin (ωt + ∅). Similarly, this value is also the acceleration
of particle Q along the x-axis which can be verified by taking the 2nd derivative of
Equation 3.25.

49
 Activity 3a:
1. A 200g block connected to a light spring for which the force constant is 5.00
N/m is free to oscillate on a friction-less, horizontal surface. The block is
displaced 5.00 cm from equilibrium and released from rest.

(a) Find the period of its motion

(b) Determine the maximum speed of the block.

Solution:

(a) Since the mass has been connected to a spring system, hence ω can be
calculated

k 5.00 N/M
ω= = = 5.00 rad/s
m 200 × 10−3 kg

The period can then be calculated by making use of Equation 3.18

2π 2π
T= =
ω 5.00 rad/s

T = �. ���

3.3 Damped Oscillations

As earlier stated, we model simple harmonic motion using trigonometry functions (sine
and cosine) as shown in fig.3.5a :

50
 a b
figure 3.5: Trigonometric representations of oscillations
However, this sort of model presents the oscillatory motion as ideal, that is, systems that
oscillate indefinitely under the action of only one force, a linear restoring force. In reality,
oscillatory system do not oscillate indefinitely and are constantly faced with non
conservative forces such us friction and air resistance that retard the motion of the system.
Consequently, the total energy of the system diminishes over time and the motion is said to
be damped. Therefore, a more accurate way of modelling of oscillatory systems in reality
is by considering them as damped oscillations as depicted in Fig. 3.5b.

Mathematically, we can model SHM systems by considering a

mass-spring system immersed in a viscous fluid as shown in fig.

As the system is oscillating, the viscous fluid provides a

retarding force which is proportional to the speed of the moving
object and acts in the direction opposite to the velocity of the
object with respect to the medium, hence:

R =− bv �ℎ��� � is a constant called

the retarding coefficient

Figure 3.6: Spring

oscillating system
The system will now be faced with both the restoration force ( F =− kx) and the
retardation force (R =− bv). Therefore the sum of the forces can be represented as:

�� = − kx −bvx = m��

51
 �� �2�
−kx − b �� = m ��2 (3.26)

Equation 3.26 is a 2nd order differential equation with the solution:

b
x = A e− 2m
t
cos (ωt + ∅) (3.27)

Where the angular frequency (ω) is :

k b 2
ω = m
− 2m
(3.28)

This result (equation 3.28) can conveniently be expressed as:

b 2
ω = ω0 2 − 2m
(3.29)

k
Where ω0 = m
represents the natural frequency of the oscillator. Natural frequency

refers to the angular frequency of a system in the absence of the retarding force (the
undamped oscillation).

52
Activity 3b:
1. A 10.6 kg object oscillates at the end of a vertical spring that has a spring
constant of 2.05 × 104 N/m. The effect of air resistance is represented by the
damping coefficient � = 3.00 �s/m.

(a) Calculate the frequency of the damped oscillation.

(b) By what percentage does the amplitude of the oscillation decrease in each
cycle?

Solutions:
(a) We can make use of Equation 3.29
2
2
b
ω = ω0 −
2m
Where �0 can be calculated,
� 2.05 × 104 �/�
�0 = = = 44.0�−1
� 10.6 ��
Therefore,
2
3.00Ns/m
ω = 44.0s−1 2 −
2(10.6 kg)
ω = 44.0s−1
ω 44.0s−1
f= =
2π 2π
f = �. �� ��

53
 (b) Here, we make use of Equation 3.27
b
x = A e− 2m t cos (ωt + ∅)
However, we’ll use the following reasoning:
b
2π
In x = A e− 2m
t
cos (ωt + ∅) over one cycle, a time T = , the amplitude
ω
�2�
−
changes from A0 to �0 � 2�� for a fractional decrease of:
��
�0 − �0 �− �� �� 3�
= 1 − �0 �− �� = 1 − �0�− 10.6×44.0
�0
1 − 0.979 98 = 0.0200
= �. ��%

3.4 Free and Forced Oscillations

Free oscillation refers to when an oscillating body is made to vibrate with its own natural
frequency, with no aid from any external forces. The frequency of freely oscillating bodies
depend on the intrinsic properties of the body (e.g. shape, elasticity). In the absence of
frictional forces, the amplitude of free oscillations would remain constant (undamped
oscillations). However, in actual practice, the amplitude of free oscillations decreases with
time due to the presence of friction force. When whole of the originally imparted energy to
the body is used up in overcoming frictional forces, the body stops executing oscillations
and comes to a rest at its mean position.
On the contrary, it is possible to compensate for the energy decrease in damped oscillations
by applying a periodic external force that does work on the system. When a body is
maintained in a state of oscillations by an external periodic force of frequency other than
the natural frequency of the body, the oscillations are referred to as forced oscillations. In
this case, positive work is done on the system if the applied force is in the direction of
motion of the system.
A common example of a forced oscillator is a damped oscillator driven by an external
force that varies periodically, such as F(t) = F0 sin ωt, where F0 is a constants and ω is the

54
angular frequency of the driving force. In this case, the resultant force of a forced
oscillating system will involve the summation of e forces (i.e. restoration, retarding and
driving forces).

�� = F0 sin ωt − kx −bvx = m��

�� �2�
F0 sin ωt −kx − b �� = m ��2 (3.30)

The general solution for equation 3.29 is rather complicated and we would rather consider
solution in special cases.

Consider an object initially at rest and a driving force beings to act on the object. The
driving force leads to increase in the amplitude of oscillation. After a sufficiently long
period of time, when the energy input per cycle from the driving force equals the amount
of mechanical energy transformed to internal energy for each cycle, a steady-state
condition is reached in which the oscillations proceed with constant amplitude. In this
particular case, the solution for equation 3.30 takes the form:

x = A cos ωt (3.31)

Where
F0

A= m
(3.32)
2 bω 2
ω2 −ω20 +
m

�
Where ω0 = �
is the natural frequency of the undamped oscillator (b = 0). Equations 3.31

and 3.32 show that forced oscillator vibrates at a frequency of the driving force and the
amplitude is constant for a given driving force since it is being driven in a steady state by
an external force.

A special situation occurs when the frequency of the driving force nears the natural
frequency of the oscillation (when ω≈ ω0 ) and this results in a dramatic increase in the
amplitude of oscillation.

55
Resonance is the term assigned to the dramatic increase in the amplitude of an oscillator
as the frequency of the driving force approaches the natural frequency of the oscillator.
Therefore at resonance:
F
�
A= m
= ��0 (3.33)
bω 2
m

The reason for the high amplitude oscillation at resonance frequency is because the energy
is being transferred to the system under the most favourable conditions.

To understand this better, consider the 1st derivative of equation 3.31, hence;

�� �
= A cos ωt
�� ��
��
We see that ��
an expression of velocity (v) of the oscillator.

We find that v is proportional to sin ωt, which is the same trigonometric function as that

describing the driving force (F0 sin ωt). In this case, we can see that the driving force (F)
and and the velocity are in phase. You may wish to recall that the rate at which work is
done on the oscillator by force (F) is the dot product of the force and velocity (P = F ∙ v).
You may also wish to recall that the dot product F ∙ v is maximum when F and v are in
phase. we can conclude that at resonance, the applied force is in phase with the velocity
and the power transferred to the oscillator is a maximum.

If we plot amplitude as a function of the driving

frequency of an oscillator with and without damping
(Figure ). From the graph, we can deduce the inverse
proportional relationship between amplitude and
damping: that is decrease in damping (b → 0) results
in increase in the resonance amplitude.

56
 Figure 3.7: amplitude as a function
of the driving frequency

Activity 3 c:
1. A block weighing 40.0N is suspended from a spring that has a force constant
of 200 N/m. The system is undamped (b=0) and is subjected to a harmonic
driving force of frequency 10.0 Hz, resulting in a forced motion amplitude of
2.00 cm. Determine the maximum value of the driving force.

Solution:

We can make use of Equation 3.32

F0
A= m
2
2 bω
ω2 − ω20 +
m

In this case, we are told that the system is undamped (b=0), hence:

F0
A= m
2
ω2 − ω20

Making F0 subject, we get

2
F0 = mA ω2 − ω20
Where the driving frequency (�2 ) can be calculated:
�2 = (2πf)2 = (2π × 10s−1 )2 = 3.95 × 103 s−2
Similarly, the natural frequency(�20 ) can be calculated:
� 200 ��
�20 = = = 49.0�−2
� 40.0�
9.81�/�2

57
 2 40.0N
F0 = mA ω2 − ω20 = 200 × 10−2 m 3.95 × 103 s−2 − 49.0�−2
9.81m/s2
= ��� �

Summary
SHM involves the study of motion of objects whose motion pattern is periodic in nature.
SHM can be easily studied through simple spring-mass systems and pendulum systems.
Analysis of both spring and pendulum systems helps us derive a general equation for
simple harmonic oscillators.
The relationship between SHM and circular motion also allows us to easily understand
SHM by modelling them as circular motion.

Further Reading
Energy of a Simple harmonic Oscillator.

58
Unit Test
1. A 0.60kg block attached to a spring with force constant 130 N/m is free to move on a
friction-less, horizontal surface as in Figure 15.1. The block is released from rest when the
spring is stretched 0.13 m. At the instant the block is released, find

(a) the force on the block

(b) its acceleration.

2. A vertical spring stretches 3.9 cm when a 10g object is hung from it. The object is
replaced with a block of mass 25 g that oscillates up and down in simple harmonic motion.
Calculate the period of motion.

3. A piston in a gasoline engine is in simple harmonic motion. The engine is running at the
rate of 3,600 rev/min. Taking the extremes of its position relative to its center point as
65.00 cm, find the magnitudes of the

(a) maximum velocity

(b) maximum acceleration of the piston.

4. As you enter a fine restaurant, you realize that you have accidentally brought a small
electronic timer from home instead of your cell phone. In frustration, you drop the timer
into a side pocket of your suit coat, not realizing that the timer is operating. The arm of
your chair presses the light cloth of your coat against your body at one spot. Fabric with a
length L hangs freely below that spot, with the timer at the bottom. At one point during
your dinner, the timer goes off and a buzzer and a vibrator turn on and off with a frequency

of 1.50 Hz. It makes the hanging part of your coat swing back and forth with remarkably
large amplitude, drawing everyone’s attention. Find the value of L.

59
Answers to Unit Activities
1. (a) -17N
(b) -28m/s
2. 0.63s
3. (a) 18.8m/s
(b)7.11 km/s2
4. 11.0 cm

60
 Unit 4: Sound Waves
4.0 Introduction
In the preceding unit (unit 3), we have carefully analyzed the characteristics of single
objects moving in an oscillatory manner. In this unit, we extend the study of a single
oscillator to the behavior of a series of oscillators that are connected to one another.
Connecting the oscillators leads to an assortment of new phenomena, including sound
waves which we are able to perceive through hearing. The knowledge from this unit will
help us understand various phenomena, e.g. why different string musical instruments are
made in varying sizes (guitar, violin, banjo,harp, e.t.c)

Learning Outcomes
By the end of this Unit, you should be able to:
 Define the different characteristics of sound (quality, frequency,intensity, pitch and
tone)
 Describe the Doppler’s effect
 Describe the fundamental tone
 Describe Harmonics
 Discuss sound waves in pipes and strings
 Describe resonance in air columns

Key Terms
Ensure that you understand the key terms or phrases used in this unit as listed below:
 Infrasonic
 Ultrasonic
 First Harmonic
 Natural Frequency
 Resonance

61
4.1 Characteristics of Sound

Sound is a from of energy transfer propagated through longitudinal waves. Sound can be
characterized by a number of factors namely: frequency, pitch, intensity, tone and Timbre.

4.1.1 Frequency
The term frequency refers to the number of complete cycles per second. The frequency of
sound is measured in hertz and has an inverse relationship with the sound’s period:
1
f=T (4.1)

There exists a range of sound frequencies which are audible to human beings. The human
audible frequency ranges from 20 Hz to 20,000 Hz (20kHz). Any sound with a frequency
20 Hz is called infrasonic while any sound with frequency above 20 kHz is called
ultrasonic. Though we are unable to hear ultrasonic sounds, they commonly occur in
nature and have various applications in different technologies. Uses of Ultrasonic sounds
include, geological mapping, medical imaging of foetus, navigation systems of ships and
submarines as well as bats and dolphins also use ultrasonic sound as a medium of
communication.
Sound being a wave is also governed by the wave equation:
v = fλ (4.2)
Where:
� is the wavelength
f is the frequency
v is the velocity
However, it should be noted that despite having a wide range of frequencies for sound, the
speed of sound in air in air is the same for all frequencies. For example, if the frequency of
a sound wave is doubled, its wavelength is halved so that the speed (v) in Equation 4.2
remains constant.

62
4.1.2 Pitch
Pitch is the characteristic that enables us to distinguish a sharp note from a hoarse one. For
example, voices of women and children tend to be of higher pitch than of men. The main
factor that affects the pitch of sound is the sound’s frequency. The relationship between the
pitch and frequency of sound is a direct proportional one (i.e. sounds of higher frequencies
will have a higher pitch). Making reference to Equation 4.2, we can therefore derive the
relationship between the pitch and wavelength as an inverse proportional relationship (i.e.
sounds with longer wavelengths will have a lower pitch).

4.1.3 Intensity
In simple terms, intensity refers to the loudness of sound. Alternatively intensity can be
defined as the quantity of energy received per second on a unit area. If the energy E passes
through and area A at time t, then intensity I is given by:
E
I=
At
(4.3)

Recalling that power (P) is given by the formula:

E
P= t
(4.4)

Equation 4.3 can then be expressed in terms of equation 4.4 to give:

Pt P
I=
At
=A (4.5)

Equation 4.5 then gives us an idea of the SI units for the intensity of sound as
Watts/(metres)2 (W/m2).

4.1.3.1 Dependence of Intensity on Amplitude

There exists a direct proportional relationship between intensity and the amplitude of a
vibrating body. The intensity (I) of a wave is proportional to the square of the wave
amplitude (A). We can therefore relate the amplitude quantitatively to the intensity I or
sound level β as follows:

1 I
A=
πf 2ρν
(4.6)

63
4.1.3.2 Dependence of Intensity on distance
You may have noticed that when you focus on listening to a sound source (person
speaking/music on a speaker), the loudness of the sound tends to decrease the further away
you move from the sound source. The reason for this reduction in intensity is simply that
the energy emitted per time by the source spreads out over a larger area. Therefore, it is
safe for us to conclude that the relationship between intensity of sound and distance from
source is an inverse proportional relationship. In particular, the intensity (I) of sound is
inversely proportional to the square of the distance (r) i.e.

1
I ∝ r2 (4.7)

The proportionality relationship in Equation 4.7 can then be written as:

P
I = 4πr2 (4.8)

Where

P is the power

r is the distance from the sound source

This result assumes that no sound is reflected, that no sound is absorbed, and that the
sound propagates outward spherically. In writing this expression, we have used the fact
that the area of a sphere of radius r is � = 4πr2 .

Now suppose we have a situation where sound waves produced by a bat are detected by
two moths at two separate points in space as shown by figure 4.1

64
 Figure 4.1: sound detection by moths at separate points
In this case, the bat acts as the source whereas the moths act as observers at positions r1 and
r2. The waves emanating from the bat propagate outward spherically, with the wave crests
forming a series of concentric spheres. If we assume no reflections of sound, and a power
output by the bat equal to P, the intensity detected by the first moth is:

P
I=
4πr21

Likewise, the second moth hears the same sound at the intensity:

P
I=
4πr22

In each case the value of the magnitude of the power of the sound is the same. Therefore
solving the intensity of the sound detected by moth 2 in terms of intensity at moth 1, we
get:

r1 2
I2 = r2
I1 (4.9)

In other words, the intensity falls off with the square of the distance; doubling the distance

reduces the intensity by a factor of four.

65
4.1.3.2.1 Intensity Level and Decibels
As a measure of loudness of sound, the term intensity level (not to be confused with
intensity) is used. Intensity level (β) is defined as follows:
I
β = (10dB)log
I0
(4.10)

In this expression, log indicates the logarithm to the base 10, and I0 is the intensity of the
faintest sounds that can be heard. Experiments show this lowest detectable intensity to be:

I0 = 10−12 W/m2

When using Equation 4.8, you’ll notice that intensity level (β) is a dimensionless quantity
since the only dimensions that enter the equation are those of intensity and they cancel out.
However, it is more convenient to label the values of intensity level with a name. The
name bel is to indicate the various values of the intensity levels (in honor of Alexander
Graham Bell). Since bel itself is rather a large unit, it is more common to measure (β) units
that are one-tenth of a bel (hence the units decibels (dB)).
To get a feeling for the decibel scale, let’s start with the faintest sounds. If a sound has an
intensity I = I0, the corresponding intensity level is:

I
β = (10dB)log = (10dB)log 1 = 0
I0
Increasing the intensity by a factor of ten makes the sound seem twice as loud. In terms of
decibels, we have:

10I0
β = (10dB)log = (10dB)log 10 = 10 dB
I0
Going up in intensity by another factor of ten doubles the loudness of the sound again, and

yields:

100I0
β = (10dB)log = (10dB)log 10 = 20dB
I0
Here we can see that the loudness of sound doubles with each increase in intensity by a
factor of 10 dB. The smallest increase in intensity level that can be detected by the human
ear is about 1 dB.

66
 Activity 4a:
A crying child emits sound with an intensity of 8.0 x10-6 W/ m2. Find
(a) the intensity level in decibels for the child’s sounds.
(b) the intensity level for this child and its twin, both crying with identical
intensities.

Solutions:
(a) Here we make use of Equation 4.10
I
β = (10dB)log
I0
8.0 × 10−6 Wm−2
β = (10dB)log = �� ��
10−12 W/m2
(b) We’ll again make use of Equation 4.10, with a minor adjustment (2 times the
intensity)
2I
β = (10dB)log
I0
I
β = (10dB) log 2 + (10dB)log
I0
β = 3.0dB + 69dB = �� ��

4.2 Doppler’s Effect

You may have noticed that as the pitch of the sound of the car or train changes depending
on whether it is approaching you or moving away from you. This phenomena is caused by
what is called the Doppler’s effect. Doppler effect is the change in pitch (or frequency) due
to the relative motion between a source and the receiver. Though the Doppler’s effect is
most easily observed with sound waves, it applies to all wave phenomena, e,g light and
other electromagnetic waves.

4.2.1 Doppler’s Effect - Moving observer

67
In this case, the sound source is stationery and the observer moves towards the sound
source as shown in figure 4.2. The radiated sound is represented by the circular patterns of
compressions moving away from the source with speed v.

Figure 4.2: Moving observer

The wavelength (λ) ( distance between the compressions) and the frequency (f) are related
using the wave equation:
v = fλ
For an observer moving toward the source with a speed u, the sound appears to have a
higher speed, hence, v+u. As a result, more compressions move past the observer in a
given time than if the observer had been at rest. To the observer, then, the sound has a
frequency, ƒ′, that is higher than the frequency of the source, ƒ . We can find the frequency
ƒ′ by first noting that the wavelength of the sound does not change (it is still λ). The speed,
however, has increased to v′ = v + u. Thus, we can solve v′ = v′=λf′
v′ = λf ′
v' v + u
f'= =
λ λ
v
we have already established that the wavelength ( λ ) remains constant, hence, λ = f .

Therefore:
v+u v+u v+u
f'= = v = f
λ v
f
u
f'= 1+v f (4.11)

We can clearly see from equation 4.11 that the frequency detected by the observer is
greater than the frequency of the sound source (f ‘ > f).

68
On the contrary, If the observer had been moving away from the source with a speed u, the
sound would appear to the observer to have the reduced speed:
v' = v − u
Repeating the above calculations, we get:
v' v − u
f'= =
λ λ
u
f'= 1−v f (4.12)

Equation 4.12 shows that in this case, the frequency of the observer becomes less than the
frequency of the source (f ‘ < f).
Combining equations 4.11 and 4.12 we can then come up with a general equation for the
Doppler’s effect for a moving observer:
u
f'= 1±
v
f (4.13)

4.2.2 Doppler’s Effect - Moving Source

In this case, the observer is stationery an the sound source (the car) moves towards the
observer as shown in figure 4.3 below:

Figure 4.3: Moving source

In this case, the Doppler effect is not due to the sound wave appearing to have a higher or
lower speed, as when the observer moves. On the contrary, the speed of a wave is
determined solely by the properties of the medium through which it propagates. Thus, once
the source emits a sound wave, it travels through the medium with its characteristic speed
v regardless of what the source is doing.

69
Consider, then, a source moving toward an observer with a speed u. If the frequency of the
source is ƒ , it emits one compression every T seconds, where T = 1 f . Therefore, during
one cycle of the wave a compression travels a distance vT while the source moves a
distance uT. As a result, the next compression is emitted a distance vT − uT behind the
previous compression. This means that the wavelength in the forward direction λ' is:
λ' = vT − uT = (v − u)T
Now, as already stated the speed of the wave remain v, hence,
v = λ' f '
Now solving for the new frequency f ’, we get
v v v
f'= = =
λ' (v − u)T 1
(v − u)
f
1
f'= 1−
u f (4.14)
v

When the source is moving away from the observer, the wavelength is increased by the
amount uT. Thus,
λ' = (v + u)T
v v v
f'= = =
λ' (v + u)T 1
(v + u)
f
1
f'= 1+
u f (4.15)
v

Just as we did with the moving observer, in a similar fashion we can come up with a
general equation for the Doppler’s effect for the moving source i.e.
1
f'= 1∓
u f (4.16)
v

Where the subtraction sign is used in situation where the source approaches the observer
and conversely the addition sign where the source moves from away from the observer.

70
4.2.3 General case for the Doppler’s Effect
The results derived earlier in this section can be combined to give the Doppler effect for
situations in which both observer and source move. Letting us be the speed of the source,
and uo be the speed of the observer, we have:
u
1± o
f'= v
u f (4.17)
1∓ s
v

Activity 4b:
A car traveling north at 21 m/s sounds its 480 Hz horn. A bicyclist travels south
on the same road with a speed of 6.7 m/s, heading directly toward the car—
though at a safe distance. What frequency does the bicyclist hear?

Solution:
We’ll make use of the general Equation 4.17
uo
1±
f'= v
us f
1∓
v
Since the source and the observer are approaching, hence
uo
1+
f'= v
us f
1−
v
6.7m/s
1+
343 m/s
f'= (480 Hz) = 520 Hz
21 m/s
1−
343 m/s

71
4.3 Waves on a String

We begin by considering a string of length L that is tied

down at both ends, as shown in figure 4.4 (a). If you pluck
this string in the middle, it vibrates as shown in figure 4.4
(b). This is referred to as the fundamental mode of
vibration for this string or the first harmonic. First
harmonic is the simplest normal mode in which the string
vibrates in a single loop. However, making reference to
figure 4.4 (c), we can see that the fundamental mode
corresponds to half a wavelength of a usual wave on a
string. One can think of the fundamental as being formed
by this wave reflecting back and forth between the walls
holding the string. If the frequency is just right, the
Figure 4.4
reflections combine to give constructive interference and
the fundamental is formed; if the frequency differs from
the fundamental frequency, the reflections result in
destructive interference and a standing wave does not
result.

4.3.1 Fundamental Frequency

The fundamental frequency (also known as just ‘the fundamental’) refers to the lowest
frequency of a periodic waveform. Using the string fixed at both ends, we can find the
frequency of the First harmonic by firstly using the fact that the full wavelength of the first
harmonic is twice the distance between the walls (2L), thus:
�1 = 2� (4.18)
If the speed of waves on the string is ν, it follows that the frequency of the first harmonic,
f1, is determined by the wave equation (ν = λf). In this case
ν = λ1 f1

72
And now solving for the frequency, we get:
ν ν
f1 = λ = 2L (4.19)
1

In this case, we notice that the frequency of the first harmonic increases with increase in
speed of the wave but decreases when the string is lengthen.

4.3.2 Higher Harmonics

The first harmonic is not the only standing wave that exists on a string. There are an
infinite number of harmonics for any given string, with frequencies that are integer
multiples of the first harmonic. However, to find the other harmonics, we must ensure that
the both ends of the string remain fixed. To better understand the higher harmonics, we
have to be introduced to two new terms, namely; Nodes and Anti-nodes. Nodes (N) refer
to the points on the standing wave that are fixed while Anti-nodes (A) on the other hand
refer to the point on the wave that has maximum displacement. In this case we see that the
first harmonic corresponds to having two nodes and and one anti-node in the sequence N-
A-N as shown in figure 4.5 (a).

Figure 4.5: Harmonics

the second harmonic can be formed by adding one more half wavelength to the first
harmonic with the sequence N-A-N-A-N as shown in figure 4.5(b). The second harmonic
has one complete wavelength between the walls ( �2 = � ),thus its frequency can be
calculated:
� �
�2 = � = � = 2�1 (4.20)
2

73
Similarly, the third harmonic contains one half wavelength more (than the second
harmonic) as shown in figure 4.4(c). Now we have one and a half wavelength in length L,
hence 3 2 �3 = � or �3 =
2�
3
. The corresponding frequency f3 can then be calculated:
� � �
�3 = � = 2� = 3 2� = 3�1 (4.21)
3
3

4.3.3 General Formula for the nth Harmonic Frequency

At this point it is easy to notice that each harmonic has a frequency that is an integer
multiple of the first-harmonic frequency. In general, the nth harmonic, with n = 1, 2, 3, . . .,
has the following frequency:

ν
fn = nf1 = n 2L (4.22)

In terms of the wavelengths, the general formula take the form:

�1 2�
�� =
�
=
�
(4.23)

The integer n in these expressions represents the number of half wavelengths in the
standing waves. In addition, notice that the difference in frequency between any two
successive harmonics is equal to the first-harmonic frequency, ƒ1

Activity 4c:
A section of drainage culvert 1.23 m in length makes a howling noise when the wind
blows across its open ends. Determine the frequencies of the first three harmonics of
the culvert if it is cylindrical in shape and open at both ends. Take v = 343 m/s as the
speed of sound in air.

Solution:
We’ll make use of Equation 4.19 for the 1st Harmonic
ν 343 m/s
f1 = = = ��� ��
2L 2(1.23m)
Subsequently, we’ll make use of Equation 4.20 and Equation 4.21 for the 2nd and 3rd
Harmonics respectively
�2 = 2�1 = 2(139 ��) = ��� ��
�32 = 3�1 = 3(139 ��) = ��� ��

74
4.4 Vibrating in Columns of Air
If you blow across the open end of a pop bottle `(as illustrated in
figure 4.6), you hear a tone of certain frequency. If you pour some
water into the bottle and blow into it, the sound you hear has a
higher frequency.
In both cases you have excited the fundamental mode of the column
of air within the bottle.When water was added to the bottle,
however, the column of air was shortened, leading to a higher Figure 4.6:
frequency—in the same way that a shortened string has a higher vibrating air
frequency columns

4.4.1 Columns closed in one end

Consider figure 4.6 where when one blows across the opening of and empty bottle, the
result is a swirling movement of air that excites rarefactions and compressions. For this
reason, the bottle opening is considered to have an anti-node for the sound wave. On
contrary, the bottom is considered to have a Node since it is closed and prevents the
movement of air. We can now generalize that any standing wave in the bottle must have a
node at the bottom and an anti-node at the top as illustrated in figure 4.7

The lowest frequency of a standing wave that is

corresponding with these conditions is shown in
figure 4.7(a), i.e. the first harmonic.If we plot
the density variation of the air for this wave, we
see that one-quarter of a wavelength fits into the
column of air in the bottle. Thus, if the length of
the bottle is L, the first harmonic (fundamental)
has a wavelength satisfying the following
relationship:
Figure 4.7: Harmonics in columns

75
 closed on one end
1
�=�
4

λ = 4L

And the frequency for the first harmonic can be calculated:

� �
�1 = �
= 4� (4.24)

Equation 2.24 implies that the fundamental frequency for columns closed in one end
simply equals half of the corresponding fundamental frequency for the wave on a string.

For higher harmonics, the next harmonic is produced simply by adding half a wavelength
from the preceding harmonic. Thus, if the fundamental is represented by N-A, the second
harmonic can be written as N-A-N-A. Since the distance from a node to an antinode is a
quarter of a wavelength, we see that three-quarters (3/4) of a wavelength fits into the bottle
for this mode. Figure 4.7 (b) is a good representation for this mode (where 3/4λ = L), thus

4
λ= L
3

Therefore the corresponding frequency can be calculated:

� � �
�
= 4 = 3 4� = 3�1 (4.25)
�
3

Notice that this is the third harmonic of the pipe; that is, its frequency is three times f1. In a
similar manner, the next-higher harmonic is written as N-A-N-A-N-A, as illustrated in
figure 4.7(c). In this case, 5/4λ = L and λ = 4/5L, hence the frequency can be calculated:
� � �
�
= 4 =5
4�
= 5�1 (4.26)
�
5

Likewise, equation 2.26 is the 5th harmonic of the pipe since its frequency is five times f1.
in a more general description, column of air that is closed at one end and open at the other
end is described by the following frequencies and wavelengths:
�
�� = ��1 =
��
(4.27)
�1 4�
�� = �
= �
(4.28)

76
In this case, equations 4.27 & 4.28 are valid for n=1,3,5,7… (odd numbers) as opposed to
waves on a string which are valid for all integers.

4.4.2 Columns Open on both ends

Standing waves can also be produced in columns of
air that are open at both ends as illustrated in figure
4.8. In this case there is an antinode at each end of the
column. Hence, the first harmonic, or fundamental,
has the sequence A-N-A, as illustrated in figure
4.8(a). . In this case, half a wavelength fits into the
Figure 4.8:Figure 4.7: Harmonics pipe. The frequency for the first harmonic can then be
in columns open on both ends calculated:

�
�1 = 2� (4.29)

Which is the same as the corresponding result for waves on a string.

The next harmonic is A-N-A-N-A, which fits one complete wavelength in the pipe. This
harmonic is shown in Figure4.8 (b), and has the frequency:

v
f = L = 2f1 (4.30)

Equation 2.30 represents the 2nd harmonic which also is the same as the corresponding
result for waves on a string. The rest of the harmonics continue in exactly the same manner
as for waves on a string, with all integer harmonics present
Therefore a more general way to represent the frequencies and wavelengths in a column of
air open at both ends are as follows:
ν
fn = nf1 = n 2L (4.31)

�1 2�
�� = �
= �
(4.32)

Where both equations are valid for n=1,2,3,4,5…. (all integers).

77
4.5 Resonance in air columns

To understand the term resonance, we first have to understand the concept of Natural
frequency. Consider that you hold the end of a string from which a small weight is
suspended as shown in figure 4.9.

If the weight is set in motion and you hold your hand still, it
will soon stop oscillating. If you move your hand back and
forth in a horizontal direction, however, you can keep the
weight oscillating indefinitely. The motion of your hand is
said to be “driving” the weight, leading to driven
oscillations. If you move your hand very slowly, the weight
will simply track the motion of your hand. Similarly, if you
oscillate your hand very rapidly, the weight will exhibit only
small oscillations. Oscillating your hand at an intermediate
frequency, however, can result in large amplitude oscillations Figure 4.9: demonstrating
for the weight. Natural frequency

To achieve the greatest amplitude, you’ll have to oscillate your hand at a specific
frequency; precisely the frequency at which the weight oscillates without being driven
(also known as the Natural frequency). In general, resonance is achieved when the driving
frequency of an oscillation equals the natural frequency of the object. At resonance, we
tend to get oscillations with the greatest amplitude.

Now to draw a relationship between standing waves and resonance, we’ll firstly have to
understand what standing waves are. A standing wave refers to one that oscillates with
time, but remains fixed in its location. The waves discussed in the above sections (wave on
a string & waves in air columns) are both examples of standing waves Standing waves can
occur at more than one frequency. It should however be noted that frequencies at which the
standing waves are produced are the natural/resonant frequencies of the cord.

78
Summary
The way we perceive the quality sound depends on its characteristics, namely, Frequency,
Pitch and Intensity. Intensity of sound is conventionally measured in the units of decibels.
The relationship between distance between the source of sound and its observer to how the
observer perceives the sound’s frequency is perfectly explained using the Doppler’s Effect.
A fundamental frequency refers to the lowest frequency of a periodic waveform and it is
also known as the first harmonic. The higher harmonics are simply integer multiples of the
fundamental harmonic.

Further Reading
Pressure variations in sound waves.

Unit Test
1. You drop a stone from rest into a well that is 7.35 m deep. How much time elapses
before you hear the splash?
2. At a busy street corner, the sound level is 75 dB. What is the intensity of sound there?
3. A street musician sounds the A string of his violin, producing a tone of 440 Hz. What
frequency does a bicyclist hear as she
(a) approaches
(b) recedes from the musician with a speed of 11.0 m/s?
4. A train sounds its whistle as it approaches a tunnel in a cliff. The whistle produces a
tone of 650.0 Hz, and the train travels with a speed of 21.2 m/s.
(a) Find the frequency heard by an observer standing near the tunnel entrance.
(b) The sound from the whistle reflects from the cliff back to the engineer in the train.
What frequency does the engineer hear.

79
Answers to Unit Activities
1. 1.22s
2. 3.2 × 10−5 Wm−2
3. (a) 454 Hz
(b) 426 Hz
4. (a) 693 Hz
(b) 736 Hz

80
 Unit 5: Physical Optics
5.0 Introduction
This section aims at trying to explain some of the ways light behaves which cannot be
adequately explained using the traditional geometric optics. This is achieved by
particularly taking into account the wave nature of light as proposed by the famous Dutch
physicist Christiaan Huygens (1629–1695).

Learning Outcomes
By the end of this Unit, you should be able to:
 Describe the conditions for interference
 Discuss Young’s double slit experiment
 Describe Thin film interference
 Describe wedge film interference
 Discuss Newton rings
 Discuss single and double slit diffraction
 Discuss Fraunhofer diffraction and Frensel diffraction
 Discuss resolving power

Key Terms
Ensure that you understand the key terms or phrases used in this unit as listed below:
 Interference
 Constructive interference
 Destructive interference
 Path difference

81
5.1 Conditions for Interference

Interference of waves refers to net displacement caused by a combination of waves as the

algebraic sum of the displacements caused by each wave individually. Waves combine to
either form a larger wave (constructive interference) or a smaller wave (destructive
interference). When dealing with light, interference is only noticeable when two conditions
are met;

i. The light should be monochromatic

- The light waves combining should have a single color and hence a single
frequency.

ii. The light must be coherent

- The light waves combining must maintain a constant phase with respect to
each other.

Sources whose relative phases vary randomly with time show no visible interference
patterns and are said to be incoherent. Examples of incoherent light sources include
incandescent and fluorescent lights. In contrast, lasers emit light that is both
monochromatic and coherent and are therefore serve as the best sources for observing
interference patterns. If two waves are in phase (i.e. have zero phase difference), they add
constructively, and the net result is an increased amplitude as illustrated in figure 5.1(a) .
Constructive interference also occurs where the two wave are out of phase by a full
wavelength (phase difference is 3600) as illustrated in figure 5.1(c).

82
 Figure 5.1: constructive and destructive interference
Conversely, if waves of equal amplitude are 180° out of phase, however, the net result is
zero amplitude and destructive interference as illustrated in figure 5.1 (b). In this case, a
180° difference in phase corresponds to waves being out of phase by half a wavelength

5.2 Young’s double slit experiment

A famous physicist by the name of Thomas Young (1773-1829) performed this experiment
which did not only demonstrate the nature of light but also provided a way of determining
the wavelength of a beam. The experiment, consists of a beam of monochromatic light that
passes through two small slits and illuminates a screen.

83
Figure 5.2 illustrates the setup for Young’s double slit experiment.
The first screen (with the single slit) serves only to produce a small
source of light that prevents the interference pattern on the distant
screen from becoming smeared out. Therefore, the key elements in
the experiment are the two slits in the second screen. The two slits are
equidistant from the single slit such that the light passing through
them has the same phase. This ensures that the two slits act as
monochromatic, coherent sources of light
The main argument for this setup is that, if light were composed of
small particles, they would simply pass straight through the two slits
and illuminate the distant screen directly behind each slit. On the
contrary, if light is a wave, each slit acts as the source of new waves,
therefore light is spread out over a large area on the distant screen
(i.e. it is not localized in small regions directly behind the slits). Figure 5.2: double slit
setup
However results for this setup show that after the light passes through the two slits and
shines on the distant screen, an interference pattern of bright and dark “fringes” is observed
as illustrated in figure 5.2. These fringes are the direct result of constructive and
destructive interference.

5.2.1 Analysis of the interference patterns on the screen

Consider the waves of wavelength λ entering the slits as shown in figure 5.3 below. The
waves spread out in all directions after passing through the slits, but they are shown in
Figure 5.3 only for three different possibilities.

84
 Figure 5.3: formation of dark and bright fringes
In figure 5.3(a), Waves from the two slits travel the same distance, so they are in phase (i.e.
a crest of one wave arrives at the same time as a crest of the other wave). The result is that
the amplitudes of the two waves add to form a larger amplitude (constructive interference),
hence a bright fridge formed at that spot.
Constructive interference also occurs when the paths of the two rays differ by one full
wavelength (or any whole number of wavelengths), as shown in figure 5.3 (b), hence, there
will be a bright line on the screen.
However, in figure 5.3 (c) we see that the crests of one wave arrive at the same time as the
troughs of the other wave, and so they add to produce zero amplitude. This happens due to
one of the wavelengths traveling an extra distance of one-half wavelength (or
3λ/2 . 5λ/2, etc) which renders the two waves out of phase when they reach the screen,
hence destructive interference and consequently a dark fringe being formed on the spot.

5.2.2 Determining exactly where the bright fringes appear on

the screen
Consider figure 5.4 below as a geometrical representation for the Young’s double slit
experiment setup.

85
 Figure 5.3: locating exactly where bright fringes appear
The viewing screen is located a perpendicular distance L from the barrier containing two
slits, S1 and S2 (Fig. 5.3). These slits are separated by a distance d (assumed to be very
small in comparison to L), and the source is monochromatic. To reach any arbitrary point
P in the upper half of the screen, a wave from the lower slit �2 ) must travel farther than a
wave from the upper slit (�1 ). The extra distance traveled from the lower slit is called the
path difference ( δ ). If we assume the rays labeled �1 and �2 are parallel (which is
approximately true if L ≫ d) then path difference (δ) is given by:

δ = r2 − r1 = d sinθ (5.1)

As earlier discussed, when the path difference is zero or equals a full wavelength, it
implies constructive interference. This information helps us generate a general formula for
locating a bright fringe on the screen as:
mλ = d sinθbright (5.2)
Where � is any integer (i.e. 0, ± 1, ± 2, ± 3, ± 4…)
� is the wavelength
d is the separation distance of two slits
� is the angle relative to the normal

86
Similarly, we also pointed out that fact that destructive interference only occurs when the
path difference between the two waves is one half of a wavelength. This information helps
us generate a general formula for locating a dark fringe on the screen as:
1
m+
2
λ = d sinθdark (5.3)

Where � is any integer (i.e. 0, ± 1, ± 2, ± 3, ± 4…)

� is the wavelength
d is the separation distance of two slits
� is the angle relative to the normal

We consider the bright fringes as peaks or maxima of light intensity, the dark fringes are
minima. The value of m is called the order of the interference fringe. Notice that the m=0
fringe occurs when θ = 0 and this is regarded as the first order fringe. positive values of m
indicate fringes above the central bright fringe; negative values indicate fringes below the
central bright fringe.

In addition, a convenient way to characterize the location of interference fringes is in terms

of their linear distance from the central fringe (distance y in fig. 5.3). If the distance to the
screen is L (and � ≫ � ),it follows that the linear distance y is given by the following
expression:

y = L tan θ (5.3)

Here we make the following assumptions:

i. L ≫ d

ii. d ≫ λ

These two assumptions also imply that the angle θ very small. Mathematically, for very
small θ, tan θ ≈ sin θ, therefore equation 5.3 becomes:

y = L sin θ (5.4)

87
5.2.3 Change of phase during reflection
An alternative arrangement for producing an interference pattern with a single light source
is known as Lloyd’s mirror. A point light source S is placed close to a mirror, and a
viewing screen is positioned some distance away and perpendicular to the mirror as
illustrate in figure 5.4 below.

Light waves can reach point � on the screen either directly from S
to � or by the path involving reflection from the mirror. The
reflected ray can be treated as a ray originating from a virtual
source S’. As a result, we can think of this arrangement as a double
slit source where the distance d between sources S and S' in Figure
5.4 is analogous to length d in Figure 5.3. Needless to say, the
arrangement in fig. 5.4 here represents an alternative way of
achieving the results from Young’s double slit experiment.

At observation with points far from the source (� ≫ �), we expect

waves from S and S' to form an interference pattern exactly like
the one formed by two real coherent sources. An interference
Figure 5.4: Phase change
pattern is indeed observed. The positions of the dark and bright
due to reflection
fringes, however, are reversed relative to the pattern created by
two real coherent sources (Young’s experiment). Such a reversal
can only occur if the coherent sources S and S’ differ in phase by
180°.We therefore conclude that a 180° phase change must be
produced by reflection from the mirror.

In general, an electromagnetic wave undergoes a 180° phase change when reflected from a
boundary leading to an optically denser medium (defined as a medium with a higher index
of refraction), but no phase change occurs when the wave is reflected from a boundary
leading to a less dense medium.

88
 Activity 5a:
A screen containing two slits 0.100 mm apart is 1.20 m from the viewing screen. Light
of wavelength λ=500nm falls on the slits from a distant source. Approximately how far
apart will adjacent bright interference fringes be on the screen?

Solution:
λ=500nm, m=1, d=0.100mm,L=1.20m , ℎow far apart=y

We can make use of Equation 5.4, hence

y = Lsin θ (i)
If we make ��� � subject in Equation 5.2, we get
��
sin � =
�
(ii)
Combining Equations (i) and (ii) we get
Lmλ 1.20m × 1 × 500 × 10−9 m
y= =
d 0.1 × 10−3 m

� = �. ����

5.3 Thin Film interference

Interference effects are commonly observed in thin films, such as thin layers of oil on
water or the thin surface of a soap bubble. The varied colors observed when white light is
incident on such films result from the interference of waves reflected from the two surfaces
of the film. These colors are the result of the constructive and destructive interference that
can occur when white light reflects from a thin film. Some colors undergo destructive
interference and are eliminated from the incident light, while others colors are enhanced by
constructive interference. Consider a film of uniform thickness t and index of refraction n.
The wavelength of light �� in the film is:

89
 �
�� = � (5.5)

Where � is the wavelength of light in free space

n is the index of refraction of the film

Consider the ray diagram in fig.5.5 representing interference

taking place in a thin film. Reflected ray 1, which is reflected from
the upper surface (A), undergoes a phase change of 180° with
respect to the incident wave. Reflected ray 2, which is reflected
from the lower film surface (B), undergoes no phase change
because it is reflected from a medium (air) that has a lower index
of refraction. Therefore, ray 1 is 180° out of phase with ray 2,
��
which is equivalent to a path difference of 2
.

We must also consider, however, that ray 2 travels an extra

distance 2� before the waves recombine in the air above surface A.
in this case, we are considering light rays that are close to normal Figure 5.5: rays passing
to the surface (If the rays are not close to normal, the path through a thin film
difference is larger than 2t.)
�
If 2t= 2� , ray 1 and ray 2 recombine in phase and the result is constructive interference. In

general, the condition for constructive interference in thin films is:

1
2t = m + 2 λn m = 0,1,2, . . . (5.6)

Equation 5.6 takes into account two conditions:

i. The path length for the two rays (i.e. mλn )

1
ii. The 1800 phase change upon reflection (hence the term 2 �� )

Now we can combine equation 5.5 and 5.6 to have a more expanded version of equation
5.6 as:

90
 1
2nt = m + 2 λ m = 0,1,2, . . . (5.7)

If the extra distance 2t traveled by ray 2 corresponds to a multiple of �� , the two waves
combine out of phase and the result is destructive interference. The general equation for
destructive interference in thin films is:

2nt = mλ m = 0,1,2, . . . (5.8)

The foregoing conditions for constructive and destructive interference are valid when the
medium above the top surface of the film is the same as the medium below the bottom
surface or, if there are different media above and below the film, the index of refraction of
both is less than n. If the film is placed between two different media, one with
� < nfilm and the other with � > nfilm , the conditions for constructive and destructive
interference are reversed. In that case, either there is a phase change of 180° for both ray 1
reflecting from surface A and ray 2 reflecting from surface B or there is no phase change
for either ray; hence, the net change in relative phase due to the reflections is zero.

91
Activity 5b:
Calculate the minimum thickness of a soap-bubble film that results in constructive
interference in the reflected light if the film is illuminated with light whose
wavelength in free space is λ=600 nm. The index of refraction of the soap film is
1.33.

Solution:
Firstly since we are dealing with constructive interference, we’ll make use of
Equation 5.7
1
2nt = m + λ
2
We know that the minimum thickness happens when m = 0, we firstly make m
subject and equate the result to zero
2nt 1
m= − =0
λ 2
λ
t=
2(2n)
600 × 10−9 m
t=
4(1.33)
� = �����

92
5.4 Newton Rings

Interference in light waves can also be observed by placing a plano-convex lens on top of a
flat glass surface as shown in fig 5.8.

Figure 5.8: formation of newton rings

The air film between the glass surfaces varies in thickness from zero at the point of contact
(O) to some nonzero value at point P. If the radius of curvature R of the lens is much
greater than the distance r and the system is viewed from above, a pattern of light and dark
rings is observed as shown in Figure 5.8(b). A series of concentric rings is seen when
illuminated from above by either white light (as shown) or by monochromatic light. These
circular fringes, discovered by Newton, hence why they are called Newton’s rings.
The interference effect is due to the combination of ray 1 (reflected from the flat plate),
with ray 2 (reflected from the curved surface of the lens). Ray 1 undergoes a phase change
of λ/2 (180°) upon reflection (because it is reflected from a medium of higher index of
refraction). Ray 2 undergoes no phase change (because it is reflected from a medium of
lower index of refraction). Therefore, the conditions for constructive and destructive
interference are similar to those of thin films, with n = 1 because the film is air. There is
no path difference and the total phase change is due only to the 180° phase change upon
reflection, the contact point at O is dark as seen in Figure 5.8(b).

93
Now, using the geometry shown in Figure 5.8(a), the expressions for the radii of the bright
and dark bands in terms of the radius of curvature R and wavelength λ can be found.
���
����� = �
(5.9)

And for bright fringes, it is represented:

1
�+ ��
�����ℎ� = �
2
(5.10)

5.4.1 Applications of Newton rings

i. Newton’s rings are used in the testing of optical lenses. In this case, Lenses producing
a pattern different to the one by newton are reground and re-polished.
ii. It is also used in determining the wavelength of a monochromatic light.
iii. Newton rings are also use in determination of the refractive index of the liquid.

5.5 Wedge film interference

If two glass plates are placed face to face with one end separated by a piece of tissue paper
or thin metal foil an air wedge will be formed between them (figure 5.9). If monochromatic
light is shone on the plates a series of straight-line fringes will be seen parallel to the line
along which they touch.

94
Ray 1 reflects at the glass-air interface; it experiences
no phase change. Ray 2 travels a distance d through the
air (n = 1.00), reflects from the air-glass interface, then
travels essentially the same distance d in the opposite
direction before rejoining ray 1. The reflection from
the air-glass interface results in a 180° phase change
(the same phase change as if the wave had traveled half
a wavelength) and hence the effective path length
(Δℓeff ) of ray 2 is:
λ
Δℓeff = 2d +
2
(5.11)
Figure 5.9: light rays passing through a
wedge

λ
If the effective path length is an integer number of wavelengths, 2d + 2 = mλ, rays 1 and

2 will interfere constructively. Therefore, dividing by the wavelength, we have the

following condition for constructive interference:
2� 1
�
+2=� � = 0,1,2, . . . (5.12)

Equation 5.12 can be represented in a more pleasant manner by making 2d subject as:
1
2d = m − 2 λ (5.13)

Similarly, if the effective path length of ray 2 is an odd half integer, there will be
� 1
destructive interference: 2
+ 2� = m + 2 λ . Dividing by the wavelength, we have the

following condition for destructive interference:

1 2� 1
2
+ �
=�+2 � = 0,1,2, . . . .

Which can be simplified as:

2d = mλ (5.14)
Noting that the distance between the plates, d, increases linearly with the distance from the
point where the glass plates touch, we can conclude that the dark and bright interference
fringes are evenly spaced as shown in fig. 5.10 below.

95
Figure 5.10: interference pattern generated by a wedge

96
5.6 Single slit diffraction

Diffraction refers to the bending of waves as they pass through a barrier or through an
opening. A familiar example of diffraction is the observation that you can hear a person
talking even when that person is out of sight around a corner. In such a situation, in order
for the person to be audible to you, the sound waves from the person bend around the
corner. Diffraction effects in light are typically small compared with those in sound waves
and water waves (the angle through which a wave bends is greater, the larger the
wavelength of the wave).

Now let us consider a common situation, that of light passing through a narrow opening
modeled as a slit and projected onto a screen. To simplify our analysis, we assume the
observing screen is far from the slit and the rays reaching the screen are approximately
parallel.

Figure 5.11: interference pattern due to a single slit

With the setup in fig. 5.11(a), a bright fringe is observed along the axis at θ = 0 , with
alternating dark and bright fringes on each side of the central bright fringe. What do you
reckon is the cause of the bright and dark fringes? Diffraction or interference?

Until now, we have assumed slits are point sources of light. In this section, we abandon
that assumption and see how the finite width of slits is the basis for understanding of this

97
form diffraction. We can explain some important features of this phenomenon by
examining waves coming from various portions of the slit as shown in Figure 5.11.

According to Huygens’s principle, each portion of the slit acts as a source of light waves.
Hence, light from one portion of the slit can interfere with light from another portion, and
the resultant light intensity on a viewing screen depends on the direction θ. Based on this
analysis, we recognize that a diffraction pattern is actually an interference pattern in
which the different sources of light are different portions of the single slit!

To analyze the diffraction pattern, let’s divide the slit into two halves as shown in
Figure 5.12 below.
Keeping in mind that all the waves are in phase as they
leave the slit, consider ray 1 and ray 3. As these two rays
travel toward a viewing screen far to the right of the figure,
ray 1 travels farther than ray 3 by an amount equal to the
�
path difference 2
sin � , where � is the width of the slit.

Similarly, the path difference between ray 2 and ray 4 is

�
also 2
sin � , as is that between ray 3 and ray 5. If this

path difference is exactly half a wavelength (corresponding

to a phase difference of 180°), the pairs of waves cancel
each other and destructive interference results.
Figure 5.12: analysis of the slit
This cancellation occurs for any two rays that originate at points separated by half the slit
width because the phase difference between two such points is 180°. Therefore, waves
from the upper half of the slit interfere destructively with waves from the lower half when:

� �
sin � =
2 2

or, if we consider waves at angle θ both above and below the dashed line in Figure 5.12
and below:

�
sin � =±
�
(5.15)

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Here, by including both positive and negative values for m, we have taken into
account the symmetry of the diffraction pattern about its midpoint. Dividing the slit into
four equal parts and using similar reasoning, we find that the viewing screen is also dark
when:

�
sin � =± 2
�
(5.16)

Likewise, dividing the slit into six equal parts shows that darkness occurs on the screen
when:

�
sin � =± 3
�
(5.17)

We can therefore deduce the general equation for destructive interference as:
�
sin ����� = �
�
m =± 1, ± 2, ± 3, …. (5.17)

5.6.1 Intensity of a single slit diffraction patterns

When analyzing diffraction patterns, equation 5.17 above helps us in identifying the
positioning of the fringes. However, the equation tells us nothing about the variation in
light intensity along the screen. The general features of the intensity distribution are shown
in both Figure 5.11 above as well as figure 5.13 below.

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A broad, central bright fringe is observed;
this fringe is flanked by much weaker bright
fringes alternating with dark fringes. The
various dark fringes occur at the values of
θdark that satisfy Equation 5.17. Each bright-
fringe peak lies approximately halfway
between its bordering dark-fringe minima.
Notice that the central bright maximum is
twice as wide as the secondary maxima.
There is no central dark fringe, represented
by the absence of m=0 in Equation 5.17.

Figure 5.13: intensity variation in a single

slit
Therefore, with the aid of figure 5.13 (a) above, analysis of the intensity variation in a
diffraction pattern from a single slit of width a shows that the intensity is given by:

� 2
sin �� sin
� = ���� �
�
5.18
�� sin
�

Where ���� is the intensity at θ = 0(central maximum)

� is the wavelength of the light used to illuminate the slit

a is the width of the slit

5.6.2 Intensity of a two slit diffraction patterns

When more than one slit is present, we must consider not only diffraction patterns due to
the individual slits but also the interference patterns due to the waves coming from
different slits.

100
 �
sin ����� = �
�
�� (1)(511 × 10−9 �)
θdark = sin−1 = sin−1 = ��. ��
� (2.2 × 10−6 �)

�
sin ����� = �
�
�� (2)(511 × 10−9 �)
θdark = sin−1 = sin−1 = ��. ��
� (2.2 × 10−6 �)

5.7 Fraunhofer vs. Fresnel diffraction Patterns

Consider a situation of light passing through a narrow opening (modeled as a slit) and
projected onto a screen. The diffraction patterns observed on the screen will differ
depending on how far the screen is placed from the slit. In a case where the screen is
placed at a finite distance from the slit, a Fresnel diffraction pattern is observed.
Conversely, where the screen is placed at an infinite distance (far away) from the slit, a
Fraunhofer diffraction pattern is observed on the screen.

101
 (a)
(b)

Figure 5.14: Frensel vs Fraunhofer diffraction

Fraunhofer diffraction can be achieved experimentally either by placing the observing
screen far from the slit or by using a converging lens to focus the parallel rays on a nearby
screen, as illustrated in figure 5.14(b).

5.8 Resolving Power

Resolution is simple terms, refers to the ability of optical instruments to visually separate
closely packed objects. To understand this limitation, consider

Figure 5.14 below, which shows two light sources far from a narrow slit of width a. The

sources can be two noncoherent point sources �1 and �2 . For example, they could be two
distant stars.

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 Figure 5.15: wave illustration of resolving power
If no interference occurred between light passing through different parts of the slit, two
distinct bright spots (or images) would be observed on the viewing screen. Because of such
interference, however, each source is imaged as a bright central region flanked by weaker
bright and dark fringes, a diffraction pattern. What is observed on the screen is the sum of
two diffraction patterns: one from �1 and the other from �2 .

If the two sources are far enough apart to keep their central maxima from overlapping as in
Figure 5.15(a), their images can be distinguished and are said to be resolved. If the sources
are close together as in Figure 5.15(b), however, the two central maxima overlap and the
images are not resolved. To determine whether two images are resolved, the following
condition known as the Rayleigh’s criterion. The Rayleigh’s criterion states that When the
central maximum of one image falls on the first minimum of another image, the images are
said to be just resolved.

Using Rayleigh’s criterion, we can determine the minimum angular separation ( ���� )
subtended by the sources at the slit in figure 5.15 for which the images are just resolved.
Recall Equation 5.15 which indicates the first minimum (dark fringe) in a single-slit
diffraction pattern, i.e.

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 �
sin � =
�

According to Rayleigh’s criterion, this expression gives the smallest angular separation for
which the two images are resolved. Because � ≪ � in most situations, sin � is small and
we can use the approximation sin � ≈ �. Therefore, the limiting angle of resolution for a
slit of width a is:

�
θmin = � 5.19

Here, θmin is in radians. Therefore the angle subtended by the two sources at the slit must
be greater than λ/� if the images are to be resolved.

However, many optical systems use circular apertures rather than slits. The diffraction
pattern for two object observed from a circular aperture is represented in figure 5.15 below:

Figure 5.16: graphical illustration of resolving power

When the sources are far apart, their images are well resolved (Fig. 5.16a). When the
angular separation of the sources satisfies Rayleigh’s criterion, the images are just resolved

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(Fig. 5.15b). Finally, when the sources are close together, the images are said to be
unresolved (Fig. 5.16c) and the pattern looks like that of a single source.

Analysis shows that the limiting angle of resolution of the circular aperture is given by:

�
���� = � 5.20

Where � is the wavelength of the light

� is the diameter of the circular aperture

An easier way to relate to the concept of Rayleigh’s criterion is a situation whereby a car is
coming towards you with its headlights on as illustrated in figure 5.19 below:

(a) (b) (c)

Figure 5.17: Rayleighs criterion on car headlights
In the first photo we see a brilliant light in the distance that may be the single headlight of
an approaching motorcycle or the unresolved image of two headlights on a car. If we are
seeing the headlights of a car, the angular separation between them will increase as the car
approaches. When the angular separation exceeds 1.22λ/�, as in the second photo, we are
able to distinguish the two headlights as separate sources of light. As the car continues to
approach, its individual headlights become increasingly distinct, as shown in the third
photo.

105
Summary
By letting light pass through different obstacles, we are able to generate various diffraction
patterns which helps in the study of the wave nature of light. Arranging our obstacles in
different ways helps us develop unique diffraction patterns which can be studied
geometrically to predict the positions of the alternating bright and dark fringes. Careful
study of a singe slit diffraction pattern leads to the conclusion that diffraction patterns are
actually an interference patterns in which the different sources of light are different
portions of the single slit which sets the basis of studying the resolving power of various
optical instruments.

106
Further Reading
The Michelson Interferometer.

Unit Test

1. Red light (λ = 734 nm2 ) passes through a pair of slits with a separation of

6.45 × 10−5 m. Find the angles corresponding to

(a) the first bright fringe

(b) the second dark fringe above the central bright fringe.

2. Light of wavelength 530 nm illuminates a pair of slits separated by 0.300 mm. If a

screen is placed 2.00 m from the slits, determine the distance between the first and second
dark fringes.

3. A Young’s interference experiment is performed with blue-green argon laser light. The
separation between the slits is 0.500 mm, and the screen is located 3.30 m from the slits.
The first bright fringe is located 3.40 mm from the center of the interference pattern. What
is the wavelength of the argon laser light?

4. Light of wavelength 620 nm falls on a double slit, and the first bright fringe of the
interference pattern is seen at an angle of 15.0° with the horizontal. Find the

separation between the slits.

Answers to Unit Activities

1. (a) 0.6520
(b)0.9780
2. 3.53mm
3. 515nm
4. 240μm

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 Unit 6: Polarization of Light
6.0 Introduction
In the preceding topic, we explored the different diffraction patterns that help to support
the notion that light should be regarded as a wave. In this topic, we further explore more
evidence of the wave nature of light. In particular, this topic looks at the electromagnetic
properties of light and how manipulation of these properties are used in the scientific
world to create everyday products like sunglasses.

Learning Outcomes
By the end of this Unit, you should be able to:
 Describe the Transverse nature of light waves
 Define Polarization
 Discuss the various types of polarization

Key Terms
Ensure that you understand the key terms or phrases used in this unit as listed below:
 Transverse waves
 Electromagnetic waves
 Electromagnetic Spectrum
 Polarization

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6.1 Transverse nature of waves.

Consider a string that is free on one end attached on the other end as shown in fig. 6.1
below. If a person holds the string on the free-end and moves it up and down, we develop a
wavelike motion on the string.

Figure 6.1: generating a Transverse wave

You’ll notice that the waves generated on the string will have two properties (i.e. the
direction of disturbance as well as the direction of propagation of the wave). In the wave
generated in fig. 6.1 you’ll see that the direction of propagation (from right to left) is
perpendicular to the direction of disturbance (up and down) of the wave. When the
direction of propagation and direction of disturbance are perpendicular, the wave generated
is called a Transverse wave.
Now making reference to the wave nature of light in particular, for a long time scientist
that backed the wave theory of light believed that the vibrations of light as a wave were in
the same direction to its propagation. However, series of experimental results confirmed
that light as a wave was transverse in nature (i.e. the direction of vibration is perpendicular
to the direction of propagation of the wave). Therefore we now know that light is indeed a
transverse wave. Light is part of the Electromagnetic waves spectrum which are all
transverse waves that have both magnetic and electric waves oscillating perpendicular to
each other as illustrated in fig. 6.2 below:

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 Figure 6.2: Electromagnetic waves

6.2 Definition of Polarization

Polarization of an electromagnetic wave simply refers to the direction which the electric
field of the electromagnetic wave is restricted to oscillate. To visually represent
polarization, consider fig. 6.3 below:

Figure 6.3: Polarization of Electromagnetic waves

Electromagnetic waves pictured in Fig. 6.3. Each of these waves has an electric field that
points along a single line. For example, the electric field in Fig. 6.3(a) points in either the
positive or negative � direction. We say, then, that this wave is linearly polarized in the �
direction. Similarly, the direction of polarization for the wave in Fig. 6.3 (b) is in the � − �
plane at an angle of 60° relative to the � direction. However, a more convenient way of
representing polarization is through the use of arrows as illustrated in fig. 6.4 below:

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 Figure 6.4: Representing polarization with the aid of arrows
The representation in fig. 6.4 brings to our attention two terms, namely, polarized light and
unpolarized light. Polarized light refers to light waves whose electric field only oscillates
in one plane only (as illustrated in fig. 6.4a). Conversely, unpolarized light refers to a
superposition of many beams, in the same direction of propagation, but each with random
polarization (as illustrated in fig. 6.4b). A common incandescent lightbulb produces
unpolarized light because each atom in the heated filament sends out light of random
polarization. Similarly, the light from the Sun is unpolarized.

6.3 Types of Polarization

i. Polarization by absorption
ii. Polarization by reflection
iii. Polarization by scattering

6.3.1 Polarization by Absorption

Absorption involves pass a beam of unpolarized light through a polarizer. A polarizer is a
material that is composed of long, thin, electrically conductive molecules oriented in a
specific direction. When a beam of light strikes a polarizer, it is readily absorbed if its
electric field is parallel to the molecules. Light whose electric field is perpendicular to the
molecules passes through the material with little absorption. As a result, the light that
passes through a polarizer is preferentially polarized along a specific direction.

111
Many years ago it was discovered accidentally that some natural crystals affect light
passing through them. Examples of such crystals include Quartz, Tourmaline and Iceland
spar. It was noticed that passing unpolarized light through such crystals affects the
brightness of the light coming out of the crystal. Slowly rotating the crystal along a line of
vision resulted in the light coming out of it becoming darker and darker until it completely
disappears at some point. When continuously rotated, the light would re-appear and would
become the brightest at a specific angle. We can then conclude that such crystals can be
regarded as polarizers. Other common examples of a polarizers are the well-known
Polaroid sheets which are used to make Polaroid sunglasses.

The most common technique for producing polarized light is to use a material that
transmits waves whose electric fields vibrate in a plane parallel to a certain direction and
that absorbs waves whose electric fields vibrate in all other directions. The plane in which
the material allows the transmission of the electric field is called the transmission axis.
Consider fig. 6.5 below showing the process of polarizing light with the aid of a polarizer

Figure 6.5: unpolarized light passing through a polarizer

The light that passes through this sheet is polarized vertically, and the transmitted electric
field vector is E0. A second polarizing sheet, called the analyzer, intercepts this beam with
its transmission axis at an angle of θ to the axis of the polarizer. The component of E0 that
is perpendicular to the axis of the analyzer is completely absorbed. The intensity of the

112
transmitted beam varies as the square of its magnitude, we conclude that the intensity I of
the (polarized) beam transmitted through the analyzer varies as:

I = I0 cos2 θ 6.1

Where I0 is the intensity of the polarized beam incident on the analyzer. The relationship
between the intensity of the beam and the angle θ as illustrated in Equation 6.1 is popularly
known as Malus’s Law. The Malus’s law applies to any two polarizing materials whose
transmission axes are at an angle θ to each other. Malus’s law shows that the intensity of
the transmitted beam is maximum when the transmission axes are parallel (θ=0 0r 180) and
is zero (complete absorption by the analyzer) when the transmission axes are perpendicular
1
to each other. Since the average value of cos2θ is 2 , the intensity of initially unpolarized

light is reduced by a factor of one-half as the light passes through a single ideal polarizer.

6.3.2 Polarization by Reflection

When an unpolarized light beam is reflected from a surface, the reflected light is
completely polarized, partially polarized, or unpolarized, depending on the angle of
incidence. In this case, full or partial polarization is dependent on the angle of incidence.
When the angle of incidence is either 00 or 900 the beam remains unpolarized. For other
angles of incidence, the reflected light is polarized to some extent, and for one particular
angle of incidence, the reflected light is completely polarized. The angle of incidence at
which this polarization occurs is called the polarizing angle θp .

Suppose an unpolarized light beam is incident on a surface, as illustrated in fig. 6.6 below
The beam can be described by two electric field components, one parallel to the surface
(represented by dots) and the other perpendicular to the first component and to the
direction of propagation (represented by orange arrows).

113
 Figure 6.6: Polarization by Reflection
It is found that the parallel component reflects more strongly than the other components,
and the result is a partially polarized beam. In addition, the refracted beam is also partially
polarized. Now suppose the angle of incidence, θ , is varied until the angle between the
reflected and refracted beams is 90° as illustrated in fig. 6.6(b). At this particular angle of
incidence (polarizing angle �� ), the reflected beam is completely polarized, with its electric
field vector parallel to the surface, while the refracted beam is partially polarized. The
polarizing angle is related to the index of refraction of the two materials on either side of
the boundary by the equation:
�1
tan �� =
�2
(6.2)

Where n1 is the index of refraction of the material in which the incident beam is traveling,
and �2 is that of the medium beyond the reflecting boundary. This expression is called
Brewster’s law, and the polarizing angle �� is sometimes called Brewster’s angle. Since n
varies with wavelength for a given substance, Brewster’s angle is also a function of
wavelength

114
6.3.3 Polarization by Scattering
When light is incident on a system of particles, such as a gas, the electrons in the medium
can absorb and re-radiate part of the light. The absorption and re-radiation of light by the
medium, called scattering, is what causes sunlight reaching an observer on Earth from
straight overhead to be polarized. Scattering effect can be observed by looking directly up
at the sky through a pair of sunglasses whose lenses are made of polarizing material. Less
light passes through at certain orientations of the lenses than at others.

The concept of polarization by scattering can be evidenced by the variations of colours in

the atmosphere as sunlight is scattered by the air molecules in the atmosphere. When light
of various wavelengths λ is incident on gas molecules of diameter � (where � ≫ �), the
relative intensity of the scattered light varies as 1 �4 . This implies that short wavelengths
(violet light) are scattered more in comparison to longer wavelengths (red light). Therefore,

when sunlight is scattered by gas molecules in the air, the short-wavelength radiation

(violet) is scattered more intensely than the long-wavelength radiation (red).

When you look up into the sky in a direction that is not toward the Sun, you see the
scattered light, which is predominantly violet. Your eyes, however, are not very sensitive
to violet light. Light of the next color in the spectrum, blue, is scattered with less intensity
than violet, but your eyes are far more sensitive to blue light than to violet light. Hence,
you see a blue sky.

However, If you look directly toward the west at sunset (or toward the east at sunrise), you
are looking in a direction toward the Sun and are seeing light that has passed through a
large distance of air. Most of the blue light has been scattered by the air between you and
the Sun. The light that survives this trip through the air to you has had much of its blue
component scattered and is therefore heavily weighted toward the red end of the spectrum;
as a result, you see the red and orange colors of sunset (or sunrise).

115
 6.7: colour variations due to polarization of sunlight by air molecules
Ultimately, the scattering of sunlight by the atmosphere produces polarized light for an
observer looking at right angles to the direction of the Sun. This observer also sees more
blue light than red. An observer looking toward the Sun sees unpolarized light that
contains more red light than blue.

Summary
Electromagnetic waves are transverse waves that have both electric and magnetic waves
oscillating perpendicular to each other. Polarization of an electromagnetic wave simply
refers to the direction which the electric field of the electromagnetic wave is restricted to
oscillate. Polarization of light can be achieved through reflection, scattering or absorption.

Further Reading
Polarization by double refraction.

116
Unit Test
1. Certain sunglasses use a polarizing material to reduce the intensity of light reflected as
glare from water or automobile windshields. What orientation should the polarizing filters
have to be most effective?

2. Why is the following situation impossible? A technician is measuring the index of

refraction of a solid material by observing the polarization of light reflected from

its surface. She notices that when a light beam is projected from air onto the material
surface, the reflected light is totally polarized parallel to the surface when

the incident angle is 41.0°.

3. Plane-polarized light is incident on a single polarizing disk with the direction of E0

parallel to the direction of the transmission axis. Through what angle should the disk be
rotated so that the intensity in the transmitted beam is reduced by a factor of

(a) 3.00,

(b) 5.00

(c) 10.0

4. Unpolarized light passes through two ideal Polaroid sheets. The axis of the first is
vertical, and the axis of the second is at 30.0° to the vertical. What fraction of

the incident light is transmitted?

117
Answers to Unit Activities
1. The polarizers should absorb light with its electric field horizontal.

�
2. tan �� = 2 �1 , the index of refraction n2 of the solid material must be larger than that
of air (n1 = 1.00). Therefore, we must havetan �� > 1. For this to be true, we must
have�� > 450 . so θp = 41.0° is not possible.

3. (a) 54.70
(b)63.4
(c)71.6
4. 0.375

118
 Glossary
Accommodation of the eye refers to the ability of the eye to see near and distant objects
by automatic adjustment of the thickness of the eye lens.

A restoring Force is one that opposes displacement of the system.

Doppler effect refers to the change in pitch (or frequency) due to the relative motion
between a source and the receiver.
Freuency is the the number of cycles completed per unit time.
Period is the time required to complete a full cycle.
Polarization refers to the direction which the electric field of the electromagnetic wave is
restricted to oscillate
Radius of curvature is the distance between the centre o curvature of the mirror and its
poles
Real Image is one in which light actually passes through the image point (can be displayed
on screens)
Virtual Image is one in which the light does not pass through the image point (cannot be
displayed on screens)

119
 References
Bueche, F. J., & Hecht, E. (1997). College Physics. |New York: The McGraw-Hill
Companies,Inc.
Duncan, T. (2007). Advanced Physics. London: Hodder Murray.
Halliday, D., & Resnick, R. (2014). Fundamentals of Physics. Danvers,MA: John Wiley
and Sons.
Physics for Rwanda Secondary Schools (Teahchers guide Book 4). (2020). Government of
Rwanda.
Serway, R. A. (2014). Physics for Scientists and Engineers. Boston, MA: Brooks/Cole.
Walker, J. S. (2016). PHYSICS. San Francisco: Pearson.

120
 Module Test
1.a Define polarization. (1 mark)
i. State three methods of polarization. (3 marks)
ii. From question (i) above, describe one method of polarization. (3marks)
iii. Unpolarized light passes through two polarizers whose transmission axes are at
an angle of 30.0° with respect to each other. What fraction of the incident
intensity is transmitted through the polarizers? (3 marks)
b. Vertically polarized light with intensity 0.55W/m2 passes through a polarizer whose
transmission axis is at an angle of 85.00 with the vertical. What is the intensity of
the transmitted light. (4 marks)
c. Explain three uses of curved mirrors. (3 marks)

2.a Derive an appropriate relation connecting the distances of an object and its image

and the focal length from the surface of a convex spherical mirror. (5 marks)

b. Two converging lenses, A and B, with focal lengths 20cm and 25 cm are placed

80.0 cm apart. An object is placed 60.0 cm in front of the first lens. Determine

(i) the position of the final image formed by the combination of the two lenses.(4

marks)

(ii) the magnification of the final image formed by the combination of the two lenses.

(4 marks)

c. A simple harmonic oscillator is represented by the equation: � = 0.7 sin (440� +

0.61). Find the values of

(i) Amplitude (1 mark)

(ii) Angular frequency (1 mark)

(iii)Frequency of oscillation (2 mark)

121
(iv) Time period of oscillation (2 mark)

(v) Initial phase (1 mark)

3.a A train on one track moves in the same direction as a second train on the adjacent
track. The first train, which is ahead of the second train and moves with a speed of
36.8 m/s, blows a horn whose frequency is 124 Hz. If the frequency heard on the
second train is 135 Hz, what is its speed? (5 marks)
b. A motorcycle and a police car are moving toward one another. The police car emits
sound with a frequency of 523 Hz and has a speed of 32.2 /s. The motorcycle has a
speed of 14.8 m/s. What frequency does the motorcyclist hear? (5 marks)
c. A person wears a hearing aid that uniformly increases the sound level of all audible
frequencies of sound by 30.0 dB. The hearing aid picks up sound having a
frequency of 250 Hz at an intensity of 3.0�10−11 �/�2 . What is the intensity
delivered to the eardrum? (6 marks)
d. You drop a stone from rest into a well that is 7.35 m deep. How much time elapses
before you hear the splash? ( 4 marks)

4.a A converging lens has a focal length of 25.5 cm. If it is placed 72.5 cm from an
object, at what distance from the lens will the image be? (4 marks)
b. An empty soda pop bottle is to be used as a musical instrument in a band (see
figure T1). In order to be tuned properly, the fundamental frequency of the bottle
must be 440.0 Hz.
i. If the bottle is 26.0 cm tall, how high should it be filled with water to produce
the desired frequency? Treat the bottle as a pipe that is closed at one end (the
surface of the water) and open at the other end. (4 marks)

122
 Figure T1
ii. What is the frequency of the next higher harmonic for this bottle? (1 marks).
c. A spring whose force constant is 80N/m hangs vertically supporting a 1kg mass at
rest. Find the distance by which the mass should be pulled down so that on being
released it may pass the equilibrium position with a velocity of 1m/s. (4 marks)
d. A particle executes simple harmonic motion of period 10 seconds and amplitude
5cm. Calculate the maximum velocity of the particle. ( 4 marks)

5.a How many dark fringes will be produced on either side of the central maximum if
green light (� = 553 nm) is incident on a slit that is 8.00 mm wide? (4 marks)
b. Monochromatic light with � = 648 nm shines down on a plano-convex lens lying
on a piece of plate glass, as shown in figure I. When viewing from above, one sees
a set of concentric dark and bright fringes, referred to as Newton’s rings. If the
radius of the twelfth dark ring from the center is measured to be 1.56 cm, what is
the radius of curvature, R, of the lens? (7 marks)

Figure T2

i. If light with a longer wavelength is used with this system, will the radius of the
twelfth dark ring be greater than or less than 1.56 cm? Explain. (2 marks)
c. A physics instructor wants to produce a double-slit interference pattern large
enough for her class to see. For the size of the room, she decides that the distance
between successive bright fringes on the screen should be at least 2.50 cm. If the

123
slits have a separation d = 0.0175 mm, what is the minimum distance from the slits
to the screen when 632.8-nm light from a He–Ne laser is used? (7 marks)

END OF QUESTION PAPER.

124