Frequency Between Media
As light travels from one
medium to another, its
frequency does not
change
Both the wave speed and
the wavelength are
changing
Note: A hertz is defined
as the number of
oscillations or repeats of a
cycle in one second.
Index of Refraction Extended
The frequency stays the same as the wave travels
from one medium to the other
v = ƒλ
ƒ1 = ƒ2 but v1 v2 so λ1 λ2
The ratio of the indices of refraction of the two
media can be expressed as various ratios
c
λ1 v1 n1 n2
λ2 v 2 c n1
n2
More About Index of
Refraction
The previous relationship can be simplified to
compare wavelengths and indices: λ1n1 = λ2n2
In air, n1 1 and the index of refraction of the
material can be defined in terms of the
wavelengths
λ λ in vacuum
n
λn λ in a medium
Snell’s Law of Refraction
From 1st lec. sin θ2 v 2 constant
sin θ1 v1
n1 sin θ1 = n2 sin θ2
θ1 is the angle of incidence
θ2 is the angle of refraction
The experimental discovery of this relationship
is usually credited to Willebrord Snell and is
therefore known as Snell’s law of refraction
Snell’s Law – Problem
A light ray of wavelength 589
nm traveling through air is
incident on a smooth, flat slab
of crown glass at an angle of
30.0° to the normal.
(A) Find the angle of refraction.
(B) Find the speed of this light
once it enters the glass.
(C) What is the wavelength of
this light in the glass?
Note: n for crown glass = 1.52
Solution
(A) Find the angle
of refraction.
(B) Find the speed
of this light once it
enters the glass.
(C) What is the
wavelength of this
light in the glass?
Huygens’s Construction for a
Plane Wave
At t = 0, the wave front is
indicated by the plane AA’
The points are
representative sources for
the wavelets
After the wavelets have
moved a distance cΔt, a
new plane BB’ can be
drawn tangent to the
wavefronts
Huygens’s Construction for a
Spherical Wave
The inner arc represents
part of the spherical
wave
The points are
representative points
where wavelets are
propagated
The new wavefront is
tangent at each point to
the wavelet
Dispersion
For a given material, the index of refraction
varies with the wavelength of the light passing
through the material
This dependence of n on λ is called
dispersion
Snell’s law indicates light of different
wavelengths is bent at different angles when
incident on a refracting material
Angle of Deviation
The ray emerges
refracted from its original
direction of travel by an
angle δ, called the
angle of deviation
The angle of deviation
depends on the
wavelength
Refraction in a Prism
Since all the colors
have different angles
of deviation, white
light will spread out
into a spectrum
Violet deviates the
most
Red deviates the least
The remaining colors
are in between
The Rainbow
A ray of light strikes a drop of water in the
atmosphere
It undergoes both reflection and refraction
First refraction at the front of the drop
Violet light will deviate the most
Red light will deviate the least
The Rainbow, 2
At the back surface the light is
reflected
It is refracted again as it returns to
the front surface and moves into the
air
The rays leave the drop at various
angles
The angle between the white light
and the most intense violet ray is
40°
The angle between the white light
and the most intense red ray is
42°
Total Internal Reflection
A phenomenon called total internal reflection
can occur when light is directed from a medium
having a given index of refraction toward one
having a lower index of refraction
Possible Beam Directions
Possible directions of
the beam are
indicated by rays
numbered 1 through 5
The refracted rays are
bent away from the
normal since n1 > n2
Critical Angle
There is a particular angle of
incidence that will result in
an angle of refraction of 90°
This angle of incidence is
called the critical angle, θC
We can use Snell’s law of
refraction to find the
critical angle. When
n2
sin θC (for n1 n2 )
n1
Critical Angle, cont.
For angles of incidence greater than the
critical angle, the beam is entirely reflected at
the boundary
This ray obeys the law of reflection at the
boundary
Total internal reflection occurs only when light
is directed from a medium of a given index of
refraction toward a medium of lower index of
refraction
Problem
For 589 nm light, calculate the critical angle
for the following materials surrounded by air:
(a) diamond, (b) flint glass, (c) ice and (d)
water.
Note: n =2.419, n
diamond =1.66 and n =1.309
flint glass ice
Solution
n2
sin θC (for n1 n2 )
n1
1 1
(a) sin 24.4
2.419
1 1
(b) sin 37.0
1.66
1 1
(c) sin 49.8
1.309
Fiber Optics
An application of internal
reflection
Plastic or glass rods are
used to “pipe” light from one
place to another
Applications include:
medical use of fiber optic
cables for diagnosis and
correction of medical
problems
Telecommunications
Types of Images
A real image is formed when light rays pass
through and diverge from the image point
Real images can be displayed on screens
A virtual image is formed when light rays do
not pass through the image point but only
appear to diverge from that point
Virtual images can not be displayed on screens
Images Formed by Flat
Mirrors
Simplest possible mirror
Light rays leave the
source and are reflected
from the mirror
Point I is called the
image of the object at
point O
The image is virtual
Images Formed by Flat
Mirrors, 2
A flat mirror always produces a virtual image
Geometry can be used to determine the
properties of the image
There are an infinite number of choices of
direction in which light rays could leave each
point on the object
Two rays are needed to determine where an
image is formed
Images Formed by Flat
Mirrors, 3
One ray starts at point
P, travels to Q and
reflects back on itself
Another ray follows the
path PR and reflects
according to the law of
reflection
The triangles PQR and
P’QR are identical
Images Formed by Flat
Mirrors, 4
To observe the image, the observer would trace
back the two reflected rays to P’
Point P’ is the point where the rays appear to
have originated
The image formed by an object placed in front
of a flat mirror is as far behind the mirror as the
object is in front of the mirror
|p| = |q|
Lateral Magnification
Lateral magnification, M, is defined as
Image height h '
M
Object height h
This is the general magnification for any type of mirror
It is also valid for images formed by lenses
Magnification does not always mean bigger, the size
can either increase or decrease
M can be less than or greater than 1
The lateral magnification of a flat mirror is 1
This means that h’ = h for all images
Properties of the Image Formed
by a Flat Mirror – Summary
The image is as far behind the mirror as the object is
in front
|p| = |q|
The image is unmagnified [M = 1]
The image height is the same as the object height
h’ = h
The image is virtual
The image is upright
It has the same orientation as the object
There is a front-back reversal in the image
Application – Day and Night
Settings on Auto Mirrors
With the daytime setting, the bright beam of reflected light is
directed into the driver’s eyes
With the nighttime setting, the dim beam of reflected light is
directed into the driver’s eyes, while the bright beam goes
elsewhere.
So, the brightness of trailing headlights does not become a
hazard.
Spherical Mirrors
A spherical mirror has the shape of a section of a
sphere
The mirror focuses incoming parallel rays to a point
A concave spherical mirror has the silvered surface
of the mirror on the inner, or concave, side of the
curve
A convex spherical mirror has the silvered surface of
the mirror on the outer, or convex, side of the curve
Concave Mirror, Notation
The mirror has a radius of
curvature of R
Its center of curvature is
the point C
Point V is the center of the
spherical segment
A line drawn from C to V is
called the principal axis of
the mirror
Spherical Aberration
Rays that are far from the
principal axis converge to
other points on the principal
axis
This produces a blurred
image
The effect is called
spherical aberration
Image Formed by a Concave
Mirror
Geometry can be used
to determine the
magnification of the
image
h' q
M
h p
h’ is negative when the
image is inverted with
respect to the object
Image Formed by a Concave
Mirror
Geometry also shows the relationship between the
image and object distances
1 1 2
p q R
This relation is called the mirror equation
If p is much greater than R, (the object is very far
from the mirror) then the image point is half-way
between the center of curvature and the center point
of the mirror
p → ∞ , then 1/p 0 and q R/2
Focal Length
Focal Point and Focal Length,
cont.
The focal point is dependent just on the
curvature of the mirror, not on the location of
the object
It also does not depend on the material from
which the mirror is made
ƒ=R/2
The mirror equation can be expressed as
1 1 1
p q ƒ
Convex Mirrors
A convex mirror is sometimes called a diverging
mirror
The light reflects from the outer, convex side
The rays from any point on the object diverge after
reflection as though they were coming from some
point behind the mirror
The image is virtual because the reflected rays only
appear to originate at the image point
Image Formed by a Convex
Mirror
In general, the image formed by a convex mirror is
upright, virtual, and smaller than the object
Sign Conventions
These sign conventions
apply to both concave and
convex mirrors
The equations used for the
concave mirror also apply to
the convex mirror
Sign Conventions, Summary
Table
Ray Diagrams
A ray diagram can be used to determine the
position and size of an image
They are graphical constructions which reveal
the nature of the image
They can also be used to check the
parameters calculated from the mirror and
magnification equations
Drawing a Ray Diagram
To draw a ray diagram, you need to know:
The position of the object
The locations of the focal point and the center of curvature
Three rays are drawn
They all start from the same position on the object
The intersection of any two of the rays at a point
locates the image
The third ray serves as a check of the construction
The Rays in a Ray Diagram –
Concave Mirrors
Ray 1 is drawn from the
top of the object parallel
to the principal axis and
is reflected through the
focal point, F
Ray 2 is drawn from the
top of the object through
the focal point and is
reflected parallel to the
principal axis
Ray 3 is drawn through
the center of curvature,
C, and is reflected back
on itself
Ray Diagram for a Concave
Mirror, p > R
The center of curvature is between the object and the
concave mirror surface
The image is real
The image is inverted
The image is smaller than the object (reduced)
Ray Diagram for a Concave
Mirror, p < f
The object is between the mirror surface and the focal point
The image is virtual
The image is upright
The image is larger than the object (enlarged)
This latter situation applies when you use a shaving mirror or a
makeup mirror, both of which are concave.
The Rays in a Ray Diagram –
Convex Mirrors
Ray 1 is drawn from the top
of the object parallel to the
principal axis and is
reflected away from the
focal point, F
Ray 2 is drawn from the top
of the object toward the
focal point and is reflected
parallel to the principal axis
Ray 3 is drawn through the
center of curvature, C, on
the back side of the mirror
and is reflected back on
itself
Ray Diagram for a Convex
Mirror
The object is in front of a convex mirror
The image is virtual
The image is upright
The image is smaller than the object (reduced)
Problem 1
A spherical mirror has a focal length of +10.0 cm.
(A) Locate and describe the image for an object
distance of 25 cm.
(B) Locate and describe the image for an object
distance of 10 cm.
(C) Locate and describe the image for an object
distance of 5 cm.
Solution (A)
1 1 1
p q ƒ
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.
Solution (B)
1 1 1
p q ƒ
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.
Solution (C)
1 1 1
p q ƒ
The image is twice as large as the object, and the positive sign
for M indicates that the image is upright.
The negative value of the image distance tells us that the image
is virtual, as expected.
Problem 2
An automobile rearview
mirror as shown in the
figure shows an image of a
truck located 10.0 m from
the mirror. The focal length
of the mirror is -0.6 m.
(A) Find the position of the
image of the truck.
(B) Find the magnification of
the image.
Solution (A)
1 1 1
p q ƒ
The negative value of q in part (A) indicates that the image is virtual, or behind
the mirror.
The magnification in part (B) indicates that the image is much smaller than the
truck and is upright because M is positive.
