Diffraction

PHY1008 Modern Physics

Class - 2
 Diffraction

Module 2 – Wave Optics (Diffraction)

Single Slit- Diffraction

 Introduction to Diffraction Patterns Diffraction
Condition for Diffraction:
A light of wavelength comparable to or larger than the width of a slit spreads out in all forward
directions upon passing through the slit

The pattern consists of a broad, intense central band

(Central Maximum) flanked by a series of narrower, less
intense additional bands (Side Maxima or Secondary
Maxima) and a series of dark bands (Minima)

Diffraction Pattern obtained when light passes through a narrow vertical slit
 Introduction to Diffraction Patterns Diffraction
Light from a small source passes by the edge of an opaque object

Diffraction pattern consists of bright and dark

fringes appears on the screen in the region
above the edge of the object

A bright spot occurs at the center due to

Diffraction pattern created by the constructive interference
illumination of a coin (penny) Circular fringes extend outward from the
shadow’s edge
 Diffraction
Single slit, double slit and multiple slit (grating)
 Diffraction Patterns from Narrow Slits Diffraction
Let us consider a light passing through a narrow opening (slit) and projected on a
screen

Assume that the observing screen is far from the slit

and the rays reaching the screen are approximately
parallel

Light enters a single slit from the left and diffract as

it propagates toward a screen (Fraunhofer
Diffraction Pattern is observed)

Bright fringe is observed along the axis at θ = 0

Alternate dark and bright fringes on each of the
central bright fringe
Fraunhofer Diffraction Pattern
 Diffraction Patterns from Narrow Slits Diffraction

We have assumed that slits are point sources of light

But finite width of slits is the basis for understanding
Fraunhofer Diffraction

We can explain some important features of this diffraction by

examining waves coming from various portions of the slit
 Diffraction Patterns from Narrow Slits Diffraction

According to Huygens's principle, each portion of the slit

acts as a source of light waves

Light from one portion of the slit can interfere with light
from another portion and the resultant light intensity on the
screen depends on the direction θ

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
 Diffraction
Analysis of Diffraction Patterns from Narrow Slits

Let us divide the slit in to two equal parts and also remember
that all the waves are in phase as they leave the slit

Consider the rays 1 and 3

Ray 1 travels farther than Ray 3 by an amount equal to the
path difference
Path difference = (a/2) sin θ
a – Width of the slit
Similarly, the path difference between rays 2 & 4 = (a/2) sin θ
and rays 3 & 5

If the path difference is exactly λ/2 (corresponding to a phase

difference of 180⁰), the pairs of waves cancel each other
(Destructive Interference occurs)
 Diffraction
Analysis of Diffraction Patterns from Narrow Slits

The destructive interference occurs (cancellation of

waves) for any two rays that originate at points separated
by half the slit width (phase difference between two such
points is 180⁰)

Waves from the upper half of the slit interfere destructively

with waves from the lower half of the slit when,

If we consider waves at angle θ both above the dashed line

and below the dashed line

(screen is dark)
 Diffraction
Analysis of Diffraction Patterns from Narrow Slits

Dividing the slit into four equal parts and using similar
reasoning, we find that the viewing screen is also dark
when

Similarly, if we divide the slit into six equal parts, we can

observe the darkness when

General condition for destructive interference is

 Diffraction
Analysis of Diffraction Patterns from Narrow Slits

This equation gives the values θdark for

which the diffraction pattern has zero
intensity (dark fringe)

 A broad, central fringe is observed

 Central bright fringe is flanked by much weaker
bright fringes alternating with dark fringes
 Various dark fringes occur at the values of θdark
that satisfy the above equation
 Each bright fringe peak lies approximately
halfway between its bordering dark-fringe minima

There is no central dark fringe (represented by the absence of m = 0)

 Fraunhofer Diffraction at a Single Slit Diffraction
 According to Fraunhofer
diffraction, the incident wavefront
must be plane. Hence, either the
light source must be at a very
large distance or a collimating
lens must be used to get plane
wavefront

 Consider a slit AB of width ‘a’ and

the monochromatic source of light
‘S’ emitting the waves incident on
a convex lens (L1)

 The source is placed at the focal

point of convex lens so that it
becomes parallel beam and is
incident normally on the slit AB
 Diffraction
 The light emerged is focused by a
convex lens on screen ‘XY’ placed at
the focal point of the lens ‘L2’ as shown
in Fig.

 According to geometrical optics, a

bright image having uniform
illumination with sharp edges should be
formed on the screen.

 However, a diffraction pattern consists

of a wide central bright band
surrounded by a few narrow bands on
either side which are alternatively dark
and bright are observed on the screen
Diffraction
Diffraction
Diffraction
 Diffraction
Case (i): Condition for central maxima
 Diffraction
Case (ii): Condition for minima
 Diffraction
 The intensity distribution due to the diffraction at a single slit is shown in Fig. The
graph consists of central maximum with minima and secondary maxima alternately
 Diffraction
Case (iii): Condition for secondary maxima
Diffraction
Diffraction
 Diffraction
From the condition for minima y = (mD)/a

-3D/a -D/a D/a 3D/a

-2D/a 2D/a

Width of central fringe = D/a + D/a

Width of central fringe = 2D/a
 Problem 1 Diffraction
Light with wavelength of 500 nm passes through a slit a = 0.2 mm wide. The diffraction pattern on a
screen D = 60 cm away. Determine the distance between the central maximum and the second order
minimum.
 = 500 nm = 500x10-9 m
D = 60 cm = 60x10-2 m
a = 0.2 mm = 0.2x10-3 m
y=?
y = mλD/a
y = (2x500x10-9x60x10-2)/(0.2x10-3)
y = 3 x 10-3 m
y = 3 mm
 Problem 2 Diffraction
In a single-slit diffraction, the second-order dark fringe is at a distance 1.40 mm from the center
of the central maximum. The screen is 80.0 cm from a slit of width 0.80 mm. Assuming that the
incident light is monochromatic, calculate the approximate wavelength of the incident light.

D = 80 cm = 80x10-2 m
a = 0.8 mm = 0.8x10-3 m
y = 1.4 mm = 1.4x10-3 m
= ?
y = mλD/a
λ = ya/mD
λ = (1.4x10-3x0.8x10-3)/(2x80x10-2)
λ = 700 nm
 Problem 3 Diffraction
A beam of monochromatic light approaches a barrier having four openings, A, B,
C, and D, of different sizes as shown below. Which opening will cause the
greatest diffraction?

asin= m
sin = m/a
sin or  is proportional to /a
The smallest opening distance causes the greatest diffraction
 Problem 4 Diffraction
Four beams of different wavelengths approaches a barrier having four openings,
A, B, C, and D, of different sizes as shown below. Which opening will cause the
greatest and smallest diffraction?

sin or  is proportional to /a

asin= m
sin = m/a
sin or  is proportional to /a
The greatest diffraction is caused by the smallest opening distance and the longest wavelength.
 Problem 5 Diffraction
What is the distance between two successive dark fringes when a light with
wavelength 15 nm shines through a slit of length 12 mm onto a wall which is
placed 2.5 m from the slit?

 = 15 nm = 15x10-9 m
D = 2.5 m
a = 12 mm = 12x10-3 m
ym = mλD/a and ym+1 = (m+1)λD/a

ym+1 - ym = (m+1)λD/a - mλD/a

ym+1 - ym = λD/a = (15x10-9x 2.5)/ (12x10-3) = 3.1x10-6 m

 Problem 6 Diffraction
Visible light of wavelength 550 nm falls on a single slit and produces its second diffraction
minimum at an angle of 45o relative to the incident direction of the light.
1.What is the width of the slit?
2. At what angle is the first order minimum produced?

asin= mλ
We are given that
λ=550 nm, m=2, and 2=45.0°
Solving the equation asin= mλ asin= mλ
sin = mλ/a
asin2= mλ  = sin-1(mλ/a)
 for first order diff (m =1)
 = sin-1(λ/a)
 = sin-1(550x10-9)/(1.56x10-6)
 = Sin-1(0.352)
 Problem 7 Diffraction
Light of wavelength 580 nm is incident on a slit having a width of 0.300 mm. The viewing screen is 2.00
m from the slit. Find the width of the central fringe.
y
For the two dark fringes that flank the central bright
fringe m = ± 1 y1

y1

Let y represents the vertical position along the viewing y

screen (measured from the point on the screen directly
behind the slit)
 Diffraction
The width of the central fringe is twice the absolute value of y1

y1

y1

The central fringe width (7.73 mm) is much greater than the width of the slit (0.3 mm)
 Problem 8 Diffraction
What if the slit width is increased by an order of magnitude to 3.00 mm? What happens to the diffraction
pattern?
y

y1

y1

The central fringe width (0.773 mm) is smaller than the width of the slit (3.00 mm)

For large values of a, the various maxima and minima are so closely spaced that only a large,
central bright area resembling the geometric image of the slit is observed.
This concept is very important in the performance of optical instruments such as telescopes
 Problem 9 Diffraction
Two slits are 0.05 m apart. A laser of wavelength 633nm is incident to the slits.
A screen is placed 2m from the slits. Calculate the position of the first and second
bright fringe? And What is the maximum number of destructive interference
spots there can be on either side of the central maximum?
 Problem 10 Diffraction
Light with a wavelength of λ = 587.5 nm illuminates a single slit which has a width of 0.75 mm.
(a) At what distance from the slit should a screen be located if the first minimum in the
diffraction pattern is to be 0.85 mm from the center of the screen? (b) What is the width of
the central maximum?
 Problem 11 Diffraction
A screen is placed 50.0 cm from a single slit, which is illuminated with light of
wavelength 680 nm. If the distance between the first and third minima in the
diffraction pattern is 3.00 mm, what is the width of the slit?
 Problem 12 Diffraction
The atoms in a crystal lie in planes separated by a few tenths of a nanometer. Can they produce a
diffraction pattern for visible light as they do for X-rays? Explain your answer with reference to Bragg’s
law

Substitute some values of d and λ and calculate sin θ

Based on the value of sin θ, explain your answer

 Diffraction
Expression for width of central maxima
 We know that the condition for central minima is a sin θ = mλ

For m = 1, sin θ = λ/a

 If ‘x’ is the half width of central bright maxima and D is the distance between the slit and the
screen then

sin θ = x/D

Therefore, x/D = λ/a

x = Dλ/a

 Hence, the width of central maxima, 2x = 2Dλ/a

 Diffraction
 If the lens (L2) is nearer to the slit or the screen is far away from the lens (L 2) then

sin θ = x/f (f is the focal length of lens L2)

 We know that,
sin θ = λ/a

Therefore, x/f = λ/a

x = fλ/a

 This means that the fringe width increases with decreasing slit width ‘a’, which means when slit
becomes narrower, the fringe becomes wide

 Also, for a given value of ‘a’, the fringe width is proportional to the wavelength ‘λ’

 If the wavelength is higher, the fringe becomes wider

 Diffraction

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