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UPH013 – Physics
 Observable Interference

Light
source
Interference pattern
(with white light)
Diffraction
Diffraction definition:

Diffraction is a phenomenon that occurs when wave encounters an obstacle or a slit. It is

defined as bending/spreading out of waves as they pass by some objects or through a
finite-width aperture.

Diffraction is the why we can hear sound even though we are not in a straight line from
the source – sound waves will diffract around doors, corners, and other barriers.

The amount of diffraction depends on the wavelength, which is

why we can hear around corners but not see around them.
Difference between interference and diffraction:
If you have two infinitely-narrow double slits, there will be just interference, but for
finite-width slits there can be both interference and diffraction effects.
Finite width slit: The width of slit ~ wavelength λ.
One finite width slit is equivalent to the N → ∞ limit of the N-slit result of
interference.

Interference pattern (with white light)

Diffraction pattern (with monochromatic light)

Diffraction is simply the N → ∞ limit of interference, there is technically no need to

introduce a new term for it. But on the other hand, a specific kind of pattern arises,
so it makes sense to give it its own name.
Difference between interference and diffraction:

 InInterference, minima are usually perfectly dark while this is not

the case for diffraction.

 Ininterference, all maxima are of same intensity but they have

varying intensity in diffraction.
 Fringe width could be equal in some cases in interference while
they are never equal in diffraction.

 In interference, interaction takes place between two separate

wavefronts originating from two coherent sources while in
diffraction, interaction takes place between secondary wavelets
originating from same wavefront.
Two classes of diffraction:

When either the source or the screen or

both are at a finite distance from the
aperture, the diffraction pattern
corresponds to the Fresnel class.

In Fraunhofer class
both the source and the
screen are at infinity.
Fresnel diffraction

Non-zero minima
Fraunhofer diffraction
 Fraunhofer diffraction Fresnel diffraction
Source of light and screen are at The source and the screen are at
infinite distance from diffracting finite distances from the diffracting
device. device.
Incident wave-front is plane The incident wave-front is either
wave-front obtained with the help cylindrical or spherical.
of convex lens.
Diffracted light is collected by a No lens is used for observation.
convex lens.
The center of diffraction pattern The center of diffraction pattern
is always bright. may be dark or bright depending
upon the number of Fresnel’s zone.
Single slit or double slit or Fresnel’s zone plate is used to get
grating is used to get diffraction diffraction pattern.
pattern.
 Fraunhofer Diffraction at single slit:

Slit is a narrow
aperture whose
width is very small
with respect to its
length.

To get a diffraction pattern the diffracted light is

focused on the screen using a lens. The lens is very
close to the slit and screen is at second focus of the lens.
 Fraunhofer Diffraction at single slit:

We need to know the intensity profile at the screen.

In other words:
We wish to know the intensity at an arbitrary point P1 at screen where all the diffracted
waves at an angle θ w.r.t. direction of initial light ray get focused using lens L2. Angle
between P1O and P0O is θ.
 Fraunhofer Diffraction at single slit:
Methodology:

Divide the slit of width ‘e’ into a

large number of segments which act Diffracted light rays
as secondary sources emitting
secondary waves.

These secondary waves travel in all

the directions and meet at a point on
the screen depending upon the angle
of diffraction and interfere there.

All the (parallel) diffracted light rays

travelling in direction θ w.r.t. the
original wave meet at a single point at
the focal plane of the lens.
For a start, choose an angle of diffraction θ, find the path
difference between consecutive light waves and find the resultant
amplitude at screen, then generalize the result.
 2 Let the width of the slit is divided into n equal
parts and the amplitude of the wave from each
part is ‘a’.
4
1  The path difference between secondary
2  wavelets from A and B in direction θ is
3 
4 
5  Corresponding phase difference is
6

The phase difference between any two

consecutive waves from these parts is

Thus there are n equal amplitude light rays with increasing phase difference w.r.t.
the first ray are travelling in same direction. They will give a resultant amplitude
R at the point where they will meet.
 Resultant amplitude on the screen

Now this problem is reduced to find the resultant

of n equal magnitude vectors with phase
difference between consecutive vectors equal to δ.

Sin

MX 1 or = MC sin

Sin

MY 𝑛 θ or = MC sin

=
𝝅𝒆 𝒔𝒊𝒏𝜽
λ
θ= 𝜽= 𝝅𝒆 𝒔𝒊𝒏𝜽
nλ
 Resultant amplitude on the screen

The resultant Rθ at a point at screen is:

A0 is amplitude of incident
is very large 𝜃 light beam, and ‘a’ is
amplitude of secondary
light ray from each of n
𝜃
segments of the slit.
A0 = n.a
𝟐
Intensity of light is square
𝟎
of its amplitude in wave
mechanics.
Intensity distribution Case I: sinα = 0
(Position of Maxima and minima) α = ± mπ
at m = 0 (α = 0)
lim  sin  
  1
  0  
This gives a maxima known as central
principal maxima.
Intensity of diffraction fringes on the screen m = ±1, ±2, ±3, … give minimum
is a function of α or θ. intensity.
The primary condition for maxima and Thus for mth minima α = ± mπ
minima is and 𝒎

This is the condition for minima.

Either Sinα = 0
or Case II: α cosα – sinα = 0
or
These two cases give the positions of This is the condition for secondary maxima.
maxima and minima as follows:
Intensity distribution
(Position of Maxima and minima)

Case II: tanα = α (= y)

Only numerical solution of

above equation are
possible that give values of
α for secondary maxima.

3 5 7 9 Exactly
    ,   ,   ,   , ...   1.43 , 2.46 , 3.47 , ...
2 2 2 2
So, roughly the condition
for secondary maxima is
or 𝒎
Intensity distribution
(Position of Maxima and minima)

Intensity of principle maxima

2
 sin  
I  I0    I0
  
Intensity of secondary maxima
2 2
 sin(3 / 2)   2 I0
I1  I 0   I 
 3 / 2 
0
 3  22
2 2
 sin(5 / 2)   2  I0
I2  I0   I 
 5 / 2 
0
 5  61.5
2 2
 sin(7 / 2)   2  I0
I3  I0   I 
 7 / 2 
0
 7  121
2 2
 sin(9 / 2)   2  I0
I4  I0    I0   
 9 / 2   9  200
Intensity distribution
(Position of Maxima and minima)

Thus in case of Fraunhoffer diffraction fro a single slit the intensities of the
successive maxima are nearly in ratio:
1 1 1 1
I 0 : I1 : I 2 : I 3 : I 4 : ...  1 : : : : : ...
22 61.5 121 200
 3 f
z 
Fringe width e   3
5 f 5
z    
2e 2
2 f   2
z 
e 3
  
3 f 2
z 
2e   f is very large
f
z 
e
and
e
z is very small.
f
  z 
e
3 3 f
   z 
tanθ = θ = z/f 2
2e

For minima e.sinθ = mλ

For secondary maxima e.sinθ = (m+1/2)λ e.sinθ = e. θ = mλ

So form principal maxima, the distance of

Facts and uses: z 
3 f
e   3
5 f 5
z    
2e 2
By the condition for minima 2 f   2
z 
e 3
e.sinθ = mλ 3 f
 
2

z 
It is clear 2e  
f
e>λ z 
e
e
Essential condition for diffraction;
finite width of the slit.   z 
f
e
3 3 f
   z 
 If e is infinitesimally small (i.e., 2
2e
e < λ), the diffraction pattern will
not be found.

Distance of from center

Known: f, e
Measurables: z
Wavelength of light can be calculated.
 Formulas at a glance (Single slit Fraunhofer diffraction)

𝟐
Resultant intensity at screen due to
single slit diffraction: 𝟎

For principal maxima α = 0 θ=0

For mth minima α = ± mπ 𝒎

For secondary maxima α = ± (m+1/2)π 𝒎

Position of mth minima from center min

Position of mth maxima from center max

 Fraunhofer Diffraction at double slit:
e – slit width
d = S1S2 – separation of
slits

𝜋𝑒 sin 𝜃
𝛼=
𝜆

The diffraction by two parallel slits is a case of diffraction as well as interference.

Two contributions -
 Diffraction due to individual slits
and
 Interference due to diffracted secondary waves from corresponding points on the
two slits.

This problem can be treated as there are two identical sources of light at positions S1
and S2 emitting waves of equal amplitude R and the resultant on the screen will be
due to interference of two waves in direction θ. The phase difference between the
waves from S1 and S2 is:
 Fraunhofer Diffraction at double slit:

Resultant amplitude R at P1:

𝟐
𝟐
𝟎

The first term is the intensity due to diffracted

beams by single slit.
𝑛
The second term corresponds to the
interference of the two diffracted beams from
two slits. Conditions of maxima and minima 𝑛
are given by, respectively:

For mth minima for single slit 𝒎

 𝟐
𝟐
𝟎
Diffraction pattern

With increasing θ, the values

of  increase faster than α.

So in a range of θ in which α
runs from 0 to  only once, 
Interference pattern
may run from 0 to  a
number of times.

Therefore, as the final pattern is

the multiplication of both the
patterns, at θ-axis the interference
pattern will be found within the
envelop of single slit diffraction
pattern.
 𝟐
𝟐
𝟎

𝟐
Diffraction
envelope

 e  d sin 

Interference
fringes ( )


 e sin  (𝑚 + 1/2)𝜋 𝑚𝜋


The combined effects of two-slits to give diffraction pattern.

This is the pattern produced when 650-nm light waves pass through
two 3.0 μm slits that are 15 μm apart.
Missing fringes in double slit diffraction
The double slit diffraction pattern is produced by product of patterns in A and B.
Therefore it is possible that at any point (at any given value of θ) the minima due to
A and maxima due to B lie simultaneously.
If this happens the
maxima due to B will This gives the absent
be absent due to spectra in the diffraction
multiplication by pattern.
zero. For e = d; n = 2m
For 2e = d; n = 3m
For 3e = d; n = 4m
…etc.
Formulas at a glance
For single slit and
double slit diffraction Single slit diffraction Double slit diffraction
𝟐 𝟐
𝟐
Intensity at screen 𝟎 𝟎

Governing parameter

Principal max α=0

Maxima
Secondary max α= π 𝒏
𝒎

α=
Minima
𝒎 𝒏

Order of missing fringes ---

Diffraction by n-slits:
(Diffraction Grating)

An arrangement consisting of
large number of parallel slits
of the same width and
separated by equal opaque
spaces is known as
diffraction grating. grating element – (e+d)

(1) Plane transmission grating (2) Plane reflection grating

If e is the with of the slit and d is the width of opaque part then (e + d) is
known as grating element.

The wavelets proceeding from all points in a slit along the direction θ, are equivalent to a
single wave of amplitude (R = A0 sinα/α) starting from the middle point of the slit, where
α = (πe sin θ/λ).
Diffraction by n-slits:
(Diffraction Grating)

If there are N slits, then we

have N emitted waves, one
each from N slits.

Hence, the problem of determining the intensity in a direction θ reduces to

finding the resultant amplitude of N vibrations each of amplitude (R = A0
sinα/α ) and having a common phase difference between consecutive diffracted
waves
 Resultant amplitude on the screen

Now this problem is reduced to find the resultant of n equal magnitude vectors with
path difference between consecutive vectors equal to δ.

Sin

MX 1 or = MC sin

Sin

MY 𝑁 θ or = MC sin

=R
Diffraction by n-slits:
(Diffraction Grating)
By the method of vector addition the resultant of N-slits is given by:
𝜋
𝛼= 𝑒 sin 𝜃
𝜆
𝜋
𝛽= 𝑒 + 𝑑 sin 𝜃
𝜆

The intensity is:

𝐬𝐢𝐧 𝜶 𝟐 𝐬𝐢𝐧(𝟐𝜷) 𝟐 𝐬𝐢𝐧 𝜶 𝟐 𝟐 𝐬𝐢𝐧(𝜷)𝒄𝒐𝒔(𝜷) 𝟐

If N = 2 𝟎 𝜶 𝐬𝐢𝐧 𝜷 𝟎 𝜶 𝐬𝐢𝐧 𝜷

𝐬𝐢𝐧 𝜶 𝟐 𝟐 The result is same as that due to

𝟎 𝜶 double slit pattern.
Due to single slit

Due to grating
Intensity distribution: 𝑰 = 𝑰𝟎
𝐬𝐢𝐧 𝜶
𝟐
.
𝐬𝐢𝐧( 𝑵𝜷)
𝟐

𝜶 𝐬𝐢𝐧 𝜷
(Diffraction Grating)
𝜋 𝜋
𝛼= 𝑒 sin 𝜃 , 𝛽 = 𝑒 + 𝑑 sin 𝜃
𝜆 𝜆

Principal maxima: or

Intensity at principal maxima:

→ →

NOTE: There are a number of principal maxima corresponding to n = 0,1,2,3…

Minima: or N

m has all the integer values except 0, N, 2N, … nN.

m = 1, 2, 3, … (N–2), (N–1)
Thus between two adjacent principal maxima there are (N–1) minima.
m = nN+1
And therefore there must be another (N–2) maxima between two adjacent principal
maxima. These are the secondary maxima.
Intensity distribution: 𝑰 = 𝑰𝟎
𝐬𝐢𝐧 𝜶
𝟐
.
𝐬𝐢𝐧( 𝑵𝜷)
𝟐

𝜶 𝐬𝐢𝐧 𝜷
(Diffraction Grating)
𝜋 𝜋
Secondary maxima: 𝛼=
𝜆
𝑒 sin 𝜃 , 𝛽 = 𝑒 + 𝑑 sin 𝜃
𝜆

As N increases, the intensity of secondary maxima

relative to principal maxima decreases and
becomes negligible when N becomes large.
 𝟐 𝟐 𝜋 𝜋
𝐬𝐢𝐧 𝜶 𝐬𝐢𝐧( 𝑵𝜷)
𝑰 = 𝑰𝟎 . 𝛼= 𝑒 sin 𝜃 , 𝛽= 𝑒 + 𝑑 sin 𝜃
𝜶 𝐬𝐢𝐧 𝜷 𝜆 𝜆
Missing spectra with a diffraction grating:

(e  d ) sin   n Maxima In case of grating

e sin   m Minima In case of single slit

If both the conditions are satisfied simultaneously, a particular maximum of

order n will be missing in grating spectrum.
(e  d ) sin  n (e  d )
 
e sin  m e

a) When d = e, then n = 2m. The even orders are missing.

b) When d = 2e, then n = 3m. The 3rd, 6th, 9th,… orders are missing.
c) When d = 3e, then n = 4m. The 4rd, 8th, 12th,… orders are missing.
.
.
.
 Grating

A diffraction grating
can be manufactured
by carving glass with
a sharp tool in a large
number of precisely
positioned parallel
lines.
Dispersive power of the grating:
The dispersive power of grating is defined as the rate of variation of angle of
diffraction with variation of wavelength (dθ/dλ).

(e  d ) sin   n
d 𝑑𝜃 𝑛
(e  d ) cos n =
d 𝑑𝜆 (𝑒 + 𝑑) cos 𝜃

or

(e  d ) sin(  d )  n(  d )
(e  d )sin   cos .d   n(  d )

(e  d ) cos .d  nd

d n

d (e  d ) cos
Resolving power of optical instrument:
The minimum separation at which the two objects look as separate is called the
“limit of resolution”.
The resolution limit of a normal human eye is 1 minute-of-arc.
Resolving power of a telescope is the reciprocal of the limit of resolution.
The image of an object by an optical instrument is the diffraction maxima of the
image due to the diffraction that happen when light waves from the object enter the
objective lens of the optical instrument.

An optical instrument is said to be able to resolve two point objects if the

corresponding diffraction patterns are distinguishable from each other.

Geometric resolution: In forming image of the object.

Spectral resolution: In forming the spectra of light.

The spectral resolving power of grating is defined as the capacity to form separate
diffraction maxima of two wavelengths which are very close to each other.
Resolving power of grating:
According to Rayleigh criterion, an optical instrument can resolve two sources
when the central maximum in the diffraction pattern of one falls over the first
minimum in the diffraction of the other and vice versa.

Similarly, in case of spectral lines of two different wavelengths, the lines will
be resolved when the central maximum due to one-wavelength falls over the
first minimum due to other and vice-versa.
Resolving power of grating:

The spectral resolving power

of grating is defined as the
capacity to form separate
diffraction maxima of two
wavelengths which are very
close to each other. This is
given by /d

At P2 for minima of
= (𝑛𝑁 + 1)𝜆

At P2 for maxima of 𝜆 + 𝑑𝜆

𝒏
Formulas at a
glance Single slit Double slit Grating

𝟐 𝟐 𝟐 𝟐
𝐬𝐢𝐧 𝜶 𝐬𝐢𝐧 𝜶 𝐬𝐢𝐧 𝜶 𝐬𝐢𝐧 𝑵𝜷
Intensity at screen 𝑰 = 𝑰𝟎 𝑰 = 𝟒𝑰𝟎 𝐜𝐨𝐬𝟐 𝜷 𝑰 = 𝟒𝑰𝟎
𝜶 𝜶 𝜶 𝒔𝒊𝒏𝜷

Governing 𝝅 𝝅
𝜶 = 𝒆 𝐬𝐢𝐧 𝜽 𝜷= 𝒆 + 𝒅 𝐬𝐢𝐧 𝜽
parameter 𝝀 𝝀

Principal max α=0

Maxima 𝛽 = ±𝑛𝜋 𝒆 + 𝒅 𝒔𝒊𝒏 𝜽𝒏 = ±𝒏𝝀
Secondary α = ±(𝑚 + 1/2)π 𝒆 + 𝒅 𝒔𝒊𝒏 𝜽𝒏 = ±𝒏𝝀 n = 1, 2, 3, ….
max 𝒆 𝐬𝐢𝐧 𝜽𝒎 = ±(𝒎 + 𝟏/𝟐)𝝀

α = ±𝑚𝜋 𝛽 = ±(𝑛 + 1/2)𝜋 N 𝒆 + 𝒅 𝒔𝒊𝒏 𝜽𝒏 = ±𝒎𝝀

Minima 𝒆 𝐬𝐢𝐧 𝜽𝒎 = ±𝒎𝝀 m =1,2,3,….,(N-2),(N-1)
𝒆 + 𝒅 𝐬𝐢𝐧 𝜽𝒏 = ±(𝒏 + 𝟏/𝟐)𝝀
m =nN-1
𝒆+𝒅
Order of missing --- 𝒏 = 𝒎
fringes 𝒆

Dispersive power --- ---

Resolving power --- --- 𝒏