NITRKL

1. DIFFRACTION

2. Diffraction: Fraunhofer, Fresnel's diffraction, single
slit (infinite beam interference), two and N slits
(Grating) Fraunhofer diffraction pattern and intensity
distribution.
Course contents:
[1] Fundamentals of Optics, F. A. Jenkins, H. E. White, Tata McGraw-Hill (2011).
[2] Optics, A. Ghatak, Tata McGraw-Hill (2011).
REFERENCES:

3. Diffraction of Light
When a narrow opaque obstacle (aperture) be placed between
an source of the light and a screen, light bends round the corners of
the aperture. This encroachment of light is called “diffraction”.
* For diffraction, the size of the aperture is small (comparable to the
wavelength).
* As a result of diffraction, the edges of the shadow (or illuminated
region) are not sharp, but the intensity is distributed in a certain way
depending upon the nature of the aperture.

4. Difference between Interference and Diffraction
Interference: It occurs between waves starting from two (or more, but
finite in numbers) coherent sources.
Diffraction: It occurs between the secondary wavelets starting from the
different points (infinite in number) of the same wave.
• Both are superposition effects and often both are present
simultaneously (like young’s experiment)
Comparison:
(a) In an interference pattern the minima are usually almost perfectly
dark while in a diffraction pattern they are not so.
(b) In an interference pattern all the maxima are of same intensity but
not in the diffraction pattern.
(c) The interference fringes are usually (although not always) equally
spaced. The diffraction fringes are never equally spaced.

5. Types of Diffraction
(a)Fresnel diffraction
(b)Fraunhofer diffraction

6. Fresnel’s Diffraction
* In this case, the source of light or screen or usually both are at
finite distance from the diffracting aperture (obstacle).
* In this case no lenses are used
* The incident wavefront is either spherical or cylindrical..

7. Fraunhofer’s Diffraction
* In this case, the source of light and screen are effectively at infinite
distance from the diffracting aperture (obstacle).
* This is achieved by placing the source and screen in the focal planes
of two lenses.
* The incident wavefront is plane.
* Fraunhofer diffraction is a limiting case of the more general Fresnel’s
diffraction.
* Much easier to calculate the intensity distribution of a Fraunhofer diffraction .

8. Fraunhofer’s Diffraction at a Single Slit
* Let a parallel beam of monochromatic light of wavelength λ be
incident normally upon a narrow slit of width AB = e
* Let diffracted light be focused by a convex lens L on a screen XY
placed in the focal plane of the lens.
* The diffraction pattern obtained on the screen consists of a central
bright band, having alternate dark and weak bright bands of
decreasing intensity on both sides.
P
O
Y
X

9. Continued……
In terms of wave theory, a plane wavefront is incident on the slit AB.
According to the Huygen’s principle, each point in AB sends out
secondary wavelets in all directions.
The rays proceeding in the same direction as the incident rays
focussed at O; while those diffracted through an angle θ are focused at
P.
Let us find out the resultant intensity at P.
Let AK be perpendicular to BK. As the optical paths from the
plane AK to P are equal, the path difference between the wavelets from
A to B in the direction θ = BK = AB sin θ = e sin θ
The corresponding phase difference = (2π/λ) e sin θ
Let the width AB of the slit be divided into n equal parts. The amplitude
of vibration at P due to the waves from each part will be the same (= a)

10. Continued……
The phase difference between the waves from any two consecutive
parts is
Hence the resultant amplitude at P is given by

11. Continued……
The resultant intensity at P is

12. Continued……
Direction of Minima:
The intensity is minimum (zero) when
or, sin α = 0 (but α ≠ 0, because for α = 0, then sinα/α = 1)
α = ± mπ where m has an integral value 1,2,3 …except Zero
This equation gives the directions of the first, second,
third……minima by putting m = 1, 2, 3, ….
P
O
Y
X

13. Continued……
Direction of Maxima:
To find the direction of maximum intensity, let us differentiate the
Intensity with respect to α and equate it to zero
dI/dα = 0
This equation is solved graphically by plotting the curves
y = α
y = tan α

14. Continued……
The 1st equation is a straight line through origin making an angle of
45 °
The 2nd is a discontinuous curve having a number of branches.
The point of intersection of the two curves give the value of α
These values are approximately given by
α = 0, 3π/2, 5π/2, 7π/2, ……..
Substituting these values of α, we can find
I0 = A2
I1 ≈ A2/22
I2 ≈ A2/61
 The intensity of the 1st maximum is about 4.96% of the central maximum.

15. Continued……
The principal maxima occurs at
α = (π e sin θ)/λ = 0, or θ = 0
i.e. the principal maxima occurs at
the same direction of light
The diffraction pattern
consists of a bright principal
maximum in the direction of
incident light, having alternately
minima and weak subsidiary
maxima of rapidly decreasing
intensity on either side of it.
The minima lie at α = ± π, ± 2π, ….

16.  The single-slit diffraction patterns corresponding to e = 0.0088,
0.0176, 0.035, and 0.070 cm, respectively. The wavelength of
the light used is 6.328 x 10–5 cm

17. Fraunhofer’s Diffraction at a Double Slit
* Let a parallel beam of monochromatic light of wavelength λ be
incident normally upon two parallel slits AB and CD, each of width e,
separated by opaque space d.
* The distance between the corresponding points of the two slits is
(e+d).
* Let the diffracted light be focused by a convex lens L on a screen XY
placed in the focal plane of the lens.
X
Y
O
P

18. Continued….
The pattern obtained on the screen is the diffraction pattern due to a
single slit on which a system of interference fringes is superposed.
By Huygen’s principle, every point in the slits AB and CD
sends out secondary wavelets in all directions.
From the theory of diffraction at a single slit, the resultant
amplitude due to wavelets diffracted from each slit in a direction θ is
We can, therefore, consider the two slits as equivalent to two
coherent sources placed at the middle points S1 and S2 of the slits, and
each sending a wavelet of amplitude (A sin α)/α in the direction of θ.

19. Continued….
Therefore, the resultant amplitude at a point P on the screen will be the
result of interference between two waves of same amplitude (A sin α)/α,
and having a phase difference δ.
Let us draw S1K perpendicular to S2K. The path difference between the
wavelets from S1 and S2 in the direction θ = S2K = (e+d) sin θ
Hence the phase difference is δ = (2π/λ) x path difference
= (2π/λ) x (e+d) sin θ
The resultant amplitude R at P can
be determined by the vector amplitude
diagram, which gives
OB2 = OA2 + AB2 + 2(OA)(AB) cos BAC
R
(A sin α)/α
(A sin α)/α
θ
O
A
B
C

20. Continued….
𝑹 𝟐
= 𝑨
𝒔𝒊𝒏 𝜶
𝜶
𝟐
+ 𝑨
𝒔𝒊𝒏 𝜶
𝜶
𝟐
+ 𝟐
𝒔𝒊𝒏 𝜶
𝜶
𝒔𝒊𝒏 𝜶
𝜶
𝒄𝒐𝒔 δ
Therefore, the resultant intensity at P is I = R2

21. Continued….
Thus the intensity in the resultant pattern depends on two factors:
(i) (sin2 α)/ α2, which gives the diffraction pattern due to each
individual slit
(ii) Cos2β, which gives interference pattern due to diffracted light
waves from the two slits.

22. Continued….
(a) The diffraction term, (sin2 α)/ α2, gives a central maximum in the
direction θ = 0, having alternately minima and subsidiary maxima of
decreasing intensity on either side.
The minima are obtained in the directions given by
sin α = 0, α = ± mπ
(π e sin θ)/λ = ± mπ
e sin θ = ± mπ (m=1,2,3,… , except zero)

23. Continued….
(b) The interference term, cos2 β, gives a set of equidistant dark and
bright fringes, as in Young’s double-slit interference experiment.
The bright fringes (maxima) are obtained in the directions given by
cos2 β = 1
β = ± nπ
(π/λ) (e+d) sin θ = ± nπ
(e+d) sin θ = ± nλ, Where n = 0,1,2,…
The various maxima corresponding to n = 0, 1,2, …are zero-order, first-
order, second-order …… maxima

24. Continued….
(c) The intensity distribution in the resultant diffraction pattern is a plot of
the product of the constant term 4 A2, diffraction term (sin2 α)/ α2, and the
interference term cos2 β.
The entire pattern may be
considered as consisting
interference fringes due to
light from both slits, the
intensities of the these
fringes being governed by
the diffraction occurring at
the individual slits.

25. The double-slit intensity distribution corresponding to e = 0.0088 cm,
 = 6.328 x 10–5 cm, and d = 0.035 cm and d = 0.07cm respectively.
e = 0.0088 cm
d =0.035 cm
 = 6.328  10-5 cm


 2
2
2
0 cos
sin
4II 
e = 0.0088 cm
d =0.07 cm
 = 6.328  10-5 cm

26. Photographs and Intensity curves for double-slit
diffraction patterns.

27. The double-slit Fraunhofer diffraction pattern
corresponding to e = 0.0088 cm and  = 6.328 x 10–5 cm.
The values of d are 0, 0.0176, 0.035, and 0.070 cm,
respectively.

28. Fraunhofer’s Diffraction at a N Slit (Grating)
Diffraction grating:
* A diffraction grating is an arrangement equivalent to a large number
of parallel slits of equal widths and separated from one another by
equal opaque spaces.
* It is made by ruling a large number of fine, equidistant and parallel
lines on an optically-plane glass plate with a diamond point.
* The ruling scatter the light and are effectively opaque while the
unruled parts transmits and act as slits.
USE:
* When there is a need to separate light of different wavelengths with
high resolution, then a diffraction grating is most often the tool of
choice

29. * The diffraction grating is an immensely useful tool for the
separation of the spectral lines associated with atomic transitions.
* Sometimes this diffraction grating is also called ‘super prism’.
* The tracks of a compact disc act as a diffraction
grating, producing a separation of the colours of
white light.
* The condition for maximum intensity is the same as that for the double
slit or multiple slits, but with a large number of slits the intensity maximum is
very sharp and narrow, providing the high resolution for spectroscopic
applications.

30. Continued….
* Let AB be the section of a plane transmission grating.
* Let e be the width of each slit and d the width of each opaque space
between the slits.
* Then (e+d) is called the “grating element”.
* The points in two consecutive slits separated by the distance (e+d)
are called the “corresponding points”.
* Let a parallel beam of monochromatic light of wavelength λ be
incident normally on the grating.
P

31. Continued….
According to Huygen’s principle, all the points in each slit send
out secondary wavelets in all directions.
According to the Fraunhofer diffraction at a single slit, the wavelets
from all points in a slit diffracted in a direction θ are equivalent to a
single wave of amplitude (A sin α)/α, where α = (π/λ) e sin θ
If N be the total number of slits in the grating, the diffracted
rays from all the slits are equivalent to N parallel rays, one each from
middle points S1, S2, S3, ….. of the slits.
After calculating the path difference and corresponding phase
difference, the resultant amplitude in the direction θ is

32. Continued….
The resultant intensity I is
The first factor A2 sin2α/ α2 gives a diffraction pattern due to the
single slit, while second factor sin2N𝜷/sin2 𝜷 gives the interference
pattern due to N slits.
Conditions of Maxima:
(e+d) sin θ = ± nλ , where n = 0,1,2,…..
Conditions of Minima:
N(e+d) sin θ = ± mλ
where m takes all integral values except 0, N, 2N, ----.nN, because
these values of m make sin 𝜷 = 0, which gives principal maxima.

33. Continued….
It is clear that m = 0 gives a principal maximum, m = 1,2,3, ….(N-1) give
minima and m = N gives again a principal maximum
Thus there are (N-1) minima between two consecutive principal
maxima.

34. `
PROBLEM-15
Q: In Fraunhofer diffraction due to a narrow slit, a screen is placed 2 m
away from the lens to obtain the pattern. If the slit width is 0.2 mm and
the first minima lie 5 mm on either side of the central maximum, find the
wavelength of light?
A: In the Fraunhofer diffraction pattern due to a single slit of width e, the
direction of minima are given by
e sin θ = ± mλ m = 1,2,3 ,---
For 1st minimum, m =1, e sin θ = λ
If θ is very small and measured in radian, then sin θ = θ
θ = λ/ e radian = (λ/ 0.02 cm) radian
Now for linear separation between the 1st minimum and the central
minimum is 0.5 cm and the distance of the screen from the slit is 2 m i.e.
θ = 0.5/200 radians
Equating values of θ = (λ/ 0.02 cm) radians = 0.5/200 radians
λ = 5000 Å

35. `
PROBLEM-16
Q: Parallel light (5000 Å) is normally incident on a single slit. The central
maxima is observed at 30 ° on both sides of the direction of the incident
light. Calculate the slit width. For what width of the slit the central maximum
would spread out to 90° from the direction of the incident light ?
A: In the Fraunhofer diffraction pattern due to a single slit of width e, the
direction of minima are given by
e sin θ = ± mλ m = 1,2,3 ,---
For 1st minimum, m =1, e sin θ = λ
sin θ = λ/e
θ = 30 °, λ = 5000 Å
So e = 1 x 10-4 cm
For θ = 90 °, e = λ/ sin θ = λ = 5000 Å