Diffraction of Light

Diffraction is a phenomenon of bending a light wave near a sharp obstacle or narrow aperture due to superposition of the infinite
number of secondary wavelets originating from the same wavefront. So interference is a special case of diffraction effect. To
observe clear diffraction, the dimension of the narrow gap should be in the range of wavelength of the light used.
Interference Diffraction
The interference occurs between two separate wave fronts The diffraction occurs between the secondary wavelets originating
originating from two coherent sources. from the exposed part of the same wave front.

The interference fringes can be equally spaced. Or may not The diffraction fringes are never equally spaced.
be equally spaced.
In interference, the maxima are of same intensity. In diffraction, the intensity of central maximum is maximum and
decreases on either side as the order of maxima increases.

There are mainly two types of diffraction: Fresnel’s diffraction Fraunhofer’s diffraction
1) Fresnel diffraction either source or screen or both are at finite source of light and the screen are at
2) Fraunhofer diffraction. distances from the obstacle/aperture infinite distances from the obstacle.
the incident wave front is either spherical the incident wave front is generally plane
or cylindrical wavefront. wavefront
the center of diffraction pattern may be The center of the diffraction pattern is
bright of dark depending upon the number always bright. Ex. Single, Double slit etc.
of Fresnel’s zones, Ex. Zone Plate
Find the resultant amplitude and phase of a particle when n simple harmonic motion of equal amplitudes and periods
but with phase increasing in arithmetic progression are imposed on the particle:
Single slit Diffraction:
Maximum

Straight line
Discontinuous curve with
number of branches
The point of intersections of the two curves give the values of α.
Not at the mid way
 Problem on Single Slit
If I shine a narrow slit of width 0.1 mm by a light of wavelength 5000Å. Calculate the position of minima and maxima. If I reduce
the size of the slit to 10 m what would happen to the position of maxima and minima. Also calculate the width and angular width
of the central maxima for both the slit widths. If the screen is kept at 2 m distance what would be the position of maxima and
minima from the centre of central maxima.
For Minima, The condition is For Maxima, The condition is
𝒅𝐬𝐢𝐧 𝜽min = ±n𝝀 𝒅𝐬𝐢𝐧 𝜽max = ±(2n+1)𝝀/2 y
𝜽
( )
So, 𝜽min = sin (in rad) So, 𝜽max = sin (in rad)
Width of the central maxima = 2*ymin of first order = 0.02 m when d = 0.1 mm
D
while it becomes 0.2 m when the slit width is reduced to 10 m
Order Width of the slit d = 0.1 mm Width of the slit d = 1 m
Minima Maxima ymin/D = ymax/D = Minima Maxima ymin/D = ymax/D =
Condition: Condition: tan(𝜽min) tan(𝜽max) Condition: Condition: tan(𝜽min) tan(𝜽max)
𝒅𝐬𝐢𝐧 𝜽min = ±𝒓𝝀 𝒅𝐬𝐢𝐧 𝜽max = 𝒅𝐬𝐢𝐧 𝜽 = ±𝒓𝝀 𝒅𝐬𝐢𝐧 𝜽 =
±(2r+1)𝝀/2 𝜽 in rad ±(2r+1)𝝀/2
𝜽min in rad 𝜽max in rad ymin in m ymax in m 𝜽 in rad ymin in m ymax in m
1 0.005 0.008 0.010 0.015 0.143 0.215 0.100 0.150
2 0.010 0.013 0.020 0.025 0.286 0.358 0.200 0.251
3 0.015 0.018 0.030 0.035 0.430 0.501 0.301 0.352
4 0.020 0.023 0.040 0.045 0.573 0.645 0.403 0.454
5 0.025 0.028 0.050 0.055 0.716 0.788 0.505 0.557
Double Slit: • From the theory of diffraction at a single slit the resultant amplitude due to wavelets
diffracted from each slit in the direction θ

• Considering two coherent sources at the middle of the slits S1 and S2,
Sending the amplitude of wavelets

The resultant amplitude on the screen at P is the resultant amplitude

of the two interfering rays.
Calculation of resultant amplitude by vector method:
Resultant Intensity pattern:

Diffraction direction

Interference direction
Interference term

Constant
Diffraction term
 Diffraction Grating
Diffraction grating: Grating is an optical component which has periodic arrangement of large number of slits of same width and
separated by equal opaque spaces. X e+d
The sum of total width of a slit (e) and an opaque space (d) is called grating element X = e + d

Figure below shows a plane grating consisting of “N” number of parallel slits.

Now, consider 𝑅 = A sin𝛼/𝛼 be the amplitude of light coming from a single slit normally, then each slit may be regarded as the
source of light with initial phase 𝛼 = (𝜋/𝜆)e sin 𝜃. The phase difference between two consecutive slits is given by 𝛾
𝛾 = (2𝜋/𝜆) (e+d)sin 𝜃.
Thus the resultant field at any point P on the screen due to superposition of N light waves is
𝐸 = 𝐸1 + 𝐸2 + 𝐸3 + ⋯ + 𝐸𝑁
𝐸 = 𝑅𝑒−𝑖𝜔𝑡 + 𝑅𝑒−𝑖𝜔𝑡𝑒𝑖𝛾 + 𝑅𝑒−𝑖𝜔𝑡𝑒2𝑖𝛾 + ⋯ + 𝑅𝑒−𝑖𝜔𝑡𝑒(𝑁−1)𝑖𝛾
𝐸 = 𝑅[1 + 𝑒𝑖𝛾 + 𝑒2𝑖𝛾 + ⋯ + 𝑒(𝑁−1)𝑖𝛾]𝑒−𝑖𝜔𝑡 = 𝑅 𝑒−𝑖𝜔𝑡(1−𝑒𝑖𝑁𝛾)/(1−𝑒𝑖𝛾) = 𝐴𝑒−𝑖𝜔𝑡
The resultant complex amplitude at P is 𝐴 = 𝑅(1−𝑒𝑖𝑁𝛾)/(1−𝑒𝑖𝛾) . A
Therefore, intensity at P is 𝐼 = 𝑘|𝐴|2. P
W
∴ 𝐼 = 𝑘𝑅2[(1−𝑒𝑖𝑁𝛾) (1−𝑒−𝑖𝑁𝛾)]/[(1−𝑒𝑖𝛾 )(1−𝑒−𝑖𝛾)]
= 𝑘𝑅2(1−cos𝑁𝛾)/(1−cos𝛾)
= 𝑘𝑅2𝑠𝑖𝑛2(𝑁𝛾/2)/𝑠𝑖𝑛2(𝛾/2)  P0
= 𝑘𝑅2𝑠𝑖𝑛2𝑁𝛽/𝑠𝑖𝑛2𝛽
∴ 𝐼 = 𝑘𝑅2𝑠𝑖𝑛2(𝑁𝛽)/𝑠𝑖𝑛2𝛽 , where 𝛾 = 2𝛽 = (2𝜋/𝜆)(e+d) sin𝜃 = 2𝜋/𝜆 X sin𝜃. W’ B
∴ 𝐼 = 𝑘A2(𝑠𝑖𝑛2𝛼/𝛼2)(𝑠𝑖𝑛2𝑁𝛽/𝑠𝑖𝑛2𝛽) --- Intensity distribution in grating experiment.
 Diffraction Grating
𝐹𝑜𝑟 𝑁 = 2, 𝐼 = 𝐼0 (𝑠𝑖𝑛2𝛼/𝛼2)(𝑠𝑖𝑛22𝛽/𝑠𝑖𝑛2𝛽)
= 𝐼0(𝑠𝑖𝑛2𝛼/𝛼2) {2sin𝛽 cos𝛽}2/𝑠𝑖𝑛2𝛽 N=2 N=3 N=4
= 4𝐼0 (𝑠𝑖𝑛2𝛼/𝛼2) 𝑐𝑜𝑠2𝛽
= 2𝐼0(𝑠𝑖𝑛2𝛼/𝛼2)(1 + cos2𝛽) }--> Double slit diffraction
N=1 /2 /3
Diffraction X sin =  2/3 
3/2 
Deduce the conditions for maxima and minima: d=0 X sin = 2 5/2 2 2
e sin = 4/3
𝐼 = 𝐼0 (𝑠𝑖𝑛 𝛼/𝛼 )(𝑠𝑖𝑛 𝑁𝛽/𝑠𝑖𝑛 𝛽)
2 2 2 2
5/3
Principle maxima:
The condition for principle maxima is sin𝛽 = 0 ⇒ 𝛽 = ±𝑛𝜋 with 𝑛 = 0, 1, 2, 3, ⋯.
In terms of path difference, the condition of principle maxima is 𝜋/𝜆 X sin𝜃 = ±𝑛𝜋 or, X 𝐬𝐢𝐧𝜽 = ±𝒏𝝀.
Under this condition, the term 𝑠𝑖𝑛2𝑁𝛽/𝑠𝑖𝑛2𝛽 reduces to 𝑁2 because Lim 𝛽→±𝑛𝜋 sin𝑁𝛽/sin𝛽 = Lim 𝛽→±𝑛𝜋 𝑁cos𝑁𝛽/cos𝛽 = ±𝑁.
So, intensity of principle maxima is 𝐼𝑝 = 𝑁2𝐼0𝑠𝑖𝑛2𝛼/𝛼2.
How many principle maxima can be observed? L'Hôpital's rule
Xsin𝜃 = 𝑛𝜆 ⇒ 𝑛𝑚𝑎𝑥 = X/𝜆 . Considering both sides and central maximum there will be (2X/𝜆 + 1) principle maxima.
Minima:
The values of 𝛽 for which sin𝑁𝛽 = 0 but 𝛽 ≠ 0 gives the positions of minima.
Thus for minima 𝑁𝛽 = ±𝑚𝜋 ⇒ 𝛽 = ±(𝑚/𝑁)𝜋, but (𝑚/𝑁)≠ 𝑛.
In terms of path difference, the condition of minima is (𝜋/𝜆)Xsin𝜃 = ±(𝑚/𝑁)𝜋 or, Xsin𝜃 = ±(𝑚/𝑁)𝜆 .
Here, 𝑚 = 1, 2, 3, ⋯ ,(𝑁−1), (𝑁+1),⋯ but 𝑚 ≠ 0, 𝑁, 2𝑁, 3𝑁, ⋯.
If N = 3, m can be 1, and 2 but not 3, because if m = 3 the path difference X𝐬𝐢𝐧𝜽 = 𝝀 which the condition for 1st order maxima n
= 1; so there will be 2 minima between the primary maxima.
Therefore, there are (𝑁 − 1) number of minima in between two principle maxima if there are N number of transparent region.
Diffraction Grating

Secondary maxima: Since there are (𝑁 − 1) number of minima in between two principle maxima, thus there must be some sort
of (𝑁 − 2) secondary maxima in between two principle maxima. When N is large, these secondary maxima are very weak.
Absent spectra:
Same we have calculated in case of Double slit
Angular Width of principle maxima:
For nth order principle maxima, X sin𝜃𝑛 = 𝑛𝜆 ----> (1).
If 𝜃𝑛 + ∆𝜃𝑛 be the angle of diffraction for the first minimum corresponding to the nth order principle maxima
then we have X sin(𝜃𝑛 + ∆𝜃𝑛) = 𝑛𝜆 + 𝜆/𝑁, OR X(sin𝜃𝑛 + ∆𝜃𝑛 cos𝜃𝑛) = 𝑛𝜆 +𝜆/𝑁 ----> (2).
[Since sin(𝜃𝑛 + ∆𝜃𝑛) = sin𝜃𝑛 cos∆𝜃𝑛 + cos𝜃𝑛 sin∆𝜃𝑛 = sin𝜃𝑛 + ∆𝜃𝑛cos𝜃𝑛]
Compare equation (1) and (2), X∆𝜃𝑛cos𝜃𝑛=𝜆/𝑁 ⇒ ∆𝜃𝑛=𝜆/(X𝑁cos𝜃𝑛). This is the half of the angular width of the nth order principle
maxima. As N increases ∆𝜃𝑛 decreases that is the principle maxima becomes more and more sharp.
 Resolving Power of Grating
Resolving power:
Resolving power of a grating is its ability to distinguish two nearby spectral lines with a small
difference in wavelength ∆𝜆.
It is defined by the relation, 𝑃 = 𝜆/∆𝜆, where ∆𝜆 is the separation between two closest wavelength
which the grating can just resolved. The smaller value of ∆𝜆 the larger the resolving power.

To obtain resolving power we use Rayleigh’s criterion.

Rayleigh’s criterion states that if a principle maximum for wavelength 𝜆 + ∆𝜆 falls on the first minima
of wavelength 𝜆 then these two wavelengths 𝜆 and 𝜆 + ∆𝜆 will be just resolved.

Let 𝜃𝑛 be the angle of diffraction of nth order principle maxima for wavelength 𝜆 + ∆𝜆 then for just
resolve the first minimum corresponding to order n for 𝜆 should also fall at this position.
So, we can write 𝑒 sin𝜃𝑛 = 𝑛(𝜆 + ∆𝜆) : this is the condition for nth order maxima of 𝜆 + ∆𝜆

1st Minima of 
𝑒 sin𝜃𝑛 = 𝑛𝜆 + 𝜆/𝑁 : this is the condition for nth order first minima of 𝜆

Comparing these two equations we get,

1st Maxima of d
𝑛 Δ𝜆 = 𝜆/𝑁 ⇒ 𝜆/∆𝜆 = 𝑛𝑁 .
Total numbers of Rulings
So, resolving power 𝑃 = 𝜆/∆𝜆 = 𝑛𝑁 .
 Problem on N Slit
If I shine a three slits with by a light of wavelength 4000 Å, calculate the position of primary and secondary maxima and minima.
If the width of the slit is 0.1 mm and the distance between the slit is 0.2 mm, calculate the missing orders spectra.

Given:
d = 0.1 mm N=3
b = 0.2 mm
𝝀 = 4000 Å
/3
e = 0.4 mm  2/3
2
4/3
For Primary/Principle Maxima, For Minima, The condition is 5/3
The condition is
e 𝐬𝐢𝐧𝜽min = ±( )𝝀 m is integer by m≠N For Secondary Maxima, is
ec𝐬𝐢𝐧𝜽min = ±n𝝀
So, 𝜽min = sin ( ) (in rad) between two minima
So, 𝜽min = sin (in rad)

For N = 3, m/N values are

1/3, 2/3 are minima between central and 1st primary maxima because for 3/3 first primary maxima occurs.
So there is one secondary maxima at 1/2
4/3, 5,3 are minima between 1st and 2nd primary maxima because for 6/3 2nd primary maxima occurs.
Therefore, a secondary maxima is at 3/2

𝑛 = (𝑑+𝑏)/𝑑, 2(𝑑+𝑏)/𝑑, 3(𝑑+𝑏)/𝑑, ⋯ are the missing orders

 Problem on N Slit

For N = 4, m/N values are

1/4, 2/4, 3/4 gives the position of minima between central and 1st primary maxima because for 3/3 first primary maxima occurs.
So, 3/8, and 5/8 gives the position of secondary maxima.
5/4, 6/4, 7/4 are minima between 1st and 2nd primary maxima because for 6/3 2nd primary maxima occurs.
11/8 and 13/8 gives the position of secondary maxima.

Problems:
P1: In a diffraction grating the width of opaque space is 2.5μm and clear space is 1.5μm. Calculate the intensity ratio of 3rd and
4th order principle maxima for 𝜆 = 500 𝑛𝑚.

P2: Draw the diffraction pattern for N = 5, b = 3d, e = 10μm and 𝜆 = 500 𝑛𝑚. Find the intensity ratio of (i) 2nd and 3rd (ii) 3rd and
4th order principle maxima.
Hint: First calculate the 𝜃 of n = 2, 3, and 4. Use the theta to calculate the alpha value 𝛼 = (𝜋/𝜆)𝑑 sin𝜃. Substitute the 𝛼 value in
the intensity equation to calculate the intensity and then the ratios. Intensity of principle maxima is 𝐼𝑝 = 𝑁2𝐼0𝑠𝑖𝑛2𝛼/𝛼2.

P3: How many principle maxima can be observed for a diffraction grating with 2540 LPI and a light source of wavelength 0.6328
m?

P4: Calculate the angular separation between 3rd and 2nd order principle maxima located in same side of central maximum for a
grating of 2000 lines/inch with used wavelength 600nm.

P5: A diffraction grating of width 5 cm resolves sodium D-lines (589nm & 589.6nm) in its 1st order. Calculate the grating element.
 Problem on N Slit Maxima (Principle) Minima Secondary Maxima
For N = 5 Value of e sin Value of e sin Value of e sin
Central Maxima 0
1/5  3/10
In a five slit experiment, if the distance
between the slit is equal to the width of 2/5  5/10
the slit. calculate the absent. Assume 3/5  7/10
that the wavelength of light used in
your experiment is equal to the grating 4/5 
element (e). 
6/5  13/10
The total number of observed principle
maxima are 2(e/)+1 7/5  15/10
If e = , only three line including the 8/5  17/10
central maxima will be observed. So 9/5 
only n = 1 will be observed in the
experiment. Absent due to b = d and e =  2
11/5  23/10
The following lines will be absent for n
12/5  25/10
is integral multiples of (b+d)/d
n=2 13/5  27/10
n=4 14/5 
n=6
Absent due to e =  3