Wave Optics

Diffraction

Physics
 Diffraction of light:
The phenomenon of bending of light around the corners and the encroachment of light
within the geometrical shadow of the opaque obstacles is called diffraction.

X
X

S• S•

Slit Y
Y
Obstacle

Screen

Diffraction at a slit Diffraction at an obstacle Screen

The wavelets from the single wavefront reach the centre O on the screen in same phase
and hence interfere constructively to give Central or Primary Maximum (Bright fringe).
2) At an angle of diffraction θ = θ1:
The slit is imagined to be divided into 2 equal halves.

The wavelets from the single wavefront diffract at an angle θ1 such that BN
is λ and reach the point P1. The pairs (0,6), (1,7), (2,8), (3,9), (4,10), (5,11) and
(6,12) interfere destructively with path difference λ/2 and give First
Secondary Minimum (Dark fringe).
3) At an angle of diffraction θ = θ2:
The slit is imagined to be divided into 4 equal parts.

The wavelets from the single wavefront diffract at an angle θ2 such that BN is 2λ and reach the
point P2. The pairs (0,3), (1,4), (2,5), (3,6), (4,7), (5,8), (6,9), (7,10), (8,11) and (9,12) interfere
destructively with path difference λ/2 and give Second Secondary Minimum (Dark fringe).
4) At an angle of diffraction θ = θ1’:
The slit is imagined to be divided into 3 equal parts.

The wavelets from the single wavefront diffract at an angle θ1’ such that BN is 3λ/2 and reach the point P1’.
The pairs (0,8), (1,9), (2,10), (3,11) and (4,12) interfere constructively with path difference λ and (0,4), (1,5),
(2,6), …… and (8,12) interfere destructively with path difference λ/2. However due to a few wavelets
interfering constructively First Secondary Maximum (Bright fringe) is formed.
 Diffraction at various angles:
•• P2 θ2

θθ ’ • P1’ θ1’
AA θ = 101 θ2
0•
•
1
2• •• PP11 θ1
•
3 λ/2
4•
•
5 λ/2 θθ2 ’ θ1 θ=0
6• 1
•O
7 λ/2λ
• I
8• λ
9• 3λ/2
10• N
11
• N
θN2 θ1 1’
θ
12
•
Plane BB λ 2λ
3λ/2
Wavefront Slit
Screen
Central Maximum is the brightest fringe.
Diffraction is not visible after a few order of diffraction.
Theory:
The path difference between the 0th wavelet and 12th wavelet is BN.
If ‘θ’ is the angle of diffraction and ‘d’ is the slit width, then BN = d sin θ
To establish the condition for secondary minima, the slit is divided into 2, 4, 6, … equal parts
such that corresponding wavelets from successive regions interfere with path difference of λ/2.
Or for nth secondary minimum, the slit can be divided into 2n equal parts.
For θ1, d sin θ1 = λ Since θn is very small,
For θ2, d sin θ2 = 2λ d θn = nλ
For θn, d sin θn = nλ θn = nλ / d (n = 1, 2, 3, ……)
To establish the condition for secondary maxima, the slit is divided into 3, 5, 7, … equal parts
such that corresponding wavelets from alternate regions interfere with path difference of λ.
Or for nth secondary minimum, the slit can be divided into (2n + 1) equal parts.
For θ1’, d sin θ1’ = 3λ/2 Since θn’ is very small,
For θ2’, d sin θ2’ = 5λ/2 d θn’ = (2n + 1)λ / 2
For θn’, d sin θn’ = (2n + 1)λ/2 θn’ = (2n + 1)λ / 2d (n = 1, 2, 3, ……)
 Width of Central Maximum:

tan θ1 = y1 / D y1 = D λ / d
or θ1 = y1 / D (since θ1 is very small) Since the Central Maximum is
spread on either side of O, the
d sin θ1 = λ width is
or θ1 = λ / d (since θ1 is very small) β0 = 2D λ / d
 Fresnel’s Distance:
Fresnel’s distance is that distance from the slit at which the spreading of light due to diffraction
becomes equal to the size of the slit.
y1 = D λ / d
At Fresnel’s distance, y1 = d and D = DF
So, DF λ / d = d or D F = d2 / λ
If the distance D between the slit and the screen is less than Fresnel’s distance DF, then the
diffraction effects may be regarded as absent.
So, ray optics may be regarded as a limiting case of wave optics.
Difference between Interference and Diffraction:
Interference Diffraction
1. Interference is due to the 1. Diffraction is due to the
superposition of two different superposition of secondary
wave trains coming from coherent wavelets from the different parts
sources. of the same wavefront.
2. Fringe width is generally constant. 2. Fringes are of varying width.
3. All the maxima have the same 3. The maxima are of varying
intensity. intensities.
4. There is a good contrast between 4. There is a poor contrast between
the maxima and minima. the maxima and minima.
Q. One of the following statements is correct. Pick Sol:
out the one.
1) Diffraction can not take place without
interference
2) Interference will not take place with out
diffraction
3) Interference and diffraction are the result of
polarization
4) The fringe width in Young’s double slit experiment
does not depends on the wave length

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Q. Light of wavelength 6000A0 is incident on a
single slit. The first minimum of the diffraction
pattern is obtained at 4 mm from the center. The
screen is at a distance of 2m from the slit. The
slit width will be
1) 0.3 mm 2) 0.2 mm 3) 0.15 mm 4) 0.1 mm
Sol:

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Q. A slit of width 'a' is illuminated by white light.
The first minimum for red light 𝜆 = 6500 A0 will fall
at 𝜃 = 30∘ Then 'a' will be
1. 3250 A0
2. 1.3 micron
3. 6.5 × 10−4 mm
4. 2.6 × 10−4 m

Sol:

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Q. A small aperture is illuminated with a parallel beam
of 𝜆 = 628 nm. The emergent beam has an angular
divergence of 20 . The size of the aperture is
1. 9 𝜇m
2. 18𝜇m
3. 27𝜇m
4. 36𝜇m

Sol:

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Q. A parallel beam of monochromatic light is incident Sol:
normally on a narrow slit. A diffraction pattern is
formed on a screen placed perpendicular to the
direction of the incident beam At the first minimum of
the diffraction pattern, the phase difference between
the rays coming from the two edges of the slit is
1. zero
𝜋
2. 2
3. 𝜋
4. 2𝜋

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Q. In a diffraction pattern due to a single slit of width Sol:
'a', the first minimum is observed at an angle 30∘ when
light of wavelength 5000 A∘ is incident on the slit. The
first secondary maximum is observed at an angle of
1
1. sin−1 4
2
Sol:sin−1
2. 3
1
3. sin−1 2
3
4. sin−1 4

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Q : A parallel beam of light of wavelength 500 nm
falls on a narrow slit and the resulting diffraction
pattern is observed on screen 1 m awaw. It is
observed that the first minimum is at a distance of
2.5 mm from the centre of the screen. Find the
widht of the slit.
Sol:

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