CBSE Class 12 · 2007 · Wave OpticsLong Answer

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State the condition under which the phenomenon of diffraction of light takes place. Derive an expression for the width of the central maximum due to diffraction of light at a single slit. A slit of width 'a' is illuminated by monochromatic light of wavelength

700

nm at normal incidence. Calculate the value of 'a' for the position of
(i) first minimum at an angle of diffraction of

3 0^{\circ}

;
(ii) first maximum at an angle of diffraction of

3 0^{\circ}

Solution

Condition:

diffraction becomes appreciable when the width of the slit (or obstacle) is comparable to the wavelength of light.

Width of central maximum:

the first minima lie at

a sin θ = λ

, i.e. for small angles

θ = \frac{λ}{a}

. If the screen is at distance

D

, the first minimum is at

y = \frac{λ D}{a}

on each side, so

Width of central maximum = 2 y = \frac{2 λ D}{a}

Numerical

(

λ = 700

nm,

θ = 3 0^{\circ}

sin 3 0^{\circ} = 0.5

):
(i) First minimum:

a sin θ = λ \Rightarrow a = \frac{λ}{sin θ} = \frac{700}{0.5} = 1400

nm.
(ii) First (secondary) maximum:

a sin θ = \frac{3 λ}{2} \Rightarrow a = \frac{3 λ}{2 sin θ} = \frac{3 \times 700}{2 \times 0.5} = 2100

nm.

a_{(i)} = 1400 nm, a_{(ii)} = 2100 nm

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CBSE Class 12 · 2007 · Wave OpticsLong Answer

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