JEE Main · 2018 · 08 Apr 2018 · Offline · Wave Optics

The angular width of the central maximum in a single slit diffraction pattern is

6 0^{\circ}

. The width of the slit is

1 μ m

. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young's fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance?
(i.e. distance between the centres of each slit.)

(A) $100 μ m$
(B) $25 μ m$ ✓ Correct
(C) $50 μ m$
(D) $75 μ m$

Answer: (B)

Solution

The angular half-width of the central diffraction maximum is

3 0^{\circ}

, and the first single-slit minimum satisfies

sin 3 0^{\circ} = \frac{λ}{b}

, so

λ = b sin 3 0^{\circ} = 1 \times 1 0^{- 6} \times \frac{1}{2} = 5 \times 1 0^{- 7}

m. Treating the two slits as a Young's pair, the fringe width is

ω = \frac{λ D}{d}

; solving for the separation with

ω = 1 0^{- 2}

m and

D = 0.5

m gives

d = \frac{λ D}{ω} = \frac{5 \times 1 0 ^{- 7} \times 0.5}{1 0 ^{- 2}} = 25 \times 1 0^{- 6}

m, that is

25 μ

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option

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JEE Main · 2018 · 08 Apr 2018 · Offline · Wave Optics

Solution

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