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Q.

In Young's double slit experiment the distance between two sources is 0.1 mm. The distance of the screen from the source is 20 cm. Wave length of light used is 5460A°. The angular position of the first dark fringe is nearly

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a

0.08°

b

0.20°

c

0.32°

d

0.16°

answer is D.

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Detailed Solution

$\begin{array}{l} dsin θ = (2 n - 1) \frac{λ}{2}; \sin θ ≃ θ = \frac{(2 n - 1) λ}{2 d} rad \\ ∴ θ = \frac{[2 (1) - 1] 5460 \times 10^{- 10}}{2 (0 .1 \times 10^{- 3})} \times \frac{180^{\circ}}{π} \\ = 0 {.159}^{\circ} = 0 {.16}^{\circ} \end{array}$

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