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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 5460 $\overset{°}{A}$ . The angular position of the first dark fringe is nearly

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0.08°

0.16°

0.20°

0.32°

answer is B.

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Detailed Solution

$θ = \frac{λ}{2 d} = \frac{5460 \times 10^{- 10}}{2 \times 10^{- 4}}$ $= 273 \times 10^{- 5} rad ∴ θ = 273 \times 10^{- 5} \times \frac{180 °}{π} = 0 {.16}^{0}$

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