In a double slit experiment the angular width of a fringe is found to be 0.20 on a screen placed 1 m away. The wave length of light used is 600 nm. The angular width of the fringe is the entire experimental apparatus is immersed in water? When refractive index of water to be 4/3

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1.5°

0.0015°

0.15°

0.015°

answer is D.

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Detailed Solution

Angular width of fringe can be given as $θ_{F} = \frac{β}{D} = \frac{(λD / d)}{D} = \frac{λ}{d}$
In water wave length of light will be $λ^{1} = \frac{λ}{μ}$
$\Rightarrow$ In water new fringe width will be $θ_{F}' = \frac{λ'}{d} = \frac{λ}{μd} = \frac{θ_{F}}{μ} = \frac{0 .2}{4 / 3} = 0 {.15}^{0}$

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