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Q.

In Young's double slit experiment, the fringe width is 12 mm. If the entire arrangement is placed in water of refractive index $\frac{4}{3}$ , then the fringe width becomes (in mm)

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a

16

b

48

c

12

d

9

answer is B.

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Detailed Solution

$\begin{array}{l} B = 12 \times 10^{- 3} \\ β^{'} = \frac{β}{μ} = \frac{12 \times 10^{- 3}}{\frac{4}{3}} \\ = 9 \times 10^{- 3} m = 9 mm \end{array}$

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