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Q.

Two beams of light are incident normally on water ( $μ =$ 4 / 3) . If one of the beams passes through a glass (µ =3 / 2) slab of height 'h' as shown in the figure, the time difference for both the beams for reaching the bottom is: (C is velocity of light is vacuum)

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a

$\frac{h}{3 c}$

b

$\frac{6 h}{c}$

c

Zero

d

$\frac{h}{6 c}$

answer is D.

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Detailed Solution

$\begin{array}{l} μ = \frac{ct}{x}; t = \frac{μx}{c} \\ t_{(w + g)} - t_{w} = μ_{ω} \frac{(x - h)}{c} + μ_{g} (\frac{h}{c}) - μ_{ω} (\frac{x}{c}) \\ = \frac{h}{c} (μ_{g} - μ_{ω}) = \frac{h}{c} (\frac{3}{2} - \frac{4}{3}) = \frac{h}{6 c} \end{array}$

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