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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

h3c

b

6hc

c

Zero

d

h6c

answer is D.

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Detailed Solution

μ=ctx;t=μxct(w+g)−tw=μω(x−h)c+μghc−μωxc=hcμg−μω=hc32−43=h6c
 

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