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(D) Let $d$ be the distance of the source from the surface of the slab and $l = 5 ~cm$ be the thickness of the slab.The speed of light in air is $c$ a

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$8$

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(D) Let $d$ be the distance of the source from the surface of the slab and $l = 5 ~cm$ be the thickness of the slab.The speed of light in air is $c$ and the speed of light in the glass slab is $v = \frac{c}{\mu}$,where $\mu = 1.6$.The time taken to travel from the source to the surface is $t_1 = \frac{d}{c}$.The time taken to travel through the glass slab is $t_2 = \frac{l}{v} = \frac{l}{(c/\mu)} = \frac{\mu l}{c}$.Given that $t_1 = t_2$,we have:$\frac{d}{c} = \frac{\mu l}{c}$$d = \mu l$$d = 1.6 	imes 5 ~cm = 8 ~cm$.

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