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(B) Let $s_1$ be the distance of the source from the surface and $c$ be the speed of light in air. The time taken to reach the surface is $T_1 = \frac

Vedclass · Accepted Answer

$8$

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(B) Let $s_1$ be the distance of the source from the surface and $c$ be the speed of light in air. The time taken to reach the surface is $T_1 = \frac{s_1}{c}$.Let $s_2$ be the thickness of the glass slab $(5 \ cm)$ and $v$ be the speed of light in glass. The time taken to travel through the slab is $T_2 = \frac{s_2}{v}$.Given that $T_1 = T_2$,we have $\frac{s_1}{c} = \frac{s_2}{v}$.Since the refractive index $\mu = \frac{c}{v}$,we can write $v = \frac{c}{\mu}$.Substituting this into the equation: $\frac{s_1}{c} = \frac{s_2}{c/\mu} = \frac{s_2 	imes \mu}{c}$.Therefore,$s_1 = s_2 	imes \mu$.Given $s_2 = 5 \ cm$ and $\mu = 1.6$,we get $s_1 = 5 	imes 1.6 = 8 \ cm$.

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