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(C) The light rays from the point source $S$ strike the top surface of the block at the critical angle $i_c$ to form the boundary of the circular brig

Vedclass · Accepted Answer

$1.36$

Vedclass · Answer

(C) The light rays from the point source $S$ strike the top surface of the block at the critical angle $i_c$ to form the boundary of the circular bright spot.From the geometry of the problem,the radius of the circular spot is $r = D/2 = 11.54 / 2 = 5.77 \ mm$.The height of the block is $h = 10 \ mm$.Using the definition of the critical angle,$\sin i_c = \frac{n_L}{n_B}$.From the right-angled triangle formed by the height $h$ and radius $r$,we have $\sin i_c = \frac{r}{\sqrt{r^2 + h^2}}$.Equating the two expressions for $\sin i_c$:$\frac{n_L}{n_B} = \frac{r}{\sqrt{r^2 + h^2}}$$n_L = n_B 	imes \frac{r}{\sqrt{r^2 + h^2}}$Substituting the given values:$n_L = 2.72 	imes \frac{5.77}{\sqrt{5.77^2 + 10^2}}$$n_L = 2.72 	imes \frac{5.77}{\sqrt{33.2929 + 100}}$$n_L = 2.72 	imes \frac{5.77}{\sqrt{133.2929}}$$n_L = 2.72 	imes \frac{5.77}{11.545} \approx 2.72 	imes 0.5 = 1.36$.Thus,the refractive index of the liquid is $1.36$.

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