Total Internal Reflection
Problems from IIT JEE
Problem (IIT JEE 2014):
A point source S is placed at the bottom of a transparent block of height 10 m and refractive index 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is,
- 1.21
- 1.30
- 1.36
- 1.42
Solution:
The rays stop coming to liquid side if the angle of incidence at the block-liquid interface is greater than the critical angle $\theta_c$ (see figure). Let $\mu_b=2.72$ be the refractive index of the block and $\mu_l$ be the refractive index of the liquid. The critical angle is given by,
\begin{align}
\label{pxb:eqn:1}
\sin\theta_c={\mu_l}/{\mu_b}.
\end{align}
From geometry,
\begin{align}
\label{pxb:eqn:2}
\sin\theta_c={r}/{\sqrt{r^2+h^2}}.
\end{align}
Solve above equations to get,
\begin{align}
\mu_l=\frac{\mu_b r}{\sqrt{r^2+h^2}}=\frac{2.72(11.54/2)}{\sqrt{(11.54/2)^2+10^2}}=1.36.\nonumber
\end{align}