Question

A light ray enters a solid glass sphere of refractive index $\mu =\surd 3$ at an angle of incidence $60°$. The ray is both reflected and refracted at the farther surface of the sphere. The angle (in degrees) between the reflected and refracted rays at this surface is ________.

Answer 1

Step 1: Given data and drawing the diagram of the situation

The angle of the incident at surface, $S_1$,$i={60}^{\circ }$

Refractive index of glass, ${\mu }_{glass}=\surd 3$

We know, the refractive index of air, ${\mu }_{air}=1$

Step 2: Find the angle $\mathit{r}$

Apply Snell's law at the surface $S_1$

$$\begin{gathered} {\mu }_{air}sini={\mu }_{glass}\sin r \\ 1\times \sin {60}^{\circ }=\sqrt{3}\times \sin r \\ 1\times \frac{\sqrt{3}}{2}=\sqrt{3}\times \sin r \\ \sin r=\frac{1}{2} \\ r={30}^{\circ } \end{gathered}$$

Step 3: Find the angle $\mathit{e}$

Again, apply Snell's law at the surface $S_2$

$$\begin{gathered} {\mu }_{glass}sinr={\mu }_{air}\sin e \\ \sqrt{3}\times \sin {30}^{\circ }=1\times \sin e \\ \sqrt{3}\times \frac{1}{2}=1\times \sin e \\ \sin e=\frac{\sqrt{3}}{2} \\ e={60}^{\circ } \end{gathered}$$

Step 4: Find the angle $\mathit{\theta }$

As the straight angle is equal to ${180}^{\circ }$

Therefore, we can write

$$\begin{gathered} \angle r+\angle e+\angle \theta ={180}^{\circ } \\ {30}^{\circ }+{60}^{\circ }+\angle \theta ={180}^{\circ } \\ \angle \theta ={90}^{\circ } \end{gathered}$$

Hence, the angle (in degrees) between the reflected and refracted rays at this surface is $\mathbf{90}$ degrees.