Question

A prism is shown in the figure with prism angle $75°$ and refractive index $\sqrt{3}$. A light ray incidents on a surface at incident angle $\theta$. Other face is coated with a medium of refractive index $n$. For $\theta \le 60°$ ray suffers total internal reflection. find value of $n^2$.

Answer 1

Step 1: Given data,

Angle of prism, $A=75°$

refractive index, $\mu =\sqrt{3}$

Angle of incident $\theta =60°$

Refractive index of medium is $n$,

Step 2: Find value of $n^2$

For Total Internal Reflection, $\sin c=\frac{n}{\mu }$

From Snell's Law,

$$\begin{gathered} \sin \theta =\mu \sin \left(75°-c\right) \\ \Rightarrow \sin 60°=\sqrt{3}\sin \left(75°-c\right) \\ \Rightarrow \frac{\sqrt{3}}{2}=\sqrt{3}\sin \left(75°-c\right) \\ \Rightarrow \frac{1}{2}=\sin \left(75°-c\right) \\ \Rightarrow \sin 30°=\sin \left(75°-c\right) \\ \Rightarrow 30°=75°-c \\ \Rightarrow c=45° \end{gathered}$$

then, Total Internal Reflection,

$$\begin{gathered} \sin 45°=\frac{n}{\sqrt{3}} \\ \Rightarrow \frac{1}{\sqrt{2}}=\frac{n}{\sqrt{3}} \\ \Rightarrow n=\frac{\sqrt{3}}{\sqrt{2}} \\ \Rightarrow n^2=\frac{3}{2} \\ \Rightarrow n^2=1.5 \end{gathered}$$

Hence, the value of $n^2$is $1.5$.