Question

A parallel beam of light is incident from air at an angle $\alpha$ on the side $\text{PQ}$ of a right angled triangular prism of refractive index $\mu =\sqrt{2}.$ Light undergoes total internal reflection at the face $\text{PR}$ when $\alpha$ has a minimum value of ${45}^{\circ }$. The angle $\theta$ of the prism is:

• A. 15.0
• B. 15
• C. 15.00

Answer 1

Answer: (A), (B), (C)

Given:
Refractive index of air, ${\mu }_{air}=1$ Refractive index of prism, ${\mu }_{prism}=\sqrt{2}$

Applying snell's law at face $\text{PQ}$
$${\mu }_{air}\sin \alpha ={\mu }_{prism}\sin r_1$$
Substituting the given data we get,
$$1\times \sin \alpha =\sqrt{2}\times \sin r_1$$
$\Rightarrow$$1\times \sin {45}^{\circ }=\sqrt{2}\times \sin r_1[\text{For,}\alpha ={45}^{\circ }]$

$\therefore r_1={30}^{\circ }$

Critical angle for glass-air interface,

$\sin {\theta }_c=\frac{1}{\mu }=\frac{1}{\sqrt{2}}$

$\therefore {\theta }_c={45}^{\circ }$



From, $\Delta \text{PLM}$:

$\theta ={180}^{\circ }-({90}^{\circ }+r_1+{90}^{\circ }-{\theta }_c)$

$\Rightarrow \theta ={\theta }_c-r_1$

$\Rightarrow \theta ={45}^{\circ }-{30}^{\circ }={15}^{\circ }$