Question

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


(correct answer + 2, wrong answer - 0.50)

• A. ${15}^{\circ }$
• B. ${22.5}^{\circ }$
• C. ${30}^{\circ }$
• D. ${37}^{\circ }$

Answer 1

The correct option is A ${15}^{\circ }$


Applying Snell’s law at surface PQ,
$\therefore 1\times \text{sin}\alpha =n\times \text{sin}\beta$
At $\alpha ={45}^{\circ },n=\sqrt{2}$
$\text{sin}\beta =\frac{1}{2}\therefore \beta ={30}^{\circ }$
$\gamma =90-(180-(90+\theta +\beta \left)\right)$
$\therefore$ Applying Snell's law at surface PR,
$\therefore \sqrt{2}\text{sin}\gamma =1$
$\therefore \text{sin}(\theta +\beta )=\frac{1}{\sqrt{2}}$
$\therefore \theta +\beta ={45}^{\circ }$
$\therefore \theta +{30}^{\circ }={45}^{\circ }$
$\therefore \theta ={15}^{\circ }$.
Thus, option A is correct.