NEET (UG) 2024 — Question 22

$$A=90^{\circ}$$

In prism, $r_{1}+c=A$

$r_{1}=90^{\circ}-c…(1)$

$\sin c=\frac{1}{\mu} \Rightarrow \cos c=\frac{\sqrt{\mu^{2}-1}}{\mu}$

$\Rightarrow$ Apply Snell's law, on incidence surface

$1 \cdot \sin 30^{\circ}=\mu \sin \left(r_{1}\right) \Rightarrow 1 \times \frac{1}{2}=\mu \times \sin \left(90^{\circ}-c\right)$

$\frac{1}{2}=\mu \times \frac{\sqrt{\mu^{2}-1}}{\mu}$

On squaring $\frac{1}{4}=\mu^{2}-1$

$\Rightarrow \mu^{2}=\frac{5}{4} \Rightarrow \mu=\frac{\sqrt{5}}{2}$