A light ray enters through a right angled prism at point $P$ with the angle of incidence $30^{\circ}$ as shown in figure. It travels through the prism parallel to its base $B C$ and emerges along the face $A C$. The refractive index of the prism is: ![NEET 2024 Physics, Ray Optics — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/616059730283de43c87e3e22/Physics/images/Ray_Optics/66bcc0aeae81d148616b2b3e/question_1__q_66bcc0aeae81d148616b2b3e__cdn-question-pool.getmarks.app__uXRrqmy4hxfqmuSNfoinD3zzFWWOUQ5ENuQ02uxIDI8.original.fullsiz__391d903e11_final_ppt_sync.png)

![NEET 2024 Physics, Ray Optics — solution figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/616059730283de43c87e3e22/Physics/images/Ray_Optics/66bcc0aeae81d148616b2b3e/solution_1__q_66bcc0aeae81d148616b2b3e__cdn-question-pool.getmarks.app__54ipYLZsM_S-6hVWxunTdPSviRFxVB-Gy6wTosJooFM.original.fullsiz__bb848d47f5_final_ppt_sync.png) In prism, $r_1+c=A$ $\begin{gathered} r_1=90^{\circ}-c \ldots(1)\\ \sin c=\frac{1}{\mu} \Rightarrow \cos c=\frac{\sqrt{\mu^2-1}}{\mu} \end{gathered}$ $\Rightarrow$ Apply Snell's law, on incidence surface $\begin{aligned} 1 \cdot \sin 30^{\circ}=\mu \sin \left(r_1\right) \Rightarrow 1 \times \frac{1}{2} & =\mu \times \sin \left(90^{\circ}-c\right) \\ \frac{1}{2} & =\mu \times \frac{\sqrt{\mu^2-1}}{\mu} \end{aligned}$ On squaring $\frac{1}{4}=\mu^2-1$ $\Rightarrow \mu^2=\frac{5}{4} \Rightarrow \mu=\frac{\sqrt{5}}{2}$

A light ray enters through a right angled… - NEET UG 2024 Physics PYQ

NEET UG 2024 — Physics Optics

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A light ray enters through a right angled prism at point $P$ with the angle of incidence $30^{\circ}$ as shown in figure. It travels through the prism parallel to its base $B C$ and emerges along the face $A C$ . The refractive index of the prism is: NEET 2024 Physics, Ray Optics — question figure

Official previous-year question

Held on 30 Apr 2024 · Verified 9 Jul 2026.

Options

A
$\frac{\sqrt{5}}{2}$
B
$\frac{\sqrt{3}}{4}$
C
$\frac{\sqrt{3}}{2}$
D
$\frac{\sqrt{5}}{4}$

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Solution

NEET 2024 Physics, Ray Optics — solution figure

In prism, $r_1+c=A$ $\begin{gathered} r_1=90^{\circ}-c \ldots(1)\\ \sin c=\frac{1}{\mu} \Rightarrow \cos c=\frac{\sqrt{\mu^2-1}}{\mu} \end{gathered}$ $\Rightarrow$ Apply Snell's law, on incidence surface $\begin{aligned} 1 \cdot \sin 30^{\circ}=\mu \sin \left(r_1\right) \Rightarrow 1 \times \frac{1}{2} & =\mu \times \sin \left(90^{\circ}-c\right) \\ \frac{1}{2} & =\mu \times \frac{\sqrt{\mu^2-1}}{\mu} \end{aligned}$ On squaring $\frac{1}{4}=\mu^2-1$ $\Rightarrow \mu^2=\frac{5}{4} \Rightarrow \mu=\frac{\sqrt{5}}{2}$

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