For the given incident ray as shown in the figure, the condition of the total internal reflection of this ray and the minimum refractive index of prism will be: ![NEET 2002 Physics, Ray Optics — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/616059730283de43c87e3e22/Physics/images/Ray_Optics/66bcbd85ae81d148616b11f8/question_1__q_66bcbd85ae81d148616b11f8__cdn-question-pool.getmarks.app__ccbnWCNC4Tn-6SKia098pLnc12C0DXJgYvGAvCxH0_k.original.fullsiz__6e6d4e56d4_final_ppt_sync.png)

For the given incident ray as shown in the… - NEET UG 2002 Physics PYQ

NEET UG 2002 — Physics Optics

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For the given incident ray as shown in the figure, the condition of the total internal reflection of this ray and the minimum refractive index of prism will be: NEET 2002 Physics, Ray Optics — question figure

Official previous-year question

Held on 30 Apr 2002 · Verified 9 Jul 2026.

Options

A
$\frac{\sqrt{3}+1}{2}$
B
$\frac{\sqrt{2}+1}{2}$
C
$\sqrt{\frac{3}{2}}$
D
$\sqrt{\frac{7}{6}}$

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Solution

NEET 2002 Physics, Ray Optics — solution figure 1 For given condition and use snell's law 1. $\sin 45^{\circ}=\mu \sin \left(90-q_C\right)$ $\begin{array}{ll} \frac{1}{\sqrt{2}}=\mu \cos \theta_C=\sqrt{\mu^2-1} \\ \Rightarrow \quad \mu^2=1+\frac{1}{2}=\frac{3}{2} \\ \Rightarrow \quad \mu=\sqrt{\frac{3}{2}} \end{array}$ NEET 2002 Physics, Ray Optics — solution figure 2 $\begin{array}{ll} \therefore & \sin \theta_c=\frac{1}{\mu} \\ \Rightarrow \quad \cos \theta_c=\sqrt{\frac{\mu^2-1}{\mu}} \end{array}$