Physics · Geometrical Optics

JEE Main 2025 — 7 April, Evening Shift — Question 50

A transparent block $A$ having refractive index $\mu=1.25$ is surrounded by another medium of refractive index $\mu=1.0$ as shown in figure. A light ray is incident on the flat face of the block with incident angle $\theta$ as shown in figure. What is the maximum value of $\theta$ for which light suffers total internal reflection at the top surface of the block?

Option A:
$\sin ^{-1}(3 / 4)$
Correct
Option B:
$\tan ^{-1}(4 / 3)$
Option C:
$\cos ^{-1}(3 / 4)$
Option D:
$\tan ^{-1}(3 / 4)$

Answer: A

Step-by-step solution

At $1^{\text {st }}$ refractions $1 \times \sin \theta=\frac{5}{4} \sin \alpha$

\begin{gathered} \sin \alpha=\frac{4}{5} \sin \theta \end{gathered}

At point of TIR $\frac{5}{4} \sin \left(\theta_{C}\right)=1$

$\sin (90-\alpha)=\frac{4}{5}$

$\cos \alpha=\frac{4}{5}$ $\sin \alpha=\frac{3}{5}$

From (i) and (ii) $\alpha=\sin ^{-1}\left(\frac{3}{4}\right)$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 7 April, Evening Shift
Subject: Physics
Chapter: Geometrical Optics
Topic: Introduction to Refraction of Light (Snell's Law)

Answer: A

Step-by-step solution

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