JEE Main 2025 Apr 7 Shift 2, Physics Q20: Critical Angle and Total Internal Reflection

JEE Main2025Apr 7, Shift 2Physics

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

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

Solution

Let $r$ be the angle of refraction inside the block at the flat entrance face. By the geometry, the ray inside then strikes the top surface at an angle $θ_{c} = 9 0^{\circ} - r$ from the normal to that surface.

For total internal reflection at the top surface, the incidence angle must reach the critical angle: $sin θ_{c} = \frac{μ _{1}}{μ _{2}} = \frac{1}{1.25} = \frac{4}{5}$ , which gives $cos θ_{c} = \frac{3}{5}$ .

Applying Snell's law at the entrance face: $μ_{1} sin θ = μ_{2} sin r = μ_{2} sin (9 0^{\circ} - θ_{c}) = μ_{2} cos θ_{c}$ .

$sin θ = 1.25 \times \frac{3}{5} = \frac{3}{4}$ , so $θ_{ma x} = sin^{- 1} (\frac{3}{4})$ .

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JEE Main 2025 Apr 7 Shift 2, Physics Q20: Critical Angle and Total Internal Reflection

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