A transparent block A having refractive index μ = 1.25 is surrounded by another medium of refractive index μ = 1.0 as shown in the figure. A light ray is incident on the flat face of the block with incident angle θ as shown in the 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) sin ⁻¹(3/4) (c) tan ⁻¹(3/4) (d) cos ⁻¹(3/4)

Answer: (b). At the vertical face: sin θ = 1.25sin r. The top face is perpendicular, so the angle of incidence there is 90° - r. Critical angle: sin θC = 1/1.25 = 4/5, cos θC = 3/5. TIR needs 90° - r ≥ θC ⇒ sin r ≤ cos θC = 3/5. sin θ ≤ 1.25×3/5 = 3/4. So the answer is (b) θmax = sin ⁻¹(3/4).

A transparent block A having refractive index μ = 1.25 is surrounded… | JEE Main 2025 (7 Apr, Shift 2)

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Total Internal Reflection

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

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Answer: (b)

At the vertical face: $\sin\theta = 1.25\sin r$ . The top face is perpendicular, so the angle of incidence there is $90^\circ - r$ .
Critical angle: $\sin\theta_C = \frac{1}{1.25} = \frac{4}{5}$ , $\cos\theta_C = \frac{3}{5}$ .
TIR needs $90^\circ - r \ge \theta_C \Rightarrow \sin r \le \cos\theta_C = \frac{3}{5}$ .
$\sin\theta \le 1.25\times\frac{3}{5} = \frac{3}{4}$ .
So the answer is (b) $\theta_{max} = \sin^{-1}(3/4)$ .

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