Critical angle of incidence for a pair of optical media is 45°. The refractive indices of first and second media are in the ratio: (a) 1: 2 (b) 1: √(2) (c) 2: 1 (d) √(2): 1

Answer: (d). sin C = (μ₂)/(μ₁) (first medium is denser). (μ₂)/(μ₁) = sin 45° = 1/√(2) ⇒ μ₁: μ₂ = √(2): 1 So the answer is (d) √(2): 1.

Critical angle of incidence for a pair of optical media is 45°. The… | JEE Main 2024 (8 Apr, Shift 1)

Q15JEE Main 2024 · 8 APR · SHIFT 1Easy

Total Internal Reflection

Critical angle of incidence for a pair of optical media is $45^\circ$ . The refractive indices of first and second media are in the ratio:

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Answer: (d)

$\sin C = \frac{\mu_2}{\mu_1}$ (first medium is denser).
$\frac{\mu_2}{\mu_1} = \sin 45^\circ = \frac{1}{\sqrt{2}} \Rightarrow \mu_1 : \mu_2 = \sqrt{2} : 1$
So the answer is (d) $\sqrt{2} : 1$ .

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