Light travels in two media M₁ and M₂ with speeds 1.5 × 10⁸ ms⁻¹ and 2.0 × 10⁸ ms⁻¹ respectively. The critical angle between them is: (a) tan ⁻¹(3/√(7)) (b) tan ⁻¹(2/3) (c) cos ⁻¹(3/4) (d) sin ⁻¹(2/3)

Answer: (a). n = c/v, so n₁ = (3×10⁸)/(1.5×10⁸) = 2 and n₂ = (3×10⁸)/(2×10⁸) = 1.5. Critical angle (from M₁ to M₂): sin C = (n₂)/(n₁) = (v₁)/(v₂) = 3/4. cos C = √(1-9/16) = √(7)/4, so tan C = 3/√(7). So the answer is (a) tan ⁻¹(3/√(7)).

Light travels in two media M₁ and M₂ with speeds 1.5 × 10⁸ ms⁻¹ and… | JEE Main 2022 (26 Jul, Shift 2)

Q25JEE Main 2022 · 26 JUL · SHIFT 2Easy

Total Internal Reflection

Light travels in two media $M_1$ and $M_2$ with speeds $1.5 \times 10^{8}\text{ ms}^{-1}$ and $2.0 \times 10^{8}\text{ ms}^{-1}$ respectively. The critical angle between them is:

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Answer: (a)

$n = \frac{c}{v}$ , so $n_1 = \frac{3\times10^8}{1.5\times10^8} = 2$ and $n_2 = \frac{3\times10^8}{2\times10^8} = 1.5$ .
Critical angle (from $M_1$ to $M_2$ ): $\sin C = \frac{n_2}{n_1} = \frac{v_1}{v_2} = \frac{3}{4}$ .
$\cos C = \sqrt{1-\frac{9}{16}} = \frac{\sqrt{7}}{4}$ , so $\tan C = \frac{3}{\sqrt{7}}$ .
So the answer is (a) $\tan^{-1}\left(\frac{3}{\sqrt{7}}\right)$ .

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