A transparent cube of side, made of a material of refractive index μ 2, is immersed in a liquid of refractive index μ₁ (μ 1 sin ⁻¹√((μ₂²)/(μ₁²) - 1) (b) θ sin ⁻¹(μ₁)/(μ₂)

Answer: (b). BCADθr90°−rEμ₂μ₁ Refraction at AB: μ₁sin θ = μ₂sin r. The normals at AB and BC are perpendicular, so the angle of incidence at E is 90° - r. TIR at E needs sin (90° - r) > (μ₁)/(μ₂), i.e. cos r > (μ₁)/(μ₂). √(1 - (μ₁²sin ²θ)/(μ₂²)) > (μ₁)/(μ₂) ⇒ sin ²θ < (μ₂² - μ₁²)/(μ₁²) = (μ₂²)/(μ₁²) - 1 So the answer is (b) θ < sin ⁻¹√((μ₂²)/(μ₁²) - 1).

A transparent cube of side, made of a material of refractive index μ… | JEE Main 2019 (12 Apr, Shift 2)

Q44JEE Main 2019 · 12 APR · SHIFT 2Difficult

Total Internal Reflection

A transparent cube of side, made of a material of refractive index $\mu$ 2, is immersed in a liquid of refractive index $\mu_1 (\mu$ 1 < $\mu$ 2). A ray is incident on the face AB at an angle $\theta$ (shown in the figure). Total internal reflection takes place at point E on the face BC.

Then $\theta$ must satisfy:

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Answer: (b)

Refraction at AB: $\mu_1\sin\theta = \mu_2\sin r$ .
The normals at AB and BC are perpendicular, so the angle of incidence at E is $90^\circ - r$ .
TIR at E needs $\sin(90^\circ - r) > \frac{\mu_1}{\mu_2}$ , i.e. $\cos r > \frac{\mu_1}{\mu_2}$ .
$\sqrt{1 - \frac{\mu_1^2\sin^2\theta}{\mu_2^2}} > \frac{\mu_1}{\mu_2} \Rightarrow \sin^2\theta < \frac{\mu_2^2 - \mu_1^2}{\mu_1^2} = \frac{\mu_2^2}{\mu_1^2} - 1$
So the answer is (b) $\theta < \sin^{-1}\sqrt{\frac{\mu_2^2}{\mu_1^2} - 1}$ .

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Total Internal Reflection question from JEE Main 2019 (12 Apr, Shift 2)

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