A ray of light is incident at a glass-water interface at an angle… | MHT CET 2026 (19 May, Shift 2)

Q3MHT CET 2026 · 19 MAY · SHIFT 2Medium

Total Internal Reflection

A ray of light is incident at a glass-water interface at an angle $(i_1)$ , it emerges finally parallel to the surface of water, then the value of the refractive index of glass $(\mu_g)$ is

$\{\mu_w =$ refractive index of water $= 4/3, \sin(90)^\circ = 1, \cos(90)^\circ = 0\}$

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Answer: (c)

Snell's law gives $\mu \sin\theta =$ constant along the whole path.
Glass to water to air: $\mu_g \sin i_1 = \mu_w \sin r = \mu_a \sin 90^\circ$ .
The ray finally grazes the water surface, so $\mu_g \sin i_1 = 1 \times 1 = 1$ . The value of $\mu_w$ cancels out.
$\mu_g = \frac{1}{\sin i}$ . So the answer is (c).

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Total Internal Reflection question from MHT CET 2026 (19 May, Shift 2)

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