The critical angle for light going from medium ' x ' to medium ' Y ' is θ. The speed of light in medium ' x ' is ' Vx '. The speed of light in medium ' Y ' is (a) Vx sin θ (b) Vx tan θ (c) (Vx)/(tan θ) (d) (Vx)/(sin θ)

Answer: (d). sin θ = (μY)/(μx) = (Vx)/(VY) (since μ ∝ 1/V). VY = (Vx)/(sin θ) So the answer is (d).

The critical angle for light going from medium ' x ' to medium ' Y '… | MHT CET 2021 (22 Sep, Shift 2)

Q29MHT CET 2021 · 22 SEP · SHIFT 2Easy

Total Internal Reflection

The critical angle for light going from medium ' $x$ ' to medium ' $Y$ ' is $\theta$ . The speed of light in medium ' $x$ ' is ' $V_x$ '. The speed of light in medium ' $Y$ ' is

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Answer: (d)

$\sin\theta = \frac{\mu_Y}{\mu_x} = \frac{V_x}{V_Y}$ (since $\mu \propto \frac{1}{V}$ ).
$V_Y = \frac{V_x}{\sin\theta}$
So the answer is (d).

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