A ray of light travelling in glass (μ = 3/2) is incident on a horizontal glass-air surface at the critical angle θc If a thin layer of water (μ = 4/3) is now poured on the glass-air surface, the angle at which the ray emerges into air at the water-air surface is
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a
60°
b
45°
c
90°
d
180°
answer is C.
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Detailed Solution
μg sin θc= μ1 sin 90° or μg sin θc=1 When water is poured, μw sin r = μg sin θc or μw sin r = 1 Again, μa sin θ=μw sin r or μa sin θ=1 or Sin θ=1 or θ=90°