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Question

A cubical block of glass of refractive index n1 is in contact with the surface of water of refractive index n2. A beam of light is incident on vertical face of the block (see figure). After refraction, a total internal reflection at the base and refraction at the opposite vertical face, the ray emerges out at an angle θ. The value of θ is given by
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A
sinθ<n21n22
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B
tanθ<n21n22
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C
sinθ<1n21n22
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D
tanθ<1n21n22
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Solution

The correct option is A sinθ<n21n22
Let us consider the angle of incidence at glass air medium be α
applying Snell's law
n1sinα=1×sinθ

sinθ=n1sinα 1

Now angle of incidence at water glass medium will be 90α

Here the ray is refracted back , total internal reflection has took place
so let the 90α be the critical angle

Applying snell's law

n1sin(90α)=n2sin90

cosα=n2n1

Substitute this in eq1

sinθ=n1n22n21n1=n22n21

This tis the maximum value of sinθ as we have assumed critical angle

Hence sinθ<n22n21

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