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A ray of light is incident on a parallel slab of thickness t and refractive index μ. If the angle of incidence θ is small, then the lateral displacement of the incident ray will be equal to,

A
t θ (μ1)μ
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B
t θμ
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C
μ t θμ1
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D
None of these
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Solution

The correct option is A t θ (μ1)μ
The lateral displacement of the incident ray is,
d=tcosr sin(ir)

=tcosr(sini cosrcosi sinr)

=t(sinicosi sinrcosr)

As, μ=sinisinrsinr=siniμ

d=t(sinicosi siniμcosr)
=tsini(1cosiμcosr)
=t i (sinii)(1cosiμcosr)

As the angle i is small, the angle r will be even smaller.
sinii1, cosi1 and cosr1
As, limθ0 sinθθ=1, limθ0 cosθ=1
d=t i(11μ)

=t θ(μ1μ) (i=θ)

Hence, (A) is the correct answer.

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