Monochromatic light is incident on a glass prism of angle A . If the refractive index of the material of the prism is , a ray, incident at an angle θ, on the face AB would get transmitted through the face AC of the prism provided :

Correct Answer is : θ>sin⁡−1[μsin⁡(A−sin⁡−1(1μ))]θ > ^{-1} ≤ft[ ≤ft( A - ^{-1} ≤ft( 1/ ) ) ]θ>sin−1[μsin(A−sin−1(μ1))]

θ > ^{-1} ≤ft[ ≤ft( A + ^{-1} ≤ft( 1/ ) ) ]

θ > ^{-1} ≤ft[ ≤ft( A - ^{-1} ≤ft( 1/ ) ) ]

Correct Answer is : θ>sin⁡−1[μsin⁡(A−sin⁡−1(1μ))]θ > ^{-1} ≤ft[ ≤ft( A - ^{-1} ≤ft( 1/ ) ) ]θ>sin−1[μsin(A−sin−1(μ1))]

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VITEEE 2025 Paper

Question : 17 of 125

Marks: +1, -0

\mu

\theta

$\theta > \cos^{-1} \left[ \mu \sin \left( A + \sin^{-1} \left( \frac{1}{\mu} \right) \right) \right]$
$\theta < \cos^{-1} \left[ \mu \sin \left( A + \sin^{-1} \left( \frac{1}{\mu} \right) \right) \right]$
$\theta > \sin^{-1} \left[ \mu \sin \left( A - \sin^{-1} \left( \frac{1}{\mu} \right) \right) \right]$
$\theta < \sin^{-1} \left[ \mu \sin \left( A - \sin^{-1} \left( \frac{1}{\mu} \right) \right) \right]$

Solution:

When

r_2 = C, \angle N_2 RC = 90^{\circ}

Where

C =

critical angle

\sin C = \frac{1}{\mu} = \sin r_2

Applying Snell's law at ' R '

\mu \sin r_2 = 1 \sin 90^{\circ}

Applying Snell's law at ' Q '

1 \times \sin \theta = \mu \sin r_1

But

r_1 = A - r_2

So,

\sin \theta = \mu \sin (A - r_2)

\sin \theta = \mu \sin A \cos r_2 - \cos A

[using (i)]

From (i)

\cos r_2 = \sqrt{1 - \sin^2 r_2} = \sqrt{1 - \frac{1}{\mu^2}}

By eq. (iii) and (iv)

\sin \theta = \mu \sin A \sqrt{1 - \frac{1}{\mu^2}} - \cos A

on further solving we can show for ray not to transmitted through face AC

\theta = \sin^{-1} \left[ \mu \sin \left( A - \sin^{-1} \left( \frac{1}{\mu} \right) \right) \right] .

So, for transmission through face AC

\theta > \sin^{-1} \left[ \mu \sin \left( A - \sin^{-1} \left( \frac{1}{\mu} \right) \right) \right] .

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