Question Details

Find the value of the angle of emergence from the prism. Refractive index of the glass is √3 .

Options

A

45°

B

90°

C

60°

D

30°

Show Answer

Correct Answer :

Option C

60°

60°

Solution :

We are asked to find the angle of emergence from a prism whose refractive index is √3.

For a thin prism the relation between the refractive index n, the prism apex angle A and the angle of minimum deviation D_{min} is

n = \frac{\sin\!\left(\frac{A + D_{min}}{2}\right)}{\sin\!\left(\frac{A}{2}\right)}

When the prism is used such that the angle of incidence equals the angle of emergence (the condition for minimum deviation), the angle of emergence is equal to the angle of deviation D_{min}.

Given n = \sqrt{3}, we substitute into the formula and solve for D_{min} assuming a common apex angle of the prism, A = 60^{\circ} (as shown in the figure).

First compute the denominator:

\sin\!\left(\frac{A}{2}\right) = \sin\!\left(\frac{60^{\circ}}{2}\right) = \sin 30^{\circ} = 0.5

Set up the equation:

\sqrt{3} = \frac{\sin\!\left(\frac{60^{\circ} + D_{min}}{2}\right)}{0.5}

Multiply both sides by 0.5:

\sin\!\left(\frac{60^{\circ} + D_{min}}{2}\right) = \frac{\sqrt{3}}{2}

We know that

\sin 60^{\circ} = \frac{\sqrt{3}}{2}

Hence

\frac{60^{\circ} + D_{min}}{2} = 60^{\circ}

Solving for D_{min} gives

60^{\circ} + D_{min} = 120^{\circ}

D_{min} = 60^{\circ}

Because the angle of emergence equals the angle of minimum deviation, the angle of emergence from the prism is also 60°.

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