A light ray enters through a right angled prism at point P with the angle of incidence 30° as shown in figure. It travels through the prism parallel to its base BC and emerges along the face AC. The refractive index of the prism is [2024]

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Solution

Q.
A light ray enters through a right angled prism at point P with the angle of incidence 30° as shown in figure. It travels through the prism parallel to its base BC and emerges along the face AC. The refractive index of the prism is [2024]

1 $\frac{\sqrt{5}}{4}$

2 $\frac{\sqrt{5}}{2}$

3 $\frac{\sqrt{3}}{4}$

4 $\frac{\sqrt{3}}{2}$

Ans.
(2)

Let the refractive index of the prism be $μ .$

Applying Snell’s law at point $P,$

$1 \sin 30 ° = μ \sin r_{1}$

$\sin r_{1} = \frac{1}{2 μ}$ ...(i)

$r_{1} + r_{2} = 90 ° \Rightarrow r_{2} = 90 ° - r_{1}$

Applying Snell’s law at point $Q,$

$μ \sin r_{2} = 1 \times \sin 90 °$

$μ \times \sin (90 ° - r_{1}) = 1 \Rightarrow μ \cos r_{1} = 1$

$μ \times \sqrt{1 - \sin^{2} r_{1}} = 1$ $;$ $μ \sqrt{1 - \frac{1}{{4 μ}^{2}}} = 1$ (Using equation (i))

$1 - \frac{1}{4 μ^{2}} = \frac{1}{μ^{2}};$ $\frac{5}{4 μ^{2}} = 1;$ $μ = \frac{\sqrt{5}}{2}$

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