A light ray enters through a right-angled pri | vedclass

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Question

(A) Let the prism angle at the top be $A = 90^{\circ}$.In the prism,the sum of the angles of refraction at the two faces is equal to the prism angle,s

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the prism angle at the top be $A = 90^{\circ}$.In the prism,the sum of the angles of refraction at the two faces is equal to the prism angle,so $r_1 + r_2 = A = 90^{\circ}$.Since the ray travels parallel to the base $BC$,the angle of refraction at the second face is the critical angle $c$,so $r_2 = c$.Thus,$r_1 = 90^{\circ} - c$.Applying Snell's law at the first surface:$1 \cdot \sin 30^{\circ} = \mu \cdot \sin r_1$$1 \cdot \frac{1}{2} = \mu \cdot \sin(90^{\circ} - c)$$\frac{1}{2} = \mu \cdot \cos c$.We know that $\sin c = \frac{1}{\mu}$,so $\cos c = \sqrt{1 - \sin^2 c} = \sqrt{1 - \frac{1}{\mu^2}} = \frac{\sqrt{\mu^2 - 1}}{\mu}$.Substituting this into the equation:$\frac{1}{2} = \mu \cdot \frac{\sqrt{\mu^2 - 1}}{\mu}$$\frac{1}{2} = \sqrt{\mu^2 - 1}$.Squaring both sides:$\frac{1}{4} = \mu^2 - 1$$\mu^2 = 1 + \frac{1}{4} = \frac{5}{4}$$\mu = \frac{\sqrt{5}}{2}$.

$A$ light ray enters through a right-angled prism at point $P$ with an angle of incidence of $30^{\circ}$ as shown in the figure. It travels through the prism parallel to its base $BC$ and emerges along the face $AC$. The refractive index of the prism is:

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