For a prism,the angle of the prism is 60^{cir | vedclass

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Question

(C) For a light ray to be refracted from the second surface,the angle of incidence at the second surface $(r_2)$ must be less than the critical angle

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(C) For a light ray to be refracted from the second surface,the angle of incidence at the second surface $(r_2)$ must be less than the critical angle $(C)$.The condition for refraction at the second surface is $r_2 The critical angle $C$ is given by $\sin C = 1/\mu = 1/\sqrt{7/3} = \sqrt{3/7}$.Thus,$\sin r_2 For a prism,$r_1 + r_2 = A = 60^{\circ}$,so $r_2 = 60^{\circ} - r_1$.Substituting this,$\sin(60^{\circ} - r_1) To find the minimum angle of incidence $(i)$,we need the maximum value of $r_1$. Since $r_2$ must be less than $C$,$r_1$ must be greater than $60^{\circ} - C$.Using $\sin C = \sqrt{3/7} \approx 0.6546$,we find $C \approx 40.89^{\circ}$.So,$r_1 > 60^{\circ} - 40.89^{\circ} = 19.11^{\circ}$.Using Snell's law at the first surface: $\sin i = \mu \sin r_1 = \sqrt{7/3} \sin(19.11^{\circ}) \approx 1.5275 	imes 0.3274 \approx 0.5$.Thus,$i > 30^{\circ}$. The minimum angle is $30^{\circ}$.

For a prism,the angle of the prism is $60^{\circ}$ and the refractive index is $\sqrt{7/3}$. The minimum possible angle of incidence so that the light ray is refracted from the second surface is (in $^{\circ}$)

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