Two equilateral-triangular prisms P_1 and P_2 | vedclass

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Question

(A) For minimum deviation in prism $P_2$,the angle of incidence at the second surface of $P_2$ must be equal to the angle of emergence,and the ray ins

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) For minimum deviation in prism $P_2$,the angle of incidence at the second surface of $P_2$ must be equal to the angle of emergence,and the ray inside $P_2$ must be parallel to the base. For an equilateral prism,this implies the angle of refraction $r_2$ at the first surface of $P_2$ is $30^{\circ}$.Applying Snell's law at the interface between $P_1$ and $P_2$ (refractive index of $P_1$ is $\mu_1 = \sqrt{3/2}$ and $P_2$ is $\mu_2 = \sqrt{3}$):$\mu_1 \sin r_2' = \mu_2 \sin r_2$$\sqrt{\frac{3}{2}} \sin r_2' = \sqrt{3} \sin 30^{\circ}$$\sqrt{\frac{3}{2}} \sin r_2' = \sqrt{3} 	imes \frac{1}{2} = \frac{\sqrt{3}}{2}$$\sin r_2' = \frac{\sqrt{3}}{2} 	imes \sqrt{\frac{2}{3}} = \frac{1}{\sqrt{2}}$$r_2' = 45^{\circ}$Since the prism $P_1$ is equilateral,the sum of refraction angles $r_1 + r_2' = A = 60^{\circ}$.$r_1 = 60^{\circ} - 45^{\circ} = 15^{\circ}$.Applying Snell's law at the first surface of $P_1$ (refractive index of vacuum is $1$):$1 \sin 	heta = \sqrt{\frac{3}{2}} \sin 15^{\circ}$$	heta = \sin^{-1}\left[\sqrt{\frac{3}{2}} \sin \left(\frac{\pi}{12}ight)ight]$Comparing this with the given expression,we find $\beta = 12$.

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