For a prism of prism angle theta=60^{circ},th | vedclass

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(C) For minimum deviation in a symmetric prism $(n_1=n_2=n)$,the angle of incidence $i$ is given by $\sin i = n \sin(	heta/2)$.Given $	heta=60^{\cir

Vedclass · Accepted Answer

$B, C$

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(C) For minimum deviation in a symmetric prism $(n_1=n_2=n)$,the angle of incidence $i$ is given by $\sin i = n \sin(	heta/2)$.Given $	heta=60^{\circ}$,$\sin i = 1.5 	imes \sin 30^{\circ} = 1.5 	imes 0.5 = 0.75 = 3/4$.At the second surface,the angle of incidence is $r_2 = 30^{\circ}$. The refraction at the second surface is $n_2 \sin r_2 = 1 \sin e$.For $n_1=n$ and $n_2=n+\Delta n$,the light enters the first surface at angle $i$ and refracts at $r_1=30^{\circ}$. It then travels to the second surface. Since the prism is symmetric,the ray hits the second surface at $r_2=30^{\circ}$.Thus,$(n+\Delta n) \sin 30^{\circ} = \sin e$.Since $e = i + \Delta e$,we have $\sin(i + \Delta e) = (n+\Delta n) \sin 30^{\circ} = n \sin 30^{\circ} + \Delta n \sin 30^{\circ} = \sin i + \Delta n \sin 30^{\circ}$.Using $\sin(i + \Delta e) \approx \sin i + \Delta e \cos i$,we get $\Delta e \cos i = \Delta n \sin 30^{\circ}$.Since $\sin i = 3/4$,$\cos i = \sqrt{1 - (3/4)^2} = \sqrt{7}/4$.$\Delta e = \Delta n \frac{\sin 30^{\circ}}{\cos i} = \Delta n \frac{0.5}{\sqrt{7}/4} = \Delta n \frac{2}{\sqrt{7}} \approx 0.756 \Delta n$.Since $0.756 Since $\Delta e = (2/\sqrt{7}) \Delta n$,$\Delta e$ is proportional to $\Delta n$. Thus,$(B)$ is correct.For $\Delta n = 2.8 	imes 10^{-3}$,$\Delta e = (2/\sqrt{7}) 	imes 2.8 	imes 10^{-3} \approx 0.756 	imes 2.8 	imes 10^{-3} \approx 2.11 	imes 10^{-3} 	ext{ rad} = 2.11 	ext{ mrad}$.This value lies between $2.0$ and $3.0$ mrad. Thus,$(C)$ is correct.

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