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(C) For a prism,at the condition of minimum deviation,the angle of incidence $i$ is related to the refracting angle $A$ and the angle of minimum devia

Vedclass · Accepted Answer

$60^{\circ}, 30^{\circ}$

Vedclass · Answer

(C) For a prism,at the condition of minimum deviation,the angle of incidence $i$ is related to the refracting angle $A$ and the angle of minimum deviation $D_m$ by the following relations:$1$. $i = \frac{A + D_m}{2}$$2$. $r = \frac{A}{2}$Given the refractive index $\mu = \sqrt{2}$ and the angle of incidence $i = 45^{\circ}$.Using Snell's law at the first surface: $\mu = \frac{\sin i}{\sin r} \implies \sqrt{2} = \frac{\sin 45^{\circ}}{\sin r} = \frac{1/\sqrt{2}}{\sin r}$.Therefore,$\sin r = \frac{1}{2}$,which gives $r = 30^{\circ}$.Since $r = \frac{A}{2}$,we have $A = 2r = 2 	imes 30^{\circ} = 60^{\circ}$.Now,using $i = \frac{A + D_m}{2}$,we get $45^{\circ} = \frac{60^{\circ} + D_m}{2}$.$90^{\circ} = 60^{\circ} + D_m \implies D_m = 30^{\circ}$.Thus,the values are $A = 60^{\circ}$ and $D_m = 30^{\circ}$.

$A$ prism of refractive index $\sqrt{2}$ and refracting angle $A$ produces a minimum deviation $D_m$. $A$ ray incident on one face at an angle of incidence $i = 45^{\circ}$ undergoes minimum deviation. The values of $A$ and $D_m$ are respectively:

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