For the angle of minimum deviation of a prism | vedclass

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(B) The refractive index of a prism is given by $\mu = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)}$.Given that the angle of minimum deviation $\delta_m = A

Vedclass · Accepted Answer

lies between $2$ and $\sqrt{2}$

Vedclass · Answer

(B) The refractive index of a prism is given by $\mu = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)}$.Given that the angle of minimum deviation $\delta_m = A$,we substitute this into the formula:$\mu = \frac{\sin((A+A)/2)}{\sin(A/2)} = \frac{\sin A}{\sin(A/2)}$.Using the identity $\sin A = 2\sin(A/2)\cos(A/2)$,we get:$\mu = \frac{2\sin(A/2)\cos(A/2)}{\sin(A/2)} = 2\cos(A/2)$.For a physical prism,the angle of incidence $i$ must satisfy $0 If $A 	o 0^o$,then $\mu 	o 2\cos(0^o) = 2$.If $A 	o 90^o$,then $\mu 	o 2\cos(45^o) = 2 	imes (1/\sqrt{2}) = \sqrt{2}$.Therefore,the refractive index $\mu$ must lie between $\sqrt{2}$ and $2$.

For the angle of minimum deviation of a prism to be equal to its refracting angle,the prism must be made of a material whose refractive index

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