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(A) Given: Angle of incidence $i = 45^o$,Angle of prism $A = 60^o$.Since the ray undergoes minimum deviation,the angle of emergence $e$ is equal to th

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$30^o, \sqrt{2}$

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(A) Given: Angle of incidence $i = 45^o$,Angle of prism $A = 60^o$.Since the ray undergoes minimum deviation,the angle of emergence $e$ is equal to the angle of incidence $i$,so $e = 45^o$.The angle of minimum deviation $\delta_m$ is given by the formula: $\delta_m = i + e - A$.Substituting the values: $\delta_m = 45^o + 45^o - 60^o = 30^o$.The refractive index $\mu$ is given by: $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.Substituting the values: $\mu = \frac{\sin((60^o + 30^o)/2)}{\sin(60^o/2)} = \frac{\sin(45^o)}{\sin(30^o)}$.Calculating the values: $\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.Thus,the angle of minimum deviation is $30^o$ and the refractive index is $\sqrt{2}$.

The angle of incidence for a ray of light at a refracting surface of a prism is $45^o$. The angle of the prism is $60^o$. If the ray suffers minimum deviation through the prism,the angle of minimum deviation and the refractive index of the material of the prism,respectively,are:

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