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(C) For a prism,the refractive index $\mu$ is given by the formula: $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.At the condition of minimum devia

Vedclass · Accepted Answer

$45$

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(C) For a prism,the refractive index $\mu$ is given by the formula: $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.At the condition of minimum deviation,the angle of incidence $i$ is equal to the angle of emergence $e$,and the angle of refraction $r_1 = r_2 = A/2$.Thus,the angle of incidence is given by $i = (A + \delta_m)/2$.Substituting this into the refractive index formula: $\mu = \frac{\sin i}{\sin(A/2)}$.Given $\mu = \sqrt{2}$ and $A = 60^{\circ}$,we have: $\sqrt{2} = \frac{\sin i}{\sin(60^{\circ}/2)}$.$\sqrt{2} = \frac{\sin i}{\sin 30^{\circ}}$.Since $\sin 30^{\circ} = 0.5$,we get: $\sin i = \sqrt{2} 	imes 0.5 = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.Therefore,$i = \sin^{-1}(1/\sqrt{2}) = 45^{\circ}$.

$A$ ray of light suffers minimum deviation when incident on a $60^{\circ}$ prism of refractive index $\sqrt{2}$. The angle of incidence is......$^{\circ}$

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