A ray of light suffers a minimum deviation wh | vedclass

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(B) For an equilateral prism,the angle of the prism $A = 60^{\circ}$.At the condition of minimum deviation,the angle of incidence $i$ is equal to the

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$45$

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(B) For an equilateral prism,the angle of the prism $A = 60^{\circ}$.At the condition of minimum deviation,the angle of incidence $i$ is equal to the angle of emergence $e$,and the angle of refraction $r_1$ is equal to $r_2 = r$.From the relation $A = r_1 + r_2$,we have $A = 2r$,so $r = A/2 = 60^{\circ}/2 = 30^{\circ}$.According to Snell's law,$\mu = \frac{\sin i}{\sin r}$.Given $\mu = \sqrt{2}$ and $r = 30^{\circ}$,we have $\sqrt{2} = \frac{\sin i}{\sin 30^{\circ}}$.$\sin i = \sqrt{2} 	imes \sin 30^{\circ} = \sqrt{2} 	imes \frac{1}{2} = \frac{1}{\sqrt{2}}$.Therefore,$i = \sin^{-1}\left(\frac{1}{\sqrt{2}}ight) = 45^{\circ}$.

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

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