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(C) For a prism,the condition for minimum deviation is $i = e$ and $r_1 = r_2 = r$.Since the prism is equilateral,the angle of the prism $A = 60^{\cir

Vedclass · Accepted Answer

$60$

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(C) For a prism,the condition for minimum deviation is $i = e$ and $r_1 = r_2 = r$.Since the prism is equilateral,the angle of the prism $A = 60^{\circ}$.Using the relation $A = r_1 + r_2$,we get $60^{\circ} = 2r$,which implies $r = 30^{\circ}$.According to Snell's Law at the first surface,$n_1 \sin i = n_2 \sin r_1$.Given $n_1 = 1$ (air) and $n_2 = \sqrt{3}$ (prism material),we have $\sin i = \sqrt{3} \sin 30^{\circ}$.Since $\sin 30^{\circ} = 0.5$,we get $\sin i = \sqrt{3} 	imes 0.5 = \frac{\sqrt{3}}{2}$.Therefore,$i = \arcsin(\frac{\sqrt{3}}{2}) = 60^{\circ}$.

$A$ monochromatic light is incident at a certain angle on an equilateral triangular prism and suffers minimum deviation. If the refractive index of the material of the prism is $\sqrt{3}$,then the angle of incidence is ... $^o$.

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