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(A) $1$. The light ray enters the prism normally,so it passes undeviated to the vertical face.$2$. The angle of incidence at the vertical face is $i =

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$53$

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(A) $1$. The light ray enters the prism normally,so it passes undeviated to the vertical face.$2$. The angle of incidence at the vertical face is $i = 	heta = 37^o$.$3$. The critical angle $C$ for the prism-air interface is given by $sinC = 1/\mu = 1/(5/3) = 3/5$.$4$. Since $\sin37^o = 3/5$,the angle of incidence $i = C$. Thus,the ray grazes the vertical surface.$5$. The ray is deviated by $90^o - i = 90^o - 37^o = 53^o$ at the vertical face.$6$. After emerging,the total deviation is $53^o$.

$A$ ray of light is incident normally on the diagonal face of a right-angled prism as shown in the figure. If $\theta = 37^o$ and the refractive index of the prism is $\mu = 5/3$,then the angle of deviation is......$^o$ $(\sin37^o = 3/5)$

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