A glass prism of refractive index 1.5 is imme | vedclass

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(C) $1$. The light beam is incident normally on face $AB$,so it enters the prism without deviation and strikes face $AC$.$2$. Let the angle of inciden

Vedclass · Accepted Answer

$\sin 	heta > \frac{8}{9}$

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(C) $1$. The light beam is incident normally on face $AB$,so it enters the prism without deviation and strikes face $AC$.$2$. Let the angle of incidence at face $AC$ be $i$. From the geometry of the prism,the angle between the normal to face $AC$ and the incident ray is equal to the angle $	heta$ of the prism at $A$. Thus,$i = 	heta$.$3$. For total internal reflection $(TIR)$ to occur at face $AC$,the angle of incidence must be greater than the critical angle $C$ for the glass-water interface.$4$. The condition for $TIR$ is $i > C$,which implies $\sin i > \sin C$.$5$. Since $i = 	heta$,we have $\sin 	heta > \sin C$.$6$. The critical angle $C$ is given by $\sin C = \frac{\mu_{	ext{water}}}{\mu_{	ext{glass}}} = \frac{4/3}{3/2} = \frac{4}{3} 	imes \frac{2}{3} = \frac{8}{9}$.$7$. Therefore,the condition for total internal reflection is $\sin 	heta > \frac{8}{9}$.

$A$ glass prism of refractive index $1.5$ is immersed in water (refractive index $\frac{4}{3}$) as shown in the figure. $A$ light beam incident normally on the face $AB$ is totally internally reflected at the face $AC$ to reach the face $BC$,if:

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