A glass prism (mu = 1.5) is dipped in water ( | vedclass

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(A) For total internal reflection $(TIR)$ to occur at the surface $AC$,the angle of incidence $i$ at the surface $AC$ must be greater than the critica

Vedclass · Accepted Answer

$\sin 	heta \ge 8/9$

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(A) For total internal reflection $(TIR)$ to occur at the surface $AC$,the angle of incidence $i$ at the surface $AC$ must be greater than the critical angle $C$ between glass and water.From the geometry of the prism,the light ray incident normally on $AB$ strikes $AC$ at an angle of incidence $i = 	heta$.For $TIR$ at $AC$,we require $i > C$,which implies $\sin i > \sin C$.Substituting $i = 	heta$,we get $\sin 	heta > \sin C$.The critical angle $C$ is given by $\sin C = \frac{\mu_w}{\mu_g}$,where $\mu_w = 4/3$ and $\mu_g = 1.5 = 3/2$.Thus,$\sin C = \frac{4/3}{3/2} = \frac{4}{3} 	imes \frac{2}{3} = \frac{8}{9}$.Therefore,for $TIR$ to occur,$\sin 	heta > 8/9$. Given the options,the condition is $\sin 	heta \ge 8/9$.

$A$ glass prism $(\mu = 1.5)$ is dipped in water $(_a\mu_w = 4/3)$ as shown in the figure. $A$ light ray is incident normally on the surface $AB$. It reaches the surface $BC$ after being totally internally reflected at the surface $AC$,if:

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