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(C) For total internal reflection,the angle of incidence $i$ must be greater than the critical angle $C$,i.e.,$i > C$.From the figure,the ray strikes

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$n > \sqrt{2}$

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(C) For total internal reflection,the angle of incidence $i$ must be greater than the critical angle $C$,i.e.,$i > C$.From the figure,the ray strikes the hypotenuse at an angle of incidence $i = 	heta = 45^o$.For total internal reflection to occur,we must have $	heta > C$.Taking the sine on both sides,$\sin 	heta > \sin C$.Since $\sin C = \frac{1}{n}$,where $n$ is the refractive index of the glass relative to air,we have $\sin 	heta > \frac{1}{n}$.Rearranging for $n$,we get $n > \frac{1}{\sin 	heta}$.Substituting $	heta = 45^o$,we get $n > \frac{1}{\sin 45^o} = \frac{1}{1/\sqrt{2}} = \sqrt{2}$.Thus,the refractive index of the glass must be $n > \sqrt{2}$.

$A$ triangular prism of glass is shown in the figure. $A$ ray incident normally to one face is totally reflected,if $\theta = 45^o$. The index of refraction of glass is

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