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(B) For total internal reflection to occur,the angle of incidence $i$ must be greater than the critical angle $i_c$.Given that the angle of incidence

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$n > \sqrt{2}$

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(B) For total internal reflection to occur,the angle of incidence $i$ must be greater than the critical angle $i_c$.Given that the angle of incidence $i = 45^{\circ}$,we have $i > i_c$,which implies $45^{\circ} > i_c$.Taking the sine of both sides,we get $\sin(45^{\circ}) > \sin(i_c)$.We know that $\sin(i_c) = \frac{1}{n}$,where $n$ is the refractive index of the prism material.Substituting this into the inequality,we get $\frac{1}{\sqrt{2}} > \frac{1}{n}$.Taking the reciprocal of both sides reverses the inequality sign,resulting in $\sqrt{2}  \sqrt{2}$.

$A$ ray of light is incident on one face of a $90^{\circ}$ prism and undergoes total internal reflection at the glass-air interface. If the angle of incidence is $45^{\circ}$,then the refractive index $n$ is:

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