A ray of light incident normally on an isosce | vedclass

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(A) From the figure,it is clear that Total Internal Reflection $(TIR)$ takes place at the surfaces $AC$ and $BC$.For $TIR$ to occur,the angle of incid

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$\sqrt{2}$

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(A) From the figure,it is clear that Total Internal Reflection $(TIR)$ takes place at the surfaces $AC$ and $BC$.For $TIR$ to occur,the angle of incidence $i$ must be greater than the critical angle $C$.Here,the angle of incidence at the surfaces is $45^{\circ}$.Therefore,$45^{\circ} > C$.Taking the sine on both sides,$\sin(45^{\circ}) > \sin(C)$.Since $\sin(C) = \frac{1}{\mu}$,we have $\frac{1}{\sqrt{2}} > \frac{1}{\mu}$.This implies $\mu > \sqrt{2}$.Hence,the least value of the refractive index is $\mu_{least} = \sqrt{2}$.

$A$ ray of light incident normally on an isosceles right-angled prism travels as shown in the figure. The least value of the refractive index of the prism must be

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