Light enters at an angle of incidence alpha i | vedclass

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Question

(A) Let $\alpha$ be the angle of incidence at the end face of the rod and $r$ be the angle of refraction. By Snell's Law,$1 \cdot \sin \alpha = n \cdo

Vedclass · Accepted Answer

$n > \sqrt{2}$

Vedclass · Answer

(A) Let $\alpha$ be the angle of incidence at the end face of the rod and $r$ be the angle of refraction. By Snell's Law,$1 \cdot \sin \alpha = n \cdot \sin r$,so $\sin r = \frac{\sin \alpha}{n}$.At the lateral surface,the angle of incidence is $i = 90^\circ - r$. For total internal reflection to occur,the angle of incidence $i$ must be greater than the critical angle $C$,where $\sin C = \frac{1}{n}$.Thus,$i > C \implies \sin i > \sin C$.Substituting $i = 90^\circ - r$,we get $\sin(90^\circ - r) > \sin C$,which simplifies to $\cos r > \frac{1}{n}$.Squaring both sides,$\cos^2 r > \frac{1}{n^2} \implies 1 - \sin^2 r > \frac{1}{n^2}$.Substituting $\sin r = \frac{\sin \alpha}{n}$,we have $1 - \frac{\sin^2 \alpha}{n^2} > \frac{1}{n^2}$.Rearranging gives $1 > \frac{1 + \sin^2 \alpha}{n^2}$,or $n^2 > 1 + \sin^2 \alpha$.Since this must hold for any angle of incidence $\alpha$,we consider the maximum value of $\sin^2 \alpha$,which is $1$ (when $\alpha = 90^\circ$).Therefore,$n^2 > 1 + 1 = 2$,which means $n > \sqrt{2}$.

Light enters at an angle of incidence $\alpha$ into a transparent rod of refractive index $n.$ For what value of the refractive index of the material of the rod will the light,once entered into it,not leave it through its lateral face,regardless of the value of the angle of incidence $\alpha$?

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