A light ray falls on a rectangular glass slab | vedclass

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(A) Let the refractive index of the glass slab be $\mu$. The angle of incidence at point $A$ is $i = 45^\circ$. Let the angle of refraction at $A$ be

Vedclass · Accepted Answer

$\sqrt{\frac{3}{2}}$

Vedclass · Answer

(A) Let the refractive index of the glass slab be $\mu$. The angle of incidence at point $A$ is $i = 45^\circ$. Let the angle of refraction at $A$ be $r$. By Snell's Law,$1 \cdot \sin(45^\circ) = \mu \cdot \sin(r)$,so $\sin(r) = \frac{1}{\mu\sqrt{2}}$.At point $B$,the angle of incidence is $r' = 90^\circ - r$. For total internal reflection to occur,the angle of incidence must be greater than or equal to the critical angle $C$,where $\sin(C) = \frac{1}{\mu}$.Thus,$\sin(90^\circ - r) \ge \frac{1}{\mu}$,which implies $\cos(r) \ge \frac{1}{\mu}$.Squaring both sides,$\cos^2(r) \ge \frac{1}{\mu^2}$,so $1 - \sin^2(r) \ge \frac{1}{\mu^2}$.Substituting $\sin^2(r) = \frac{1}{2\mu^2}$,we get $1 - \frac{1}{2\mu^2} \ge \frac{1}{\mu^2}$,which simplifies to $1 \ge \frac{3}{2\mu^2}$.Thus,$\mu^2 \ge \frac{3}{2}$,or $\mu \ge \sqrt{\frac{3}{2}}$. The minimum refractive index is $\sqrt{\frac{3}{2}}$.

$A$ light ray falls on a rectangular glass slab as shown in the figure. If total internal reflection occurs at the vertical face of the slab at point $B$,the refractive index of glass is

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