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(C) The light patch is formed due to rays that undergo total internal reflection at the liquid-air interface. Rays incident at the critical angle $	h

Vedclass · Accepted Answer

$\frac{\pi h^2}{\left(\mu^2-1\right)}$

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(C) The light patch is formed due to rays that undergo total internal reflection at the liquid-air interface. Rays incident at the critical angle $	heta_C$ emerge at an angle of $90^\circ$ to the normal.From the geometry of the problem,we have $	an 	heta_C = \frac{r}{h}$.Using Snell's Law at the critical angle:$\mu \sin 	heta_C = 1 \cdot \sin 90^\circ = 1$$\sin 	heta_C = \frac{1}{\mu}$Since $\sin 	heta_C = \frac{1}{\mu}$,we can find $	an 	heta_C$ using the relation $	an 	heta_C = \frac{\sin 	heta_C}{\sqrt{1 - \sin^2 	heta_C}} = \frac{1/\mu}{\sqrt{1 - 1/\mu^2}} = \frac{1}{\sqrt{\mu^2 - 1}}$.Equating the two expressions for $	an 	heta_C$:$\frac{r}{h} = \frac{1}{\sqrt{\mu^2 - 1}}$$r = \frac{h}{\sqrt{\mu^2 - 1}}$The area of the circular patch is $A = \pi r^2 = \pi \left( \frac{h}{\sqrt{\mu^2 - 1}} ight)^2 = \frac{\pi h^2}{\mu^2 - 1}$.

Rays from a point source of light situated at height $h$ below the liquid surface having refractive index $\mu$,form a circular patch of light of radius $r$ on the surface. The area of the patch is

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