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(B) For the light source to be invisible from above,the light rays emitted from the source must undergo total internal reflection at the water-air int

Vedclass · Accepted Answer

$\frac{h}{\sqrt{\mu^2-1}}$

Vedclass · Answer

(B) For the light source to be invisible from above,the light rays emitted from the source must undergo total internal reflection at the water-air interface.Let $r$ be the radius of the opaque disc. The light rays reaching the edge of the disc must strike the surface at an angle equal to the critical angle $i_c$.From the geometry of the problem,$	an i_c = \frac{r}{h}$.We know that for total internal reflection,$\sin i_c = \frac{1}{\mu}$.Using the trigonometric identity $	an i_c = \frac{\sin i_c}{\sqrt{1-\sin^2 i_c}}$,we get:$	an i_c = \frac{1/\mu}{\sqrt{1-(1/\mu)^2}} = \frac{1/\mu}{\sqrt{(\mu^2-1)/\mu^2}} = \frac{1}{\sqrt{\mu^2-1}}$.Equating the two expressions for $	an i_c$:$\frac{r}{h} = \frac{1}{\sqrt{\mu^2-1}}$Therefore,$r = \frac{h}{\sqrt{\mu^2-1}}$.

$A$ point source of light is placed at the bottom of a vessel,which is filled with water of refractive index $\mu$ to a height $h$. If a floating opaque disc has to be placed exactly above it so that the source is invisible from above,the radius of the disc should be:

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