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-

• A

  $\frac{\mathrm{h}}{\sqrt{\mu-1}}$
• Correct answer
  B

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

  $\frac{\mathrm{h}}{\mu^{2}-1}$
• D

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

Solution

r should be such that rays beyond it got totally internally reflected For this $\theta>C$ or $\sin \theta>\sin C$ also $\mu=\frac{1}{\sin C} \therefore \frac{r}{\sqrt{h^{2}+r^{2}}}>\frac{1}{\mu}$ In limiting case $\frac{\mathrm{r}}{\sqrt{\mathrm{h}^{2}+\mathrm{r}^{2}}}=\frac{1}{\mu}$ ving we get $=\frac{\mathrm{h}}{\sqrt{\mu^{2}-1}}$

Hence, the correct answer is option: 'B'.