All lights are switched off, except for a bright point-light source kept at the bottom of a swimming pool filled with clear water of refractive index $4 / 3$. As a result, only a circular patch of $6 \mathrm{~m}$ diameter of the water surface is visible to spectators standing around the swimming pool. Which of the following gives the nearest value of the depth of the pool?

• A

  $1.6 \mathrm{~m}$
• B

  $2.0 \mathrm{~m}$
• Correct answer
  C

  $2.6 \mathrm{~m}$
• D

  $3.0 \mathrm{~m}$

Solution

For the given situation, the ray diagram is shown below.

Refractive index of water, $\mu_{w}=\frac{4}{3}$ $$ r=\frac{D}{2}=\frac{6}{2}=3 \mathrm{~m} $$ The given situation is possible only for the case of total internal reflection. If $i_{C}$ be the critical angle, then $\sin i_{C}=\frac{1}{\mu_{w}}$ $\Rightarrow \quad \sin i_{C}=\frac{1}{4 / 3}$ $\Rightarrow \quad \sin i_{C}=\frac{3}{4}$ $\Rightarrow \quad \frac{r}{\sqrt{r^{2}+h^{2}}}=\frac{3}{4} \quad\left[\because \sin i_{C}=\frac{r}{\sqrt{r^{2}+h^{2}}}\right]$ $\Rightarrow \quad \frac{3}{\sqrt{3^{2}+h^{2}}}=\frac{3}{4}$ $\Rightarrow \quad 9+h^{2}=16$ $\Rightarrow \quad h^{2}=7, h=\sqrt{7}=2.6 \mathrm{~m}$

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