Question: A catfish is$\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$ below the surface of a smooth lake. (a) What is the diameter of the circle on the surface through which the fish can see the world outside the water? (b) If the fish descends, does the diameter of the circle increase, decrease, or remain the same?

The depth of the catfish in the smooth lake is $2.00\mathrm{m}$.

We can draw a ray diagram for the given problem. From the ray diagram and Snell’s law of refraction, we can get the formula for the diameter of the circle. Using the given values, we can get the answer for the diameter of the circle.

Formula:

${\mathrm{n}}_1{\mathrm{sin\theta }}_1={\mathrm{n}}_2{\mathrm{sin\theta }}_2$

From the above diagram, we can write that

$$\begin{gathered} \tan \left(90-{\mathrm{\theta }}_{\mathrm{c}}\right)=\frac{\mathrm{Depth}}{\mathrm{Radius}} \\ \mathrm{Diameter}=2\times \mathrm{Radius} \\ \tan \left(90-{\mathrm{\theta }}_{\mathrm{c}}\right)=\frac{\mathrm{Depth}}{\left({\displaystyle \frac{\mathrm{Diameter}}{2}}\right)} \\ {\mathrm{cot\theta }}_{\mathrm{c}}=\frac{\mathrm{Depth}}{\left({\displaystyle \frac{\mathrm{Diameter}}{2}}\right)} \\ {\mathrm{tan\theta }}_{\mathrm{c}}=\frac{\mathrm{Diameter}}{2\times \mathrm{Depth}} \\ \mathrm{Diameter}=2\times \mathrm{Depth}\times {\mathrm{tan\theta }}_{\mathrm{c}} \end{gathered}$$

We have,

${\mathrm{n}}_{\mathrm{w}}=1.33\mathrm{and}{\mathrm{n}}_{\mathrm{a}}=1$

Fromthefigure, we can write that

For air, the refractive angle will be${90}^{\circ }$.

So according to Snell’s law,

$$\begin{gathered} n_w\sin {\theta }_c=n_a\sin {\theta }_a \\ \sin {\theta }_c=\frac{n_a}{n_w}\sin {\theta }_a \\ \sin {\theta }_c=\frac{1}{1.33}\sin {90}^{\circ } \\ \sin {\theta }_c=0.75 \\ {\theta }_c={\sin }^{-1}0.75 \\ {\theta }_c=48.7^{\circ } \end{gathered}$$

$$\begin{gathered} {\mathrm{n}}_{\mathrm{w}}{\mathrm{sin\theta }}_{\mathrm{c}}={\mathrm{n}}_{\mathrm{a}}{\mathrm{sin\theta }}_{\mathrm{a}} \\ {\mathrm{sin\theta }}_{\mathrm{c}}=\frac{{\mathrm{n}}_{\mathrm{a}}}{{\mathrm{n}}_{\mathrm{w}}}{\mathrm{sin\theta }}_{\mathrm{a}} \\ {\mathrm{sin\theta }}_{\mathrm{c}}=\frac{1}{1.33}\times \sin {90}^{\circ } \\ {\mathrm{sin\theta }}_{\mathrm{c}}=0.75 \\ {\mathrm{\theta }}_{\mathrm{c}}={\sin }^{-1}0.75 \\ {\mathrm{\theta }}_{\mathrm{c}}=48.7^{\circ } \end{gathered}$$

We have

$$\begin{gathered} \mathrm{Diameter}\mathrm{of}\mathrm{Circle}=2\times \mathrm{Depth}\mathrm{of}\mathrm{Catfish}\times {\mathrm{tan\theta }}_{\mathrm{c}} \\ \mathrm{Diameter}\mathrm{of}\mathrm{Circle}=2\times 2.00\mathrm{m}\times \tan 48.7^{\circ } \\ \mathrm{Diameter}\mathrm{of}\mathrm{Circle}=4.56\mathrm{m} \end{gathered}$$

We have

$\mathrm{Diameter}\mathrm{of}\mathrm{circle}=2\times \mathrm{Depth}\times \mathrm{tan\theta }$

From the above equation, we can write that

$\mathrm{Diameter}\mathrm{of}\mathrm{circle}\propto \mathrm{Depth}$

From this, we conclude that if the depth at which the catfish is present below the water surface increases, the diameter of the circle through which the fish can see the outside world also increases.