91影视

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鈥檚 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:

${\mathrm{n}}_1{\mathrm{蝉颈苍胃}}_1={\mathrm{n}}_2{\mathrm{蝉颈苍胃}}_2$

From the above diagram, we can write that

$$\begin{gathered} \tan \left(90-{\mathrm{胃}}_{\mathrm{c}}\right)=\frac{\mathrm{Depth}}{\mathrm{Radius}} \\ \mathrm{Diameter}=2脳\mathrm{Radius} \\ \tan \left(90-{\mathrm{胃}}_{\mathrm{c}}\right)=\frac{\mathrm{Depth}}{\left({\displaystyle \frac{\mathrm{Diameter}}{2}}\right)} \\ {\mathrm{肠辞迟胃}}_{\mathrm{c}}=\frac{\mathrm{Depth}}{\left({\displaystyle \frac{\mathrm{Diameter}}{2}}\right)} \\ {\mathrm{迟补苍胃}}_{\mathrm{c}}=\frac{\mathrm{Diameter}}{2脳\mathrm{Depth}} \\ \mathrm{Diameter}=2脳\mathrm{Depth}脳{\mathrm{迟补苍胃}}_{\mathrm{c}} \end{gathered}$$

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${90}^{鈭�}$.

So according to Snell鈥檚 law,

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

$$\begin{gathered} {\mathrm{n}}_{\mathrm{w}}{\mathrm{蝉颈苍胃}}_{\mathrm{c}}={\mathrm{n}}_{\mathrm{a}}{\mathrm{蝉颈苍胃}}_{\mathrm{a}} \\ {\mathrm{蝉颈苍胃}}_{\mathrm{c}}=\frac{{\mathrm{n}}_{\mathrm{a}}}{{\mathrm{n}}_{\mathrm{w}}}{\mathrm{蝉颈苍胃}}_{\mathrm{a}} \\ {\mathrm{蝉颈苍胃}}_{\mathrm{c}}=\frac{1}{1.33}脳\sin {90}^{鈭�} \\ {\mathrm{蝉颈苍胃}}_{\mathrm{c}}=0.75 \\ {\mathrm{胃}}_{\mathrm{c}}={\sin }^{-1}0.75 \\ {\mathrm{胃}}_{\mathrm{c}}=48.7^{鈭�} \end{gathered}$$

We have

$$\begin{gathered} \mathrm{Diameter}\mathrm{of}\mathrm{Circle}=2脳\mathrm{Depth}\mathrm{of}\mathrm{Catfish}脳{\mathrm{迟补苍胃}}_{\mathrm{c}} \\ \mathrm{Diameter}\mathrm{of}\mathrm{Circle}=2脳2.00\mathrm{m}脳\tan 48.7^{鈭�} \\ \mathrm{Diameter}\mathrm{of}\mathrm{Circle}=4.56\mathrm{m} \end{gathered}$$

We have

$\mathrm{Diameter}\mathrm{of}\mathrm{circle}=2脳\mathrm{Depth}脳\mathrm{迟补苍胃}$

From the above equation, we can write that

$\mathrm{Diameter}\mathrm{of}\mathrm{circle}鈭�\mathrm{Depth}$

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.