Question

A fish is in the water at a depth of 12 cm, the refractive index of water is 4/3, find the Area of the surface through which fish can see outside.

• A. $156c{m}^2$
• B. $575c{m}^2$
• C. $598c{m}^2$
• D. $582c{m}^2$

Answer 1

The correct option is D $582c{m}^2$

$\begin{array}{c}h=12cm\\ \mu =4/3\end{array}$

$\begin{array}{cc}\tan i& =\frac{r}{h}\\ r& =h\tan i\end{array}$ $r=h\frac{\sin i}{\cos i}$

$r=h\frac{\sin i}{\sqrt{1-{\sin }^{2}i}}-\left(i\right)$

$\begin{array}{c}\text{We know that}\\ \text{critical angle}\\ \sin C=\frac{1}{\mu }\sin i\end{array}$

$\sin i=\frac{3}{4}$

$\begin{array}{cc}& \text{putting in i}\\ & \begin{array}{cc}r& =\frac{12\times \frac{3}{4}}{\sqrt{1-\frac{9}{16}}}\\ r& =\frac{9\times 4}{\sqrt{7}}\end{array}\end{array}$

$\begin{array}{cc}\text{Area}=& \pi r^2\\ =& \frac{22}{7}\times \frac{9\times 4\times 9\times 4}{7}\\ & =582{cm}^2\end{array}$