There is a point source of light at a depth 8m below the surface of water.  An observer in air sees the image of the source.  Line of sight of observer makes an angle  $53°$ With the normal to surface. (use:  $\sqrt{7}=2.65 and\mu =$ of water=4/3)

$$\begin{gathered} \tan i=\frac{x}{y},x=y\tan i,dx=y{\sec }^{2}idi.\tan r=\frac{x\text{'}}{y\text{'}} \\ x\text{'}=y\text{'}\tan r.d{x}^{\text{'}}=y\text{'}{\sec }^{2}rdr.dx=dx\text{'}\Rightarrow y{\sec }^{2}idi=y\text{'}{\sec }^{2}rdr \\ {y}^{\text{'}}=y\frac{{\sec }^{2}i}{{\sec }^{2}r}\frac{di}{dr}=y\frac{{\cos }^{2}r}{{\cos }^{2}i}\frac{di}{dr} \_\_\_\_\_\_\left(i\right) \end{gathered}$$

From snell’s Law:
 $\mu \sin i=\sin r\mu \cos idi=\cos rdr\frac{di}{dr}=\frac{\cos r}{\cos i}\frac{1}{\mu }$
So from eq. (i)
$y\text{'}=\frac{y}{\mu }\left(\frac{{\cos }^{3}r}{{\cos }^{3}i}\right)For\left(P\right)y\text{'}=\frac{8}{4}\left(\frac{{\cos }^3{53}^{°}}{{\cos }^{3}37°}\right)=2.53$

For (Q) x’-x=y’tan r-y tan i
For (R)  $\sin c=\frac{1}{\mu }=\frac{3}{4}\tan c=\frac{r}{y}\Rightarrow r=y\tan c=8\times \frac{3}{\sqrt{7}}$

$For \left(S\right)p_{out}=\frac{P}{4\pi }2\pi (1-\cos C)=\frac{P}{2}(1-\frac{\sqrt{7}}{4})\frac{P_{out}}{P}\times 100=50(1-\frac{\sqrt{7}}{4})$