91Ó°ÊÓ

Depth of the source with respect to surface$=80cm=0.8m$

We can draw a ray diagram for the given problem. From the ray diagram, we can get the formula for the diameter of the circle. The value of critical angle can be found by using Snell’s law of refraction. Inserting it in that formula will give the diameter of the circle at the surface through which light emerges from the water.

Ray Diagram for the situation:

From the above diagram, we can say that

$\begin{array}{c}tan(90-{\mathrm{θ}}_C)=\frac{Depth}{RadiusofCircle}\\ DiameterofCircle=2×radiusofcircle\\ RadiusofCircle=\frac{DiameterofCircle}{2}\\ tan(90-{\mathrm{θ}}_C)=\frac{Depth}{\left(\frac{DiameterofCircle}{2}\right)}\end{array}$

$\begin{array}{c}cot{\mathrm{θ}}_C=\frac{Depth}{\left(\frac{DiameterofCircle}{2}\right)}\\ tan{\mathrm{θ}}_C=\frac{\left(\frac{DiameterofCircle}{2}\right)}{Depth}\\ tan{\mathrm{θ}}_C=\frac{DiameterofCircle}{2×Depth}\\ DiameterofCircle=2×Depth×tan{\mathrm{θ}}_C\end{array}$

We have

$n_w=1.33$And$n_a=1$

From the figure, we can say that for air, the refractive angle is.

So according to Snell’s Law,

$\begin{array}{c}{n}_{1}sin{\mathrm{θ}}_1=n_{2}sin{\mathrm{θ}}_2\\ n_{w}sin{\mathrm{θ}}_C=n_{a}sin{\mathrm{θ}}_a\\ sin{\mathrm{θ}}_C=\frac{n_a}{n_w}sin{\mathrm{θ}}_a\\ sin{\mathrm{θ}}_C=\frac{1}{1.33}×sin90°\end{array}$

$\begin{array}{c}sin{\mathrm{θ}}_C=0.75\\ {\mathrm{θ}}_C=48.7°\end{array}$

We have

$\begin{array}{l}DiameterofCircle=2×DepthoftheSource×tan{\mathrm{θ}}_C\\ DiameterofCircle=2×0.8×tan(48.7)\\ DiameterofCircle=1.82m\end{array}$