Answer:
a) 404 m² b) apparent height = 7.5 m
Explanation:
This question is about refraction and total internal refraction.
Here I will take refractive index of air and water
[tex]n_{air}=1\\ n_{water}=1.33=4/3[/tex]
Now let's look at the diagram I have attached here
At some angle A, the light from the ring (yellow point) under water will be totally internally refracted (B = 90°), which means that rays of light (yellow arrow) that make large enough angle A will not be able to escape from the water. Since we assumed that the ring is a point, there will be a critical cone of angle A with the ring at its apex which traces a circle of radius R on the surface of water, which, beyond this radius, no light could escape.
According to snell's law
[tex]\frac{sin(B)}{sin(A)} = \frac{n_{water}}{n_{air}} = 4/3[/tex]
At critical angle B = 90°
[tex]\frac{3)}{4}sin(B) = [tex]\frac{3}{4} sin(90^\circ ) = 0.75 = sin(A)[/tex]
Therefore
[tex]A = 48.6^\circ[/tex]
With this, we can find the radius of the circle (refer to my diagram)
[tex]h* tan (A) = R\\R =11.3 m [/tex]
And with that we can find the area
[tex]A = \pi R^2=404\ m^2[/tex]
Additional Problem
For apparent depth from above, we can think that, since we are accustomed to seeing light at the speed of c in air, our brain interpret light from any source to be traveling at c. This causes light that originated under water, which has the speed of
[tex]v_{water} = \frac{c}{n_{water}} = 0.75c[/tex]
to appear as if it has traveled with the same duration as light with speed c
In order for this to happen our brain perceive shortened length which is the apparent depth.
To put it in mathematical term
[tex]t_{travel}=\frac{h_{apparent}}{v_{water}} =\frac{h}{c}[/tex]
So we get apparent depth
[tex]h_{apparent}=0.75h = 7.5\ m[/tex]
