Question

A swimmer is 10 m below the surface of a pool. an observer is 10 m above the pool on the diving platform. The observer shines a flashlight that emits light that has an angular spread of 120 degrees. If the flashlight is shined directly at the surface,

1. What is the diameter of the lighted area at the surface as seen by the observer?

2. What is the diameter of the lighted area at the surface as seen by the swimmer under water?

3. What is the angular spread of the flashlight under the water?

Answer 1

1) so diameter of lighted region is equal to the arc made by flash light on pool surface .so

So r = 10m

And

So Diameter

2) Now while light travel in water angular spread decrease by factor of refractive index beacuse when light travels in rarer to denser medium it bends toward normal so,

Here n is refractive index

And light travel more optical path in water which is then

So then diameter now again

Same as previous.

3)so under the water angular spread is now

Water n = 4/3

So,