# Ray Refraction: Finding the depth of a pool

1. Feb 16, 2009

### KEØM

1. The problem statement, all variables and given/known data
A 4.0m wide swimming pool is filled to the top. The bottom of the pool becomes completely
shaded in the afternoon when the sun is 20 degrees above the horizon. How deep is the pool?

2. Relevant equations
$$\theta$$ = arcsin(n2/n1)
n1sin$$\theta$$1= n2sin$$\theta$$2

3. The attempt at a solution
The main thing that I don't understand about this problem is that if the bottom of the pool
is shaded then that means light is not reaching the bottom right? I have only seen problems where something under the surface emitting light has total internal reflection.
Could somebody please explain this to me.

2. Feb 16, 2009

### Delphi51

Sketch a ray coming up from the far end of the bottom of the pool to the surface at the near end, refracting (angle change) at the surface and heading off at an angle of 20 degrees above the water/ground. Beginning with the 20 degrees and using the Law of Refraction (Snell's Law) you can deduce the angle in the water and then the depth.

3. Feb 17, 2009

### KEØM

Ok, so using snells law I found the angle to be 45 degrees and then using trigonometry I found the height to be 4m. Is this right? I don't have the answer to the problem so I don't know.

Thanks a bunch for your help delphi.