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Refraction in a Swimming Pool.

1. Homework Statement
A 4.0-m-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. Homework Equations
Snell's law
(n1)sin(x1) = (n2)sin(x2)
Index of Refraction of Air = 1
Index of Refraction of Water = 1.33


3. The Attempt at a Solution

I drew a picture with the sun 20 degrees above the horizontal, which meant that x1 is 70 degrees relative to the normal. I know n1 is the index of refraction of Air, 1. I solved for x2, using n2 = index of refraction of water = 1.33.

(n1)sin(x1) = (n2)sin(x2)
(1)sin(70) = (1.33)sin(x2)

I solved for x2 and got 45 degrees for the light angle in water relative to the normal.

Now that I know x2, I tried to solve for the depth of the pool d by using the width w = 4.0 m.

I got tan(x2) = (4 m)/d
d = (4 m)/tan(45) = 4 m.

I don't think my result makes sense. It seems too simple. How am I supposed to figure out the depth of the pool? I've never experienced a problem where the bottom was completely shaded. Can someone please explain to me what I'm doing incorrectly? Thanks for your help.
 
Last edited:

Answers and Replies

Doc Al
Mentor
44,827
1,083
You have solved the problem correctly--whether you know it or not. (Correct your units--it's meters, not centimeters.)

Imagine the sun is coming from the right, hitting the water. The light refracts in accordance with Snell's law. Now consider the light that just hits the right-most section of the pool water. If that light bends enough so that it hits the wall of the pool, not the bottom, then surely none of the other light will shine on the pool bottom. (Draw yourself a picture and you'll understand it better.) Make sense?
 
286
0
edit: Doc Al beat me to it.
-------------
Everything is fine, all the way to (4m)/tan(45 degrees)
How'd you get 4 centimeters?? tan(45 degrees) = 1

Otherwise, everything is good.
 
Thanks I think I understand it now. I just messed up on the units.
 

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