How Do You Calculate Pool Depth Using Refraction Principles?

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Homework Help Overview

The problem involves calculating the depth of a swimming pool using principles of refraction, specifically applying Snell's law. The scenario describes a pool that is 4.0 meters wide and filled to the top, with sunlight at an angle of 20 degrees above the horizon, leading to a question about the depth based on the shading of the pool's bottom.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply Snell's law to determine the angle of refraction and subsequently calculate the depth of the pool. They express confusion regarding the simplicity of their result and question the validity of their approach, particularly in relation to the shading of the pool's bottom.

Discussion Status

Some participants affirm the original poster's calculations up to a certain point, suggesting that the approach is fundamentally correct. There is a focus on clarifying the interpretation of the problem and ensuring proper unit usage. The discussion indicates a productive direction towards understanding the implications of the refraction and shading in the context of the problem.

Contextual Notes

Participants note a potential misunderstanding regarding unit conversion, specifically between meters and centimeters, which may have contributed to the original poster's confusion about the result. The discussion also highlights the importance of visualizing the scenario to better grasp the refraction principles at play.

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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?

Homework Equations


Snell's law
(n1)sin(x1) = (n2)sin(x2)
Index of Refraction of Air = 1
Index of Refraction of Water = 1.33

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:
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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?
 
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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