Solve Depth of Pool w/ Snell's Law

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Homework Help Overview

The problem involves calculating the depth of a pool based on the visibility of a stone at the bottom, using Snell's Law to relate angles of incidence and refraction. The scenario describes a person standing a certain distance from the pool and observing half of the stone submerged in water.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the angles involved in Snell's Law and the geometry of the situation, questioning the assumption of a 45-degree angle based on visibility. There are attempts to calculate angles and dimensions using trigonometric relationships, with some participants expressing confusion about the setup.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the angles involved and the implications for calculating the pool's depth. Some guidance has been offered regarding the application of Snell's Law and trigonometric principles, but no consensus has been reached.

Contextual Notes

There are indications of confusion regarding the measurement of angles from the normal to the refracting surface and the implications of visibility on angle assumptions. Participants are also navigating the challenge of having multiple unknowns in their calculations.

BadatPhysicsguy
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Homework Statement


MGxXqsJ.png

A stone lies to the very edge at the bottom of a pool. The pool is filled with water to the top. The person standing three meters away from the pool is 1 meter tall and he can see exactly the half of the stone. Calculate the depth of the pool.

Homework Equations


Snell's law: n * sin(x) = n2 * sin(y)

The Attempt at a Solution


QHBwIM4.png

Okay.. Since I can see half of the stone, the angle should be 45 degrees, right? I should be able to calculate x degrees and therefore the bottom of the triangle (where he stands). With the bottom, I can calculate the bottom line of the triangle in the pool and then the depth. So I do that.

sin(45) * 1.33 = 1.00 * sin(x)
1.33 is n for water. 1 is n for air. x = 70.05 degrees.

So the inside of the triangle is approx 20 degrees. I use tan.

tan 70 = 1/adjacent => 1/(tan 70) = adjacent => 0.36397. So the bottom of that triangle is 0.36397 meters, but that doesn't make sense, it should be atleast 3 meters because he is standing that far away from the pool. What am I doing wrong?
 
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Angles of incidence and refraction in Snell's law are measured from the normal to the refracting surface.
 
Bystander said:
Angles of incidence and refraction in Snell's law are measured from the normal to the refracting surface.
Hello! Thank you for your answer. I did draw a normal perpendicular to the water's surface. But if is measured from the normal to the refracting surface aren't both angles 90 degrees then?
 
BadatPhysicsguy said:

Homework Statement


[ IMG]http://i.imgur.com/MGxXqsJ.png[/PLAIN]
A stone lies to the very edge at the bottom of a pool. The pool is filled with water to the top. The person standing three meters away from the pool is 1 meter tall and he can see exactly the half of the stone. Calculate the depth of the pool.

Homework Equations


Snell's law: n * sin(x) = n2 * sin(y)

The Attempt at a Solution


[ IMG]http://i.imgur.com/QHBwIM4.png[/PLAIN]
Okay.. Since I can see half of the stone, the angle should be 45 degrees, right? I should be able to calculate x degrees and therefore the bottom of the triangle (where he stands). With the bottom, I can calculate the bottom line of the triangle in the pool and then the depth. So I do that.

sin(45) * 1.33 = 1.00 * sin(x)
1.33 is n for water. 1 is n for air. x = 70.05 degrees.

So the inside of the triangle is approx 20 degrees. I use tan.

tan 70 = 1/adjacent => 1/(tan 70) = adjacent => 0.36397. So the bottom of that triangle is 0.36397 meters, but that doesn't make sense, it should be at least 3 meters because he is standing that far away from the pool. What am I doing wrong?
Seeing half the stone has nothing to do with a 45° angle.

Basically, a ray from the center of the stone should just graze the edge of the pool as it exits the water, then continue on to the eye.
 

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SammyS said:
Seeing half the stone has nothing to do with a 45° angle.

Basically, a ray from the center of the stone should just graze the edge of the pool as it exits the water, then continue on to the eye.

Hello and thank you for your answer.

But if the ray comes from the very center of the stone and it is truly in the edge of the pool, shouldn't it be "like" 45 degrees from the pool's wall down to the center of the stone?

And if the ray just grazes the edge of the pool, I then get this:
QeeP5gu.png


which would make by original attempt not work. Basically the only thing I need now is either x or y (degrees) and with it I can calculate the side (depth). But using Snell's law, now I only know the n's, not the angles. So I have two unknowns. How should I proceed?
 
BadatPhysicsguy said:
Hello and thank you for your answer.

But if the ray comes from the very center of the stone and it is truly in the edge of the pool, shouldn't it be "like" 45 degrees from the pool's wall down to the center of the stone?

And if the ray just grazes the edge of the pool, I then get this:
QeeP5gu.png


which would make by original attempt not work. Basically the only thing I need now is either x or y (degrees) and with it I can calculate the side (depth). But using Snell's law, now I only know the n's, not the angles. So I have two unknowns. How should I proceed?
Do you know trigonometry?

You can get sin(z) by using the Pythagorean theorem. Otherwise, use a scientific calculator to find the angle z

Then use Snell's Law , etc.
 
SammyS said:
Do you know trigonometry?

You can get sin(z) by using the Pythagorean theorem. Otherwise, use a scientific calculator to find the angle z

Then use Snell's Law , etc.
Hello, yes I forgot that! I calculated z degrees and then got x to be 45 degrees (making y also 45). So the depth is the same as the base of the triangle, approx 4 meters.
 
BadatPhysicsguy said:
Hello, yes I forgot that! I calculated z degrees and then got x to be 45 degrees (making y also 45). So the depth is the same as the base of the triangle, approx 4 meters.
Yes, approximately 45° , so approximately 4 meters deep.
 

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