Solve Depth of Pool w/ Snell's Law

In summary, using the Pythagorean theorem and Snell's law, it was determined that the depth of the pool is approximately 4 meters.
  • #1
BadatPhysicsguy
39
0

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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  • #2
Angles of incidence and refraction in Snell's law are measured from the normal to the refracting surface.
 
  • #3
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?
 
  • #4
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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  • #5
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?
 
  • #6
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.
 
  • #7
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.
 
  • #8
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.
 

FAQ: Solve Depth of Pool w/ Snell's Law

1. What is Snell's Law and how does it relate to the depth of a pool?

Snell's Law is a principle in physics that describes the relationship between the angle of incidence and refraction when light travels from one medium to another. In the context of a pool, Snell's Law can be used to calculate the depth of the water based on the angle at which light refracts from the surface.

2. How is Snell's Law used to solve for the depth of a pool?

To solve for the depth of a pool using Snell's Law, you need to know the angle of incidence of light from the surface of the water and the refractive index of the water. With this information, you can use the formula n1sinθ1 = n2sinθ2, where n1 and θ1 represent the refractive index and angle of incidence in air, and n2 and θ2 represent the refractive index and angle of refraction in water. By rearranging the formula, you can solve for the depth of the pool.

3. What factors can affect the accuracy of using Snell's Law to calculate the depth of a pool?

The accuracy of using Snell's Law to calculate the depth of a pool can be affected by several factors such as the presence of impurities in the water, the temperature of the water, and the curvature of the pool's surface. These factors can alter the refractive index of the water and the angle of refraction, leading to a less accurate calculation of the pool's depth.

4. Can Snell's Law be used to determine the depth of any body of water?

Yes, Snell's Law can be used to determine the depth of any body of water as long as the refractive index and angle of incidence of light can be measured or estimated. However, it is important to note that the accuracy of the calculation may vary depending on the factors mentioned in the previous question.

5. Are there any other methods for determining the depth of a pool besides using Snell's Law?

Yes, there are other methods for determining the depth of a pool, such as using a depth sounder or physically measuring the depth with a measuring tape. These methods may be more accurate and reliable, especially in cases where the water is not clear or the surface of the pool is not flat.

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