Solving the Pool Depth with Refraction in Water

[jegues]

Homework Statement

A child is standing at the edge of a swimming pool and looks straight down into the pool, where he sees a coin resting on the bottom. The coin appears to be 2.0m directly below the water's surface. How deep is the pool at the location of the coin? The index of refraction of water is 1.33.

Homework Equations

The Attempt at a Solution

See figure attached for my attempt.

Still not sure how to finish this one off.

 

[Delphi51]

I don't think you can assume x is the same for both object and image.
It is important to use the n=1.33, so you need to consider the rays and angles in the air. It really takes two rays from the object and their refracted rays to determine the location of the image. Unfortunately you aren't given the angles of two rays.

Perhaps that gives you freedom to choose two rays arriving at the eye and appearing to converge at an image 2 m beneath the surface. If so, you can use the refraction formula to calculate the two rays in the water and determine the point where they converge (the location of the coin).

 

[jegues]

I don't think you can assume x is the same for both object and image.
It is important to use the n=1.33, so you need to consider the rays and angles in the air. It really takes two rays from the object and their refracted rays to determine the location of the image. Unfortunately you aren't given the angles of two rays.

Perhaps that gives you freedom to choose two rays arriving at the eye and appearing to converge at an image 2 m beneath the surface. If so, you can use the refraction formula to calculate the two rays in the water and determine the point where they converge (the location of the coin).

What refraction formulae would that be?

 

[Delphi51]

n1*sin(θ1) = n2*sin(θ2) controls the bending of light at the surface between the water and air. Law of Refraction or Snell's Law.

 

[jegues]

n1*sin(θ1) = n2*sin(θ2) controls the bending of light at the surface between the water and air. Law of Refraction or Snell's Law.

Okay, but how do we get θ1 and θ2? We aren't given any other information so I'm not seeing how it can be done.

 

[Delphi51]

I just reread the question and noticed that the person is looking "straight down" into the water. That makes it easier. The two rays coming up from the coin are symmetric, so only one need be calculated. Looks like this:

Of course the angles are too large for an ordinary eye to catch the rays, but that doesn't matter. I think you can use any angle you like (answer won't change) for θ2, calculate the corresponding θ1 and then the depth of the coin given the image is 2 m down.

 

[jegues]

I just reread the question and noticed that the person is looking "straight down" into the water. That makes it easier. The two rays coming up from the coin are symmetric, so only one need be calculated. Looks like this:

Of course the angles are too large for an ordinary eye to catch the rays, but that doesn't matter. I think you can use any angle you like (answer won't change) for θ2, calculate the corresponding θ1 and then the depth of the coin given the image is 2 m down.

Thanks for the picture delphi, that made the situation a lot more clear!

Anyways here's how I'm thinking going about solving for the actual depth.

See figure attached.

I made an angle called θ3 which is simply 90 - θ2.

This angle should be that angle that is inside the triangle of the image. (See first triangle in figure)

From here I can calculate x.

But here's the part I'm not sure about.

Can I assume that x stays the same? because if so then I will take this x to my triangle containing the actual depth (h) and solve for h.

If I can't work with this assumption, I'm stumped as to how to solve it.

Any ideas?

 

[Delphi51]

So x is the distance from the center vertical line to the point where θ1 and θ2 have their vertex? Yes, that horizontal distance will be the same for the object and the image. Your approach looks good - carry on!

This looking straight down situation is not what I was thinking of in my first post based on your diagram. When looking not straight down, the horizontal distance to the image is not be the same as the horizontal distance to the object. Example here: http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/refr2.html#c1

 

[SammyS]

The child is looking straight down, so θ₁ and θ₂ are small.

For small θ, tan(θ) ≈ sin(θ) ≈ θ

 

[jegues]

So x is the distance from the center vertical line to the point where θ1 and θ2 have their vertex? Yes, that horizontal distance will be the same for the object and the image. Your approach looks good - carry on!

This looking straight down situation is not what I was thinking of in my first post based on your diagram. When looking not straight down, the horizontal distance to the image is not be the same as the horizontal distance to the object. Example here: http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/refr2.html#c1

Working with the first triangle we first define theta3,

$$\theta_{3} = 90 - \theta_{2} = 50^{o}$$

Now,

$$x = \frac{2}{tan\theta_{3}}$$

Then,

$$h = \frac{2}{tan\theta_{1}tan\theta_{3}} = 3.14$$

Which is incorrect. What am I doing wrong?

 

[SammyS]

The child is looking straight down, so θ₁ and θ₂ are small.

For small θ, tan(θ) ≈ sin(θ) ≈ θ

Hello jegues.

The equations in you Original Post are good.

Find the ratio: tan(θ₁) to tan(θ₂)

Change tan(θ₁) to sin(θ₁) & tan(θ₂) to sin(θ₂).

Now, use Snell's Law to find another expression for the ratio: sin(θ₁) to sin(θ₂).