Calculating the Depth of a Cylinder Using Refraction

• Const@ntine
In summary, the conversation discusses the problem of finding the depth of a non-translucent cylindrical container filled with water and open at the top. The container is exposed to sunlight at a 28.0 degree angle with the horizontal, causing the light to not reach the bottom of the container. Using the formula n1*sinθ1 = n2*sinθ2, the refractive angle is found to be 41.5 degrees. By creating a drawing and using the angle and diameter of the container, the depth is calculated to be 3.39 meters, which is a limiting case as anything higher would still satisfy the requirements.
Const@ntine

Homework Statement

A non-transclucent container in the form of a cylinder, has a diameter of 3.00 m, has its top part open, and is filled with water. When the sun created a 28.0 degree angle with the horizontal, the light doesn't illuminate the bottom of the container. What's the depth of the container?

Homework Equations

n1*sinθ1 = n2*sinθ2

The Attempt at a Solution

First things first, the 28.0 degree angle is the one with the horizontal, so the angle of prolapse, is going to be 62.0 degrees (angle between the vertical and the ray of light).

Then, what I did was take the known formula, and with the given Refractive Indexes (nwater = 1.333 & nair = 1.0002293). I ended up with the refraactive angle being 41.5 degrees. And then I got stuck.

I thought about tying it in with the speed of light, and finding the hypotenouse of the hypothetical triangle, but I don't know the time. I tried to figure out how to use the diameter but I can't come up with anything.

Then, I figured that I could take an unorthodox route, and imagine the container as being empty. In that case, the light reaches the bottom of the container, and with a few calculations I'll find the depth. Turns out the number comes out wrong.

So, any tips? Any kind of help is appreciated!

PS: The answer according to the book is 3.39 m.

Darthkostis said:
In that case, the light reaches the bottom of the container
That is strange. The refraction is towards more vertical, so I would expect the light not to reach the bottom.

Thing to do is make a drawing showing the angles. the 3.39 m is correct.

BvU said:
That is strange. The refraction is towards more vertical, so I would expect the light not to reach the bottom.

Yeah, my bad on that one. I was looking to the next drawing and mixed the two up.

BvU said:
Thing to do is make a drawing showing the angles. the 3.39 m is correct.

I made a couple of them, but I'm still stuck on how to incorporate the diameter, or how to proceed. It's pretty much this, isn't it ? :

I'm probably missing a crucial element, but I'm stuck on this one. I moved on to other exercises, but I still need to finish this one, otherwise I won't find peace!

Looks good. How much is d if the refracted ray from the left corner of the can just reaches the bottom (instead of the wall) ?

BvU said:
Looks good. How much is d if the refracted ray from the left corner of the can just reaches the bottom (instead of the wall) ?

So you mean this?

tan(41,5) = δ / d <=> d = δ / tan(41,5) = 3.00 m / 0.885 = 3.39 m

The result is correct, but I'm not sure I get the reasoning behind it. The angle of the sun above the horizontal is so small, that the ray doesn't reach the bottom, right? So how can I take the refracted angle and use it in a triangle that is compromised of the full depth and diameter of the container?

I agree that the wording of the exercise (if you transcribed it literally) is a bit vague. the word 'just' is missing: a can of 10 m depth would also satisfy the requirements.
The 3.39 m is a limiting case: anything higher is OK, anything less an a few rays reach the bottom directly.

BvU said:
I agree that the wording of the exercise (if you transcribed it literally) is a bit vague. the word 'just' is missing: a can of 10 m depth would also satisfy the requirements.
The 3.39 m is a limiting case: anything higher is OK, anything less an a few rays reach the bottom directly.
Well, it's a translation, but the exercise says "the lights stops illuminating the bottom", so I figured it stopped at a random point, say, d/3 or something. So, essentially, the ray reaches the bottom, hence why I'm able to use the classic tanx = ... Still, the numbers were rather small so I got a bit confused there for a minute, since I couldn't really make the drawing do what the exercise was speaking about.

Either way, thanks a ton for the help!

You're welcome. I think you got it just fine, so don't waste more time on this exercise

Const@ntine

1. What is refraction in a cylinder?

Refraction in a cylinder is the bending of light as it passes through a cylindrical object, such as a lens or a glass tube. This occurs because the surface of the cylinder is curved, causing the light rays to change direction.

2. How does refraction in a cylinder affect vision?

Refraction in a cylinder can correct vision problems such as astigmatism, where the cornea or lens of the eye is not perfectly spherical. The cylindrical shape of the corrective lens helps to focus light onto the retina, improving vision.

3. What factors affect refraction in a cylinder?

The shape and curvature of the cylinder, as well as the properties of the material it is made of, can affect refraction. The angle at which light enters the cylinder and the wavelength of light are also important factors.

4. How is refraction in a cylinder calculated?

The amount of refraction in a cylinder is calculated using the power of the cylinder, which is measured in diopters. The power is determined by the difference in curvature between the two surfaces of the cylinder.

5. What are some real-world applications of refraction in a cylinder?

Refraction in a cylinder is used in the production of eyeglasses, contact lenses, and camera lenses. It is also important in the design of telescopes, microscopes, and other optical instruments.

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