# Homework Help: Refraction in Cylinder.

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1. May 7, 2017

### Techno_Knight

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

n1*sinθ1 = n2*sinθ2

3. 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.

2. May 7, 2017

### BvU

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.

3. May 7, 2017

### Techno_Knight

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

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!

4. May 7, 2017

### BvU

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) ?

5. May 7, 2017

### Techno_Knight

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?

6. May 7, 2017

### BvU

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.

7. May 7, 2017

### Techno_Knight

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!

8. May 7, 2017

### BvU

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