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Total internal reflection, underwater light

 
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Aug19-08, 05:13 PM   #1
 
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Total internal reflection, underwater light


1. The problem statement, all variables and given/known data
A point source of light is at the bottom of a koi pond, at a depth of 0.525 meters. What is the radius of the circle of light formed on the water's surface? Take the index of refraction of water to be 1.33. Hint: Some of the light emitted experiences total internal reflection inside the water.

2. Relevant equations
sin(crit angle) = n2/n1
n1>n2
n for air = 1.00
n for water = 1.33

3. The attempt at a solution
So the critical angle for water to air is:
arcsin(1/1.33) = 48.7534666 degrees

This is where it all ends. There is no formula for distance or light radius. I am completely stuck at this point and cannot find anywhere in my book where I could derive the answer. I would deeply appreciate some help with this one!
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Aug19-08, 05:34 PM   #2
 
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Draw yourself a diagram of the light source with rays emanating in all directions. For various ray angles, what happens when the light reaches the water surface? At what angle with the vertical will they be totally reflected. Use a bit of trig to figure out the radius of the circle of light.
Aug19-08, 05:47 PM   #3
 
Wouldn't the radius of light only be a point at the surface if it is only a point at the bottom? I'm afraid I'm still not grasping this.
Aug19-08, 05:50 PM   #4
 
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Total internal reflection, underwater light


Did you draw a diagram of what's happening with the light rays?
Aug19-08, 05:53 PM   #5
 
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Quote by gigli View Post
Wouldn't the radius of light only be a point at the surface if it is only a point at the bottom?
Think of the point source of light sending out zillions of light rays in all directions. What happens to a light ray that goes straight up? One that goes 5 degrees from the vertical? 10 degrees? Etc.
Aug19-08, 05:53 PM   #6
 
Yes. When the angle is not quite to the critical angle it shines out of the water. when it reaches the critical angle it shines across the surface of the water, but none shines into the air, and when it is past the critical angle no light shines out of the water only deeper back into the water.

I think I see why it stays in the water and all that, but I fail to see how there is a radius of light at the surface, when it is only a point form the source. Sorry if I am being rather thick.
Aug19-08, 06:00 PM   #7
 
Ooooooooooooooooh. I got it now. Thank you so much.

The light was shining a cone straight up, but past 48 degrees it was reflecting back into the water. I was thinking it was a laser beam aimed at 48 degrees.

Thanks again! I got the right answer
Aug19-08, 06:01 PM   #8
 
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Quote by gigli View Post
Yes. When the angle is not quite to the critical angle it shines out of the water. when it reaches the critical angle it shines across the surface of the water, but none shines into the air, and when it is past the critical angle no light shines out of the water only deeper back into the water.
Good. Now imagine a line going straight up from the light source to the water's surface. That spot on the surface will be the center of a circle. Think of all those rays of light hitting the surface. What's the farthest from the center that the light can reach when it passes the surface?
I think I see why it stays in the water and all that, but I fail to see how there is a radius of light at the surface, when it is only a point form the source.
It's not a point from the source, it's a "point source" of light. That just means that the light source is small enough to treat as being a single point, instead of some extended light source. Imagine it as a small light bulb. (As compared, perhaps, to a 3 foot long flourescent light tube--which would not make a good "point source".)
Aug19-08, 06:02 PM   #9
 
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Quote by gigli View Post
The light was shining a cone straight up, but past 48 degrees it was reflecting back into the water. I was thinking it was a laser beam aimed at 48 degrees.
Excellent. I knew it would click, sooner or later.
Aug19-08, 06:03 PM   #10
 
That last little bit really clarified what I was looking for. Thanks for the help Doc Al. Guess I'll try and get cracking on another stumper.
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