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Find the smallest diameter of an opaque object

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1. Homework Statement
A glass block 3cm thick and of index of refraction 1.5 is placed on a luminous point. Find the smallest diameter of an opaque object which will prevent any light from passing out of the top of the block.


2. Homework Equations
Snell's Law


3. The Attempt at a Solution

My main problem is not understanding the question. The question had no diagram attached. I assume there is no distance between the luminous point and glass block. This makes calculations difficult? However, I have checked the back of the book and there is a single correct answer.
 

berkeman

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When light goes from air into a higher index material at an angle, what determines at what grazing angle you get total external reflection (no light going from air into glass)? It seems like that would be how you will get a limited cone of light in the block. Then you would project that cone up 3cm to the top surface of the glass block to figure out what the radius of the cone is there.
 
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Could you perhaps explain it more clearly maybe also with a diagram? Thanks.
 

Andrew Mason

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1. Homework Statement
A glass block 3cm thick and of index of refraction 1.5 is placed on a luminous point. Find the smallest diameter of an opaque object which will prevent any light from passing out of the top of the block.
As you have stated this problem, there is no opaque object that will block light passing through the top. If the luminous point is in contact with the glass block, the light will be visible at any point on the surface of the glass block.

AM
 
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As you have stated this problem, there is no opaque object that will block light passing through the top. If the luminous point is in contact with the glass block, the light will be visible at any point on the surface of the glass block.

AM
Why? Again, a small diagram would help.
 

Doc Al

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Could you perhaps explain it more clearly maybe also with a diagram? Thanks.
This site has a few diagrams that might help you: http://www.physics.louisville.edu/public/courses/phys111/davis/notes/lo_tir.html"

As you have stated this problem, there is no opaque object that will block light passing through the top. If the luminous point is in contact with the glass block, the light will be visible at any point on the surface of the glass block.
Why do you say that?
 
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Andrew Mason

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Doc Al

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A luminous point in contact with the block is equivalent to a light source on the block's surface emitting light in all directions (at least all directions from 0 to 180 degrees). There is no refraction or bending of the light.
Right... at the bottom surface of the block. But all the action (refraction, total internal reflection) takes place at the top surface.
 
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As you have stated this problem, there is no opaque object that will block light passing through the top. If the luminous point is in contact with the glass block, the light will be visible at any point on the surface of the glass block.

AM
So you are saying the question either contains not enough information or has no right answer?

Do you think they should have specified the distance between the luminous object and the glass?
 
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When light goes from air into a higher index material at an angle, what determines at what grazing angle you get total external reflection (no light going from air into glass)? It seems like that would be how you will get a limited cone of light in the block. Then you would project that cone up 3cm to the top surface of the glass block to figure out what the radius of the cone is there.
So total external reflection is light going from a lower index of refraction to a higher one. With the light reflecting off the higher index of refraction.

Since the light will only get closer to the normal after refraction, how can you get this type of reflection?
 

berkeman

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So total external reflection is light going from a lower index of refraction to a higher one. With the light reflecting off the higher index of refraction.

Since the light will only get closer to the normal after refraction, how can you get this type of reflection?
I think Doc Al's approach is probably the correct one instead. I have a hard time dealing with the light source being right at the bottom surface, so it may be correct that the light makes it in fairly uniformly from that point. Anyways, the TIR approach probalby would limit the area more than some external reflection approach at the bottom surface anyway.

Do you understand Doc Al's hint about using TIR at the upper glass-air boundary?
 

Andrew Mason

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Ok. The question is about the critical angle for total internal reflection - the minimum angle at which all incident light is reflected back into the glass. You just have to find the radius of the circle such that light from the source strikes the upper surface at an angle that is greater than the critical angle.

The question implicitly asks you to assume that the light that is reflected from the surface back into the block does not bounce around in the glass or reflect again from the bottom so that it is able to escape through the top surface.

AM
 
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Now I finally understand the problem. All the action is indeed at the top of the surface. The light source is precisely at the bottom of the block. But the opaque object they are after should precisely be at the top of the block. At the start, I always had the impression that it was at the bottom of the block. The opaque object should block all the light coming in at an angle which is less than the critical angle. When all these refracting light it blocked, the rest will just reflect internally. So the dimension of the top surface is unimportant hence not specified which is okay. The answer came put to be 5.36cm which matches the back of the book. Thanks for the help.

Although I still don't understant how total external reflection would work (in general). Light refracting off a smaller index will bend towards the normal so how can reflection of this type occur?
 
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Doc Al

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Although I still don't understant how total external reflection would work (in general).
I assume you mean total internal reflection? Did you look at the link I provided?
Light refracting off a smaller index will bend towards the normal so how can reflection of this type occur?
Just the opposite. When light traverses a boundary from higher to lower index, the refraction is away from the normal. (A consequence of Snell's law.)
 

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