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Help about scattering of light


N S

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Some time before, after learning about the refraction and scattering of light in prisms takes place, I got moreover confused and could not understand why light does not scatter in rectangular glass slabs. Here is what I think should happen based on what I have learnt so far (attatched). I know it is completely incorrect, but I have tried in vain to get an understandable explanation of why it should not happen this way from my teacher. Only red and violet colours are represented as the light bends in the glass slab. Please help me out by giving a

simple account of what is incorrect in this diagram.

 



It seems like I could not post the image. It was a "png" file. Any way in which I can upload it? When I try to copy and paste it, I get a message that I am "not allowed to use that image extension on this community."

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Did you mean scatter or refract? Scattering generally results in the light changing angle after an interaction at some point, and that change is variable in some way. Refraction is a systematic change in the angle at a surface or interface. Light doesn't scatter in glass because there's nothing that scatters it. That's the point of using it. You can get scattering off of impurities within the glass.

 

 

 

How to upload images http://www.scienceforums.net/topic/73369-uploading-images/

 

(It's possible png is not supported, though. I'm not sure.)

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What I meant was, since we say light scatters into seven (visible) colours when it lands on a glass prism at an angle, it should scatter in a rectangular glass block too.

And here is the diagram in which I seem to have made a mistake.

 

refraction.png

Edited by N S
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Put another way, all the light does in the glass is shift, because the system is symmetric. The refraction at one interface is reversed at the other one. A prism's surfaces are not parallel, so this symmetry is broken.

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I know that this might be a little irritating, but could you please explain it in an easier manner?

 

note that the angle of deviation(after the double refraction) for each light ray of different colors is zero.
You may verify this by doing a simple calculation by using snell's law, for different wavelengths of light.

I believe that the colours are obeying Snell's Law all the time. The difference in the deviation of red and violet every time is evident.

 

Put another way, all the light does in the glass is shift, because the system is symmetric. The refraction at one interface is reversed at the other one. A prism's surfaces are not parallel, so this symmetry is broken.

Since the light has scattered once and red and violet are now apart, in whatever way they further refract, they will have to go in different directions with or without crossing each other. Please explain the logic or force that prevents the light from getting thus scattered after meeting parallel surfaces as in this case.

Edited by N S
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Since the light has scattered once and red and violet are now apart,

 

That's true for an infinitely thin ray, but for a real beam, there is overlap. This is refraction, not scattering. Those terms have different meanings.

 

in whatever way they further refract, they will have to go in different directions with or without crossing each other. Please explain the logic or force that prevents the light from getting thus scattered after meeting parallel surfaces as in this case.

 

The second surface undoes the change that happened at the first. The outgoing beam is parallel to the incoming beam.

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I know that this might be a little irritating, but could you please explain it in an easier manner?

 

I believe that the colours are obeying Snell's Law all the time. The difference in the deviation of red and violet every time is evident.

As swansont has answered, the refraction at one surface is reversed at the other side.

Have you done the calculations?

The light ray reforms on the other side, parallel to the original light ray if and only if the sides of the prism is parallel. Though there might be difference in width..

Say, you can also try doing this experiment. I've done it in school last year, and nope, no rainbow was formed on the other side when white light is shone on the rectangular prism. In fact, if it did, I shouldn't be able to use the optical pin method.

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Thanks swansont and Mellinia, I think I get what you are trying to explain.

 

Say, you can also try doing this experiment. I've done it in school last year, and nope, no rainbow was formed on the other side when white light is shone on the rectangular prism. In fact, if it did, I shouldn't be able to use the optical pin method.

I don't have the apparatus for this experiment, but can you tell me whether different colours of light are seen inside the triangular glass prism and rectangular glass slab?

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The different colors of light cannot be seen anywhere since the glass slab I used for the snell's law experiment was transparent. No scattering means I can't see the light ray in the slab. Kinda same like it's impossible for you to see laser beam paths in clear vacuum.

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