A question on dispersion of Prism

In summary, a white light beam is dispersed by the prism. The refracted beams will have different directions. My question is, will their reverse extension lines intersect into one point, or not? If it will, where is the point? And the proof?f
  • #1
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As the Figure shown, a white light beam is dispersed by the prism. The refracted beams will have different directions. My question is, will their reverse extension lines intersect into one point, or not? If it will, where is the point? And the proof? Thanks a lot.
 

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  • #2
My intuition says that the extension of all rays will not meet at exactly the same point due to the chromatic aberration. Currently I don't have enough time to derive a formal algebraic proof, but I think you can follow the following reasoning.
Draw three rays and assume that all colors have the same refractive indices. Assume that the extension of the rays do meet at one point.
But in reality, the refractive index of materials always depends on the wavelength and since this dependency is arbitrary, you can change, e.g. the index for red ray only, to some other value causing the red ray to miss the intersection point with the other two rays. Therefore, generally there is no single intersection point for all rays.
 
  • #3
My intuition says that the extension of all rays will not meet at exactly the same point due to the chromatic aberration. Currently I don't have enough time to derive a formal algebraic proof, but I think you can follow the following reasoning.
Draw three rays and assume that all colors have the same refractive indices. Assume that the extension of the rays do meet at one point.
But in reality, the refractive index of materials always depends on the wavelength and since this dependency is arbitrary, you can change, e.g. the index for red ray only, to some other value causing the red ray to miss the intersection point with the other two rays. Therefore, generally there is no single intersection point for all rays.
Thank you for your kind reply. In my opinion, all of the output rays comes from the same ray, if it is like the imaging of an object from a plane mirror, where the reverse rays of all of the reflected will meet exactly the same point, the observer seeing the colored rays from the prism may think all of the rays comes from one point or ray. However, I have no proofs either.
 
  • #4
the observer seeing the colored rays from the prism may think all of the rays comes from one point or ray.
If the original ray is actually a ray (or slit), the observer will only see one narrow band of wavelengths at each angle of the eye. The only way for the observer to 'reconstruct' an image of the source will be to move the eye from side to side and to use a sort of binocular approach to estimate a point where all the various coloured rays come from.
I concur with blue_leaf77 and I don't think that there will be a single point through which all the rays appear to have come from / through. If you wanted to do the exercise yourself, it wouldn't be difficult to show what happens. Draw the thing out with a neat prism and a source and, using Snell's Law, calculate the angles of refraction of blue, yellow and red rays (just choose representative wavelengths), coming from the arrival point on the entry face. Do the same thing at the exit face. and you should get three rays (lines) emerging from the exit face. If those lines produced backwards pass through a point then your idea may be correct. Have fun with a ruler and protractor! :smile:
 
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  • #5
I concur with blue_leaf77 and I don't think that there will be a single point through which all the rays appear to have come from / through.
Thank you for your help. You and blue_leaf77 are right. I have simulated it with mathematica and find the inverse rays (I used two) really do not go through a single point.
 
  • #6
I have simulated it with mathematica
And why not? Pencil and paper are old hat and only used by old gimmers for many problems. Can be handy though!
You could be really smartarse and do it all with trigonometry and algebraic jiggery pokery.
 
  • #7
(I used two)
You have to use at least three rays of different wavelength to prove that they don't intersect at one point.
 
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  • #8
You have to use at least three rays of different wavelength to prove that they don't intersect at one point.
Yes, you are right, I made a mistake. I have added the third rays and found they don't intersect at one point. Thank you for pointing it.
 

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