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Shouldn't I see a rainbow ?

  1. Jul 5, 2011 #1
    I was just wandering, why when I look through a glass or a mirror at 45 degrees related to it's surface normal I don't see a rainbow/blurry image?

    Here is a picture that hopefully, describes better what I mean.

    5904309193_c92c7f7386.jpg
    Dispersion by Cristi .eXPV, on Flickr
     
  2. jcsd
  3. Jul 5, 2011 #2

    Claude Bile

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    Probably because the deflection caused by the dispersion of the glass is small compared to the spatial extent of the light source.

    Claude.
     
  4. Jul 5, 2011 #3

    I like Serena

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    The rays that leave the glass are parallel.
    Another incoming ray that is parallel, will split the same way, and merge with the other rays, making distinction impossible.
    It's only at the edge of the glass that you might see some colors. :)
     
  5. Jul 5, 2011 #4
    Let's say we look at an object that reflects only 3 components in visible spectrum at a large distance from one to another (i.e. red, yellow, blue). What I should see in the mirror are 3 images of the object, corresponding to each color, that are slightly displaced.

    I looked in a mirror as thick as 5 mm but I couldn't notice any dispersion .. except the multiple reflection (one on the surface and others in the glass .. I was able to see up to five reflections .. not visible in the picture .. to dimmer for my camera).
    5905741397_581cf88f80.jpg
    SP_A0122 by Cristi .eXPV, on Flickr

    I think the angle of refraction changes to little over the spectrum to make it noticeable. Maybe I can put a small mirror in a glass of water and see if anything changes. But how am I gonna make a white beam of light. There are going to be impurities in the water, imperfect surface ... a headache, everything will be blurry. I'll see what I can do.

    http://en.wikipedia.org/wiki/Snell%27s_law" [Broken] doesn't take in to account the wavelength. Is there another one that does ?
     
    Last edited by a moderator: May 5, 2017
  6. Jul 5, 2011 #5

    I like Serena

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    There's a variant of Snell's law that does:

    855b336741150b679007dd522db12eef.png
     
  7. Jul 5, 2011 #6
    It does, it's just that n is dependent of the wavelength.
     
  8. Jul 6, 2011 #7
    Yes I noticed .. I didn't pay attention when I first read the article.
     
  9. Jul 6, 2011 #8
    Got it .. I placed a small mirror in a pot and filled it with water. The pot is 200+ mm tall and I was able to see dispersion. It's not very clear in the picture, but you get the idea.
    5908611308_1af1ce05ab_b.jpg
    Photo-0010 by Cristi .eXPV, on Flickr

    So you would need a mirror 25 cm thick in order to see clear dispersion. :P
     
  10. Jul 6, 2011 #9

    I like Serena

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    Nice! :smile:

    I looked up the refractive indices for red and blue.
    For glass they are 1.520 resp. 1.525, so that a difference of 0.5%.
    With glass about 200 mm thick, that would mean a dispersion in the order of 1 mm.
    So that seems to match!
     
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