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Prove of refractive index

  1. Jul 18, 2005 #1
    Can anyone show me a proof of "refractive index= (real depth)\(apparent depth) "?
    I found the proof in my book has a mistakes and I found some contradicts to this equation.
    Last edited: Jul 18, 2005
  2. jcsd
  3. Jul 18, 2005 #2


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    Why do you say there are mistakes and contradictions?
  4. Jul 18, 2005 #3
    The book said used a pair of similar triangles to infer it but the triangles are not similar.
    Do you want a picture? If you want, I can upload it now.
  5. Jul 18, 2005 #4
    Hope it helps you bring me out of the troubles.
  6. Jul 18, 2005 #5


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    Ugh, basic geometry :(

    I think they are considered similar triangles.
  7. Jul 18, 2005 #6
    is it correct?
  8. Jul 18, 2005 #7


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    yah im pretty sure it is... hopefuly someone will verify though... its 3am here :D
  9. Jul 18, 2005 #8
    But, I think the equation should be correct as I saw it in many books.
    We need the someone's help;
  10. Jul 18, 2005 #9


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    What DEFINITION of 'refractive index' does your book give? (It is possible to use "refractive index= (real depth)/(apparent depth)" as the definition.)
  11. Jul 18, 2005 #10
    index of refraction.
  12. Jul 30, 2005 #11
    if the two triangles were similar, wouldn't r = i, or 90 degrees - r = i?

    not sure if giving the ratio of the sines will show this. :grumpy:
  13. Jul 30, 2005 #12
    i derived this equation before. the formula is just an approximation for small angle case, where [itex]sin\theta=tan\theta[/itex] for small angle. i.e. it is for the case where you almost look vertically down to the object from above. you can derive it easily by drawing slender triangles. very easy.

    if you get stuck i can post a proper solution with the drawing.
  14. Jul 30, 2005 #13


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    The triangles are NOT similar. That derivation is just extremely poorly worded (or was written by someone who knew the result and "made up" the proof). What it is actually using is the small angle approximation that sniffer mentions.

    Anyway, I believe Primary's doubts have long been resolved.
  15. Jul 30, 2005 #14
    According to the definition these triangles áre similar because when you shrink one side the become congruent which is apparently enough to qualify them as "similar" (I have looked this up.).

    Frankly I thought the explanation was pretty straightforward...
    Last edited: Jul 30, 2005
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