Is the Equation for Refractive Index Accurate?

AI Thread Summary
The discussion centers on the accuracy of the equation for refractive index, specifically "refractive index = (real depth)/(apparent depth)." Participants debate the validity of using similar triangles to derive this equation, with some asserting that the triangles are not actually similar. There are claims of mistakes in textbook proofs, leading to confusion about the definition and derivation of the refractive index. The conversation also touches on the small angle approximation and its implications for the equation's accuracy. Overall, the participants seek clarification and verification of the mathematical principles involved.
primarygun
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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.
 
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Why do you say there are mistakes and contradictions?
 
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.
 
http://paintedover.com/uploads/show.php?loc=23&f=physics.jpg
Hope it helps you bring me out of the troubles.
 
Ugh, basic geometry :(

I think they are considered similar triangles.
 
is it correct?
 
yah I am pretty sure it is... hopefuly someone will verify though... its 3am here :D
 
But, I think the equation should be correct as I saw it in many books.
We need the someone's help;
 
primarygun said:
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.

What DEFINITION of 'refractive index' does your book give? (It is possible to use "refractive index= (real depth)/(apparent depth)" as the definition.)
 
  • #10
index of refraction.
 
  • #11
Pengwuino said:
yah I am pretty sure it is... hopefuly someone will verify though... its 3am here :D

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.
 
  • #12
i derived this equation before. the formula is just an approximation for small angle case, where sin\theta=tan\theta 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.
 
  • #13
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.
 
  • #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...
 
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