# Prove of refractive index

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:

Pengwuino
Gold Member
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.

Hope it helps you bring me out of the troubles.

Pengwuino
Gold Member
Ugh, basic geometry :(

I think they are considered similar triangles.

is it correct?

Pengwuino
Gold Member
yah im 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;

HallsofIvy
Homework Helper
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.)

index of refraction.

Pengwuino said:
yah im 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. :grumpy:

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.

Gokul43201
Staff Emeritus