# Optics; Deriving the index of refraction from Snell's law

## Homework Statement

There is a diagram in the problem statement so here is a link to the image of the problem:
http://imgur.com/KDrRsyO

## Homework Equations

Snell's Law:

$n_{1} * sin(\theta_{1}) = n_{2} * sin(\theta_{2})$

## The Attempt at a Solution

My attempt using Snell's law fails because I can't see how it is possible that n ever becomes squared. It looks like either some sort of vector addition or some use of Pythagorean theorem but I haven't been able to determine how to use those ideas to solve the problem.

I can see some use of special triangles given the angles are 60 degrees, but it's a bit overwhelming and my trig skills are a bit rusty.

Thanks.

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TSny
Homework Helper
Gold Member
Start by writing out Snell's law for each of the two refractions.

tiny-tim
Homework Helper
hi steve233!

hint: θ2 + θ3 = … ?

Hmm... So here is another attempt:

1st refraction:
$n_{1} * sin(\theta_{1}) = n * sin(\theta_{2})$

2nd refraction:
$n * sin(\theta_{3}) = n_{1} * sin(\theta)$

Here I assume that the refractive index is the same when the light is outside of the prism and inside of the prism.

I think I'm wrong here but, $\theta_{2} + \theta_{3} = 60^{\circ}$.
I think this is wrong because it seems like $90^{\circ} - \theta_{2}$ = $60^{\circ} \Longrightarrow \theta_{2} = 30^{\circ}$.

I'm still stuck on how things become squared though...

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TSny
Homework Helper
Gold Member
Hmm... So here is another attempt:

1st refraction:
$n_{1} * sin(\theta_{1}) = n * sin(\theta_{2})$

2nd refraction:
$n * sin(\theta_{3}) = n_{1} * sin(\theta)$
Ok. You can assume that outside the prism you just have air where ##n_1 ≈ 1.00##. You also know the value of ##\theta_1##. So, you have three unknowns: ##\theta_2##, ##\theta_3## and ##n##. [Edit: That is, you need to eliminate ##\theta_2## and ##\theta_3## so that you can express ##n## in terms of ##\theta##.]

Here I assume that the refractive index is the same when the light is outside of the prism and inside of the prism.
I'm not following you here. ##n_1## is for air and ##n## is for the prism material.

I think I'm wrong here but, $\theta_{2} + \theta_{3} = 60^{\circ}$.
Not sure how you got that, but I think it's right! So, that's your third equation.
I think this is wrong because it seems like $90^{\circ} - \theta_{2}$ = $60^{\circ} \Longrightarrow \theta_{2} = 30^{\circ}$.
I don't see how you are deducing that. In fact, I don't think there is any way to find the values of ##\theta_2## and ##\theta_3## individually since they would depend on ##n##.

I'm still stuck on how things become squared though...
The square will show up in doing the algebra of solving for ##n## in terms of ##\theta##.

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$n_{1}$ is indeed the refractive index of air, I should have mentioned that.

So the equations become:

(1) $\sqrt{3} / 2 = n * sin(\theta_{2})$
(2) $n * sin(\theta_{3}) = sin(\theta)$
(3) $\theta_{2} + \theta_{3} = 60^{\circ}$

$\theta_{3} = 60^{\circ} - \theta_{2}$
$n * sin(\theta_{3}) = n * sin(60^{\circ} - \theta_{2}) = n * (sin(60)cos(\theta_{2}) - cos(60)sin(\theta_{2}))$

$\theta_{2} = arcsin(\sqrt{3}/(2 * n))$

So,
$n * (sin(60)cos(\theta_{2}) - cos(60)sin(\theta_{2}))$
$n * ((\sqrt{3} / 2) * cos(arcsin(\sqrt{3}/(2 * n)) - (1/2) * sin(arcsin(\sqrt{3}/(2 * n))) = sin(\theta)$

This seems a little complicated... I know there is a simple replacement for cos(arcsin x) but I don't know anything that simply replaces sin(arcsin x). Am I on the wrong track here?

Thanks again for the help.

tiny-tim
Homework Helper
hi steve233!

sin(arcsin(x)) = x

cos(arcsin(x)) = √(1-x2)

Ah yes... more evidence that I need to brush up on my trig :)
I'll give this a shot and post the update soon.

Thanks!

Edit: Success! Thanks for all of the help TSny and tiny-tim.
I'm a bit new to these forums, is there some way to "upvote"?

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tiny-tim