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

1. Jan 26, 2013

### steve233

1. The problem statement, all variables and given/known data

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

2. Relevant equations

Snell's Law:

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

3. 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.

2. Jan 26, 2013

### TSny

Start by writing out Snell's law for each of the two refractions.

3. Jan 26, 2013

### tiny-tim

hi steve233!

hint: θ2 + θ3 = … ?

4. Jan 26, 2013

### steve233

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...

Last edited: Jan 26, 2013
5. Jan 26, 2013

### TSny

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 $\theta$.]

I'm not following you here. $n_1$ is for air and $n$ is for the prism material.

Not sure how you got that, but I think it's right! So, that's your third equation.
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 $n$.

The square will show up in doing the algebra of solving for $n$ in terms of $\theta$.

Last edited: Jan 26, 2013
6. Jan 26, 2013

### steve233

$n_{1}$ 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.

7. Jan 26, 2013

### tiny-tim

hi steve233!

sin(arcsin(x)) = x

cos(arcsin(x)) = √(1-x2)

8. Jan 26, 2013

### steve233

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"?

Last edited: Jan 26, 2013
9. Jan 27, 2013

### tiny-tim

(just got up :zzz:)

not needed … saying "thanks" is enough!