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

• steve233
In summary, the conversation discusses a problem involving Snell's law and the use of special triangles to solve for the refractive index in a prism. The summary includes equations and attempts at solving the problem, with helpful hints and clarifications from other users. Eventually, the correct solution is found through the use of basic trigonometry principles.
steve233

## 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.

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

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...

Last edited:
steve233 said:
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##.

Last edited:

$n_{1}$ 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.

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"?

Last edited:
(just got up :zzz:)

not needed … saying "thanks" is enough!

## 1. What is Snell's law?

Snell's law, also known as the law of refraction, is a formula that describes the relationship between the angle of incidence and the angle of refraction when a light ray passes through a boundary between two different materials.

## 2. How is Snell's law used to derive the index of refraction?

By rearranging Snell's law equation (n1sinθ1 = n2sinθ2), we can solve for the index of refraction (n) of the second material using the known values of the angle of incidence (θ1) and the angle of refraction (θ2).

## 3. What is the index of refraction?

The index of refraction (n) is a measure of how much a material bends light as it passes through it. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the material.

## 4. What factors can affect the index of refraction?

The index of refraction can be affected by the density, composition, and temperature of the material. It also depends on the wavelength of light passing through it.

## 5. Why is the index of refraction important in optics?

The index of refraction is important in optics because it determines how light behaves when it passes through different materials. It is essential for understanding and predicting the behavior of light in lenses, prisms, and other optical devices.

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