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

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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:

[itex]n_{1} * sin(\theta_{1}) = n_{2} * sin(\theta_{2})[/itex]

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.
 

Answers and Replies

  • #2
TSny
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Start by writing out Snell's law for each of the two refractions.
 
  • #3
tiny-tim
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hi steve233! :smile:

hint: θ2 + θ3 = … ? :wink:
 
  • #4
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Hmm... So here is another attempt:

1st refraction:
[itex] n_{1} * sin(\theta_{1}) = n * sin(\theta_{2}) [/itex]

2nd refraction:
[itex] n * sin(\theta_{3}) = n_{1} * sin(\theta) [/itex]

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, [itex] \theta_{2} + \theta_{3} = 60^{\circ} [/itex].
I think this is wrong because it seems like [itex] 90^{\circ} - \theta_{2} [/itex] = [itex] 60^{\circ} \Longrightarrow \theta_{2} = 30^{\circ}[/itex].

I'm still stuck on how things become squared though...
 
Last edited:
  • #5
TSny
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Hmm... So here is another attempt:

1st refraction:
[itex] n_{1} * sin(\theta_{1}) = n * sin(\theta_{2}) [/itex]

2nd refraction:
[itex] n * sin(\theta_{3}) = n_{1} * sin(\theta) [/itex]
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, [itex] \theta_{2} + \theta_{3} = 60^{\circ} [/itex].
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 [itex] 90^{\circ} - \theta_{2} [/itex] = [itex] 60^{\circ} \Longrightarrow \theta_{2} = 30^{\circ}[/itex].
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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  • #6
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Ah ok, very helpful. Thanks.

[itex] n_{1} [/itex] is indeed the refractive index of air, I should have mentioned that.

So the equations become:

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

[itex] \theta_{3} = 60^{\circ} - \theta_{2} [/itex]
[itex] n * sin(\theta_{3}) = n * sin(60^{\circ} - \theta_{2}) = n * (sin(60)cos(\theta_{2}) - cos(60)sin(\theta_{2})) [/itex]

[itex] \theta_{2} = arcsin(\sqrt{3}/(2 * n)) [/itex]

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

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
tiny-tim
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hi steve233! :smile:

sin(arcsin(x)) = x

cos(arcsin(x)) = √(1-x2) :wink:
 
  • #8
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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:
  • #9
tiny-tim
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(just got up :zzz:)

not needed … saying "thanks" is enough! :smile:
 

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