Find index of refraction of the prism for 2 wavelengths

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SUMMARY

The discussion focuses on calculating the index of refraction for a prism using Snell's Law, specifically for two different wavelengths based on given angles of refraction. Participants clarify that the problem does not require the actual wavelengths but rather assumes different wavelengths lead to different indices of refraction. By applying Snell's Law, participants derive equations to find the refractive indices (na and nb) using the angles of refraction (12° and 20.5°) and the geometry of the prism. The solution emphasizes the relationship between the angles and the refractive indices without needing additional wavelength information.

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  • Understanding of Snell's Law and its application in optics
  • Basic knowledge of geometric relationships in prisms
  • Familiarity with the concept of refractive index
  • Ability to manipulate trigonometric equations
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Ed Boon
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Homework Statement


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Homework Equations


Snell's Law: na*sin(theta of a) = nb*sin(theta of b)
wavelength = index of refraction / n
n = c / v

The Attempt at a Solution


The only information given in the problem is the 2 angles of refraction and of course n of air is 1, I do not understand how to get the index of refraction for both wavelengths with just the angles given and nothing more. They do not state the wavelengths of the rays. I assume it is solvable because it is in the text but if someone could maybe lead me in any direction I would appreciate it.
Thanks
 
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It's not asking for their wavelengths, you just have to assume the wavelengths are different so that the refractive index can be different.
It's simply a geometry question, find the angle from the normal for each of the exit rays and then just use Snell's law to calculate 'n' for each of them.
 
mgb_phys said:
It's not asking for their wavelengths, you just have to assume the wavelengths are different so that the refractive index can be different.
It's simply a geometry question, find the angle from the normal for each of the exit rays and then just use Snell's law to calculate 'n' for each of them.

Thats what I was thinking so you will have na*sin(theta of a) equals to both nb*sin(0b) and nc*sin(0c) (b and c the 2 refractions) so if u set the 2 equal to each other:
nb*sin(20.5) = nc*sin(12)
dont you need more info to get the exact values instead of just a ratio?
thanks for the quick response
 
Assuming the prism is in air, anf consideringthe refraction at the exit face - isn't it just
na sin(internal angle ) = 1 * sin(12+external angle)
nb sin(internal angle ) = 1 * sin(12+8.5+external angle)

And the angle from normal inside the prism is just the angle from normal to the horizontal dashed line outside ( and is equal to the prism wedge angle)
na sin(x) = sin (12 + x)
nb sin(x) = sin(20.5 + x)
It should be possible to rearrange this to get exact values of na and nb without knowingx
 

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