Find index of refraction of the prism for 2 wavelengths

Click For Summary

Homework Help Overview

The discussion revolves around finding the index of refraction of a prism for two different wavelengths based on given angles of refraction. The problem is situated within the context of optics and involves applying Snell's Law.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between angles of refraction and the refractive index, questioning the sufficiency of the provided information. Some suggest that the problem can be approached geometrically and through Snell's Law, while others express uncertainty about deriving exact values without additional data.

Discussion Status

There is an ongoing exploration of how to apply Snell's Law to the problem, with participants offering different perspectives on the assumptions regarding the wavelengths and the angles involved. Some guidance has been provided regarding the geometric interpretation of the angles, but no consensus has been reached on the necessity of additional information.

Contextual Notes

Participants note that the problem does not specify the wavelengths of the rays, which raises questions about how to proceed with the calculations. The assumption that the refractive index may differ for the two wavelengths is also under discussion.

Ed Boon
Messages
8
Reaction score
0

Homework Statement


phys222.jpg


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
 
Physics news on Phys.org
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
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
6K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 11 ·
Replies
11
Views
3K