Snell's law and optical filters.

  • Thread starter Greger
  • Start date
  • #1
46
0
Hi,

I recently did a experiment in which I measured the intensity of light out of a optical filter at different angles of incidence.

The optical filter was designed such that only light of wavelength 405 nm is transmitted. The wavelength of incident light (the laser I used) was 402 nm.

My first measurement was 0 degree's which meant the laser was perpendicular to the filter. I obtained an intensity peak at 10 degree's which had a width of around 10 degree's. So the intensity at 0 and 20 degrees was small.

Now I was asked to calculate the band-width. I think I can do this using Snell's rule.

[itex]\frac{sin(\theta_1)}{sin(\theta_2)} =\frac{\lambda_1}{\lambda_2}[/itex]

I know that at 10 degree's the wavelength into the filter is 402 nm and the wavelength inside it 405 nm (since all of the light is transmitted) so I can calculate the angle of refraction (I get 10.075).

My question is, how can I calculate the wave length light inside the filter at angles 0 and 20 degrees (when the intensity drops)?

At first I was thinking that I could keep the angle of refraction constant, then just substitute the new angle of incidence in and calculate the wavelength inside, but I know that's not right (the angle of refraction wouldn't be constant).

I was wondering, would the difference between the angle of incidence and refraction be the same? Like for the incidence angle 10, the angle of refraction is 10.075 so the difference is 0.075. Would this be the same for the incidence angle 20? If so then it's possible to calculate the wavelength inside the medium with incidence angles 0 and 20, but if not, I'm not sure how else I could do it as I wouldn't know the angle of refraction.

Thank you
 

Answers and Replies

  • #3
46
0
Thanks for your reply,

I am required to determine the band width using the method I described in my first post.

The second article you posted does show similar results to what I have obtained, however I am asking how, from the information I currently have, can I calculate λ2. (Recall the information I have is the incident angle and wavelength, I am not sure about the angle of refraction)
 
  • #4
662
307
I don't know if I am correct but:-
.......I know that at 10 degree's the wavelength into the filter is 402 nm and the wavelength inside it 405 nm (since all of the light is transmitted) so I can calculate the angle of refraction (I get 10.075).
You can get the refractive index of the fibre by this and then and use it to calculate angle of refraction.
My question is, how can I calculate the wave length light inside the filter at angles 0 and 20 degrees (when the intensity drops)?
Wavelength of light depends only on the speed of light in the medium and shouldn't be affected by either angle of incidence or intensity.
I was wondering, would the difference between the angle of incidence and refraction be the same?
Nope, that doesn't make any sense to me. Their ratio on the other hand should be constant (refractive index).
Hope this helps.
 

Related Threads on Snell's law and optical filters.

  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
8
Views
19K
Replies
4
Views
3K
  • Last Post
Replies
3
Views
8K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
7
Views
446
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
10
Views
1K
Top