Critical angle

  • Thread starter jaron
  • Start date
  • #1
23
0

Homework Statement


the critical angle for a ray of light leaving an unknown material (surrounded by air) is
32.4'(degrees celcius). the speed of light in the material is:


Homework Equations


niSin(theda)c = nRSin(theda)R
(i am not sure how to make theda symbols on the computer)


The Attempt at a Solution


niSin32'/Sin32' = 1Sin90'/Sin32'
ni = 1/0.53
ni = 1.89x10^8 m/s
thats the answer that makes the most sense to me. but the back of my book has a different answer:
1.61x10^8 m/s (i have no idea how they got this)

i have a test on this stuff tomorrow
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,223
31
Well you can find the refractive index with respect to air given the critical angle..

What are the ratios you can use to find the refractive index?
 
  • #3
344
1
hmm ni =/= 1.89*10^8 ... I guessing that's just a typo.

1/0.53... is right. I'm not sure how you got your answer but I got 1.61*10^8.

n=c/v

It might just be a math error.
 

Related Threads on Critical angle

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
3K
R
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
3
Views
7K
  • Last Post
Replies
7
Views
9K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
6
Views
4K
Replies
6
Views
5K
Top