• Support PF! Buy your school textbooks, materials and every day products Here!

Index of Refraction: n1 * sin(theta1) = n2 * sin(theta2)

  • Thread starter Moony-
  • Start date
  • #1
4
0

Homework Statement


A beam of light is emitted 8.0 cm below the surface of an unknown liquid. The light beam is angled in a way that it strikes the surface 7.0 cm from the point directly above the light source. The light beam is totally internally reflected. What is the index of refraction of the unknown liquid?


Homework Equations


n1sinΘ1° = n2sinΘ2°

n = c/velocity in medium


The Attempt at a Solution


tanΘ = 7/8
Θ = 41.8°

Attempt 1:
n1sin41.2° = n1sin41.2°
Cancels out...

Attempt 2:
n1sinΘ° = 1sin90°
Two variables!

Attempt 3:
n1sin41.2° = 1sin90°
n1 = 1.518166
n1 = 1.5
Could do this, although I don't even know whether it is correct, since it does not use the actual critical angle.
 
Last edited:

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
5
Attempt 2:
n1sinΘ° = 1sin90°
Two variables!

Here angle is 41.2 degree. That itself is the critical angle. Find n1.
 
  • #3
4
0
Attempt 2:
n1sinΘ° = 1sin90°
Two variables!

Here angle is 41.2 degree. That itself is the critical angle. Find n1.
Can you assume that for sure, though? It could be that the 41.2° is bigger than the actual critical angle, which is why it reflects... =|
 
  • #4
rl.bhat
Homework Helper
4,433
5
Even if the angle of incidence slightly more than critical angle, the light will be totally internally reflected. If the angle 41.2 degree is more than critical angle, it is not possible to determine the refractive index of the medium.
 
  • #5
4
0
So that would mean that there's no guarantee whether 41.2° really is the critical angle? And Attempt 3 would only work if 41.2° is the critical angle, right?
 
  • #6
rl.bhat
Homework Helper
4,433
5
Yes. Bur in your attempt 3, how did you get the value of n1= .02?
 
Last edited:
  • #7
4
0
Ah, sorry. I made a little calculator mistake with brackets... No wonder the value for n1 looked odd....
 
  • #8
54
0
You have calculated that when angle of incidence is 41.2 deg, then the light is totally internally reflected. But it is impossible to know from the given data what will happen if angle of incidence is less than 41.2 deg. This means that we cannot know the exact critical angle. At the most we can say that it is not more than 41.2 deg.
Therefore the answer you got in the third attempt (i.e. 1.5) is the minimum possible value of n. Actual value may be equal to this or more than.
Can you find why 1.5 is the minimum possible value? Why not maximum?
 

Related Threads for: Index of Refraction: n1 * sin(theta1) = n2 * sin(theta2)

Replies
2
Views
5K
Replies
3
Views
812
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
6
Views
3K
Replies
3
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
5
Views
8K
Top