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

  1. Dec 15, 2008 #1
    1. The problem statement, all variables and given/known data
    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?


    2. Relevant equations
    n1sinΘ1° = n2sinΘ2°

    n = c/velocity in medium


    3. 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: Dec 16, 2008
  2. jcsd
  3. Dec 16, 2008 #2

    rl.bhat

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

    Here angle is 41.2 degree. That itself is the critical angle. Find n1.
     
  4. Dec 16, 2008 #3
    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... =|
     
  5. Dec 16, 2008 #4

    rl.bhat

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    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.
     
  6. Dec 16, 2008 #5
    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?
     
  7. Dec 16, 2008 #6

    rl.bhat

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    Yes. Bur in your attempt 3, how did you get the value of n1= .02?
     
    Last edited: Dec 16, 2008
  8. Dec 16, 2008 #7
    Ah, sorry. I made a little calculator mistake with brackets... No wonder the value for n1 looked odd....
     
  9. Dec 16, 2008 #8
    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?
     
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