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Total internal reflection in prism

  1. Mar 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Light is incident normally on the short face of a 30-60-90 (degree) prism. A drop of liquid is placed on the hypotenuse of the prism.


    If the index of the prism is 1.68, find the maximum index that the liquid may have if the light is to be totally reflected.
    2. Relevant equations

    sin ik = n1/n2

    3. The attempt at a solution

    sin 60 = n1/1.68

    n1= 1.45

    it said that my answer is close enough only differ in significant figure. however, they didn't accept my answer. anyone how to solve this problem?? thanks
     
  2. jcsd
  3. Mar 19, 2009 #2

    Doc Al

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    Your answer looks fine to me.
     
  4. Mar 19, 2009 #3
    i was doing this problem in mastering physic..
    and it turned out

    "Very close. Check the rounding and number of significant figures in your final answer."

    any clues about using mastering physics regarding to rounding the answer???
     
  5. Mar 19, 2009 #4

    Redbelly98

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    Since the calculation gives 1.45492..., it is possible that Mastering Physics made a rounding error and thinks it should be 1.46.
     
  6. Mar 20, 2009 #5
    OMG..!!
    thanks..
    it works..

    mastering physics sometimes is really killing me with its rounding thingies.. :(
     
  7. Mar 20, 2009 #6

    Doc Al

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    Good catch, RB!

    You will probably not be surprised to learn that you are not the first to complain about problems with mastering physics. :yuck:
     
  8. Mar 20, 2009 #7

    Redbelly98

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    It was just a hunch.

    If sin(60°)=0.8660... is rounded to 0.87, one gets 1.4616→1.46
     
  9. Mar 20, 2009 #8

    Doc Al

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    That's where I was going to go. I wondered if the 60° was meant to be "exact" or really to only 2 sig figs (or 1 :wink:). But if that were true, then the answer should have been rounded to 1.5, which won't fly with Mastering Physics...
     
  10. Mar 20, 2009 #9
    alright... another questions.. still regarding to internal reflection.. :)

    YF-33-42.jpg

    A light ray in air strikes the right-angle prism shown in the figure (Intro 1 figure) angle B=32.0. This ray consists of two different wavelengths. When it emerges at face AB, it has been split into two different rays that diverge from each other by 8.50 degree .

    Find the index of refraction of the prism for each of the two wavelengths.

    so far,, what i did is...

    for wavelength one...
    n1 sin theta1 = n2 sin theta2
    1 . sin 90 = n2 . sin 12
    n2 = sin 90 / sin 12 = 4.809

    for wavelength two...
    n1 sin theta1 = n2 sin theta2
    1 . sin 90 = n2 sin (12+8.5)
    n2 = sin 90/ sin 20.5 = 2.85

    anyone knows where the mistakes are...??
    thanks
     
  11. Mar 20, 2009 #10

    Doc Al

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    This problem has nothing to do with total internal reflection, just plain old refraction. The first refraction is trivial--the rays pass straight through. You only have to worry about the second surface, where it goes from glass to air. What's the angle of incidence at that surface? What are the angles of refraction?
     
  12. Mar 20, 2009 #11

    Redbelly98

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    Yes, they (Mastering Physics) have botched this one. It's not uncommon in physics problems to give nice round figures and assume they're "exact", as they seem to have done with the prism angles here.

    A commercial prism will typically have better than 0.05 degrees (and sometimes better than 0.01 degrees) tolerance. Not that introductory physics students should be expected to know that!
     
  13. Mar 20, 2009 #12
    ahh..i thought it was total internal reflection since the ray coming in 90 degree,, but.. for total internal reflection 90 degree should be for the angle of refraction. :biggrin:

    so, we dont need to consider from air to glass?
    yess...yess...
    blame mastering physics :wink:
     
  14. Mar 20, 2009 #13

    Doc Al

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    Nothing to consider. The ray passes straight through (as shown in the diagram) since the angle of incidence is 0 degrees.

    All the action takes place at the glass to air surface.
     
  15. Mar 20, 2009 #14
    so, for the first wavelength
    the angle of incidence is 32 degree, and the angle of refraction is 12 ??

    for the second wavelength
    the angle of incidence is 32 degree, and the angle of refraction is 20.5 ??

    am i on the right track??
     
  16. Mar 20, 2009 #15

    Doc Al

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    Your angles are off. Remember that the angles are measured from the normal to the surface, so you have to figure them out.
     
  17. Mar 20, 2009 #16
    which one is not correct??
    the angle of the incidence or the angle of refraction??
     
  18. Mar 20, 2009 #17

    Doc Al

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    None of them are correct. :frown:
     
  19. Mar 20, 2009 #18
    LOL
    :biggrin:

    ic,,is this because i took the normal wrongly..??
     
  20. Mar 20, 2009 #19

    Doc Al

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    Beats me, since you didn't show your normal on the diagram. Looks like you took the angle of incidence measured from parallel to the surface instead of normal, and the angles of refraction measured from the horizontal line. :uhh:
     
  21. Mar 20, 2009 #20
    LOL...
    how did you know?? :shy:

    ic,, i'll work on it :)
     
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