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Optics Question

  1. Aug 1, 2004 #1
    The Question:
    A 45-45-90 prism is immersed in water. What is the minimum index or fefraction the prism must have if it is to reflect totaly a ray incident normally on one of its faces.

    Now, we have found this question a bit hard to understand.
    We assume that light is hitting one of the sides at 90 degrees (normally) and so will pass into the prism without refraction. Thus, it hits the other side of the prism at an angle of incidence of 45 degrees.

    This is where we get stuck, we are assuming that it is totaly internaly reflecting and by using snells law nSin(45) = Sin(90) we find that n, the refractive index of the prism is 1.88

    This complies with
    the fact that to be totaly internaly reflected the light must travel from a high refractive index (1.88) to a low refractive index (1 for air)

    If we are on the right track, please let us know!
     
  2. jcsd
  3. Aug 1, 2004 #2

    Doc Al

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    Staff: Mentor

    You didn't get n=1.88 using that equation. I assume you meant to write Snell's law as: [itex]n_1 sin\theta_1 = n_2 sin\theta_2[/itex]. And, yes, the critical angle is found by setting [itex]\theta_2 = 90[/itex].

    If you assumed going from n=1.88 to air (n=1) I am very curious how you got your answer. (The prism is immersed in water, not air.)
     
  4. Aug 1, 2004 #3
    I think it should be n1*Sin(45) = n2*sin(90), not nSin(45) = Sin(90)

    which is

    n1*sin(45) = 1.333
    n1 = 1.333/sin 45

    n1 is approx 1.88
     
  5. Aug 2, 2004 #4
    Yes, i made a few erros in that post. Had to make it to the next class of uni.

    Yes, the prism is in water. relinquished managed to type in what i failed to. He got the same answer to.

    I just for some reason replaced water with air when i typed this up.

    So, would 1.88 be the minumum index of refraction?
     
  6. Aug 2, 2004 #5

    Doc Al

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    Staff: Mentor

    I knew you must have, since your answer was correct.

    Yep.
     
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