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Why should i=e for minimum deviation in a triangular prism?

  1. Jan 23, 2016 #1


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    When I was solving questions on Ray optics I encountered a question wherein I had to prove that when a ray of light undergoes minimum deviation through a triangular prism then both the angles of refraction that is the first one at the incident point and the second one at the emergent point should be equal.

    Since I was not able to prove this, I looked into the solution for this problem and it was written that for minimum deviation to take place in a triangular prism the angles of incidence that is the angle which the incident ray makes with the normal at that point and the angle of emergence that is the angle which the emergence ray makes with the normal at the point of emergence should be equal. I could not access why this would happen
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  3. Jan 23, 2016 #2


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    It is possible to express the deviation as function of i, and then find the minimum with standard analysis.
    There is also a symmetry argument: i and e are exchangeable, the change in angle as function of i has to be symmetric around the point of i=e. As the angle is continuous, it will have a minimum or maximum there, and it is not hard to prove it is not a maximum.
  4. Jan 23, 2016 #3


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    A bit elucidating what @mfb said, the deviation angle is given by ##\delta## is given by
    \delta = \theta_{i1}+\arcsin \left((\sin \alpha) \sqrt{n^2-\sin^2\theta_{i1}} - \cos\alpha \sin\theta_{i1}\right) - \alpha
    where ##\theta_{i1}## is the incident angle on the first prism side and ##\alpha## is the prism's apex angle. Try to differentiate the above equation with respect to ##\theta_{i1}## and find the extrema from it. You can try if you want but it is obviously going to involve a train of chain rules.
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