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Homework Help: Corner Reflector

  1. Nov 20, 2007 #1
    [SOLVED] Corner Reflector

    1. The problem statement, all variables and given/known data

    A corner reflector is to be made from a triangular prism with index of refraction n = 1.86, as shown in the diagram below. What is the maximum angle, with respect to the normal to the front surface of the prism, θ, such that total reflection will occur?

    2. Relevant equations

    [tex]n_{air}sin\theta_1 = n_{prism}sin\theta_2[/tex]

    3. The attempt at a solution

    I tried doing
    [tex] 1 * sin\theta_1 = 1.86 * sin\ 90[/tex]

    solving for [tex] \theta_1 [/tex] gives arcsin (1.86) which is undefined.

    Any help appreciated

    Here is the graphic for the problem

    Attached Files:

    Last edited: Nov 20, 2007
  2. jcsd
  3. Nov 20, 2007 #2


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    I think the limiting effect is the second inner reflection. As I recall total reflection occurs when the angle of refraction passes the point where one of the rays must be parallel to the surface.

    In your attempt you have I believe the wrong angle set at 90deg. In the inner reflection the angle of the "refracted" beam outside the back of the prism will be 90deg in the limiting case. Thence you get minimum inner angle of arcsin(1/1.86). Any smaller angle will not reflect but refract with the beam in air less than 90deg from normal.

    You'll still have to take this minimum inner angle for second reflection and determine the maximum angle at which the beam enters the prism using geometry inside and the index of refraction for the initial refraction.
    Last edited: Nov 20, 2007
  4. Nov 20, 2007 #3
    Could you expand on the last part, I'm still not getting it
  5. Nov 22, 2007 #4


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    Case 1: The light enters the prism 0deg from normal. It hits one back pane at 45deg reflects, hits the other back pane at 45deg reflects then leaves the prism.

    Case 2: The light enters the prism at 10deg from normal. It refracts to something less than 10deg normal to first surface. Let's just say 9deg.

    It then hits the first back pane at 45+9=54 deg, reflects and hits the second back face at 45-9 = 36deg. If this is less than the minimum angle I mentioned then it will not reflect 100% but rather the light will pass through the back surface.

    You must figure this second angle in terms of the original angle of entry, theta. The bigger the value of theta the smaller this second reflections angle from normal.
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