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Angle of incidence and reflection of a parabola?

  1. Oct 13, 2006 #1
    I've been playing around with parabolic mirror calculations in my free time, and right now I'm trying to derive the equation for the line that models the angle of reflection off of a point. I understand how to find the angle of incidence and that the reflection will have the same angle off the tangent line to that point, but I can't figure out how to get the equation for the reflection line. Any help would be great.

    PS Basically what I'm doing is just finding the focus through geometric methods and seeing the intersection of these reflection lines. Thanks.
  2. jcsd
  3. Oct 13, 2006 #2
    Okay I'm working on it, but I just spent time and found a false answer.

    I said the andle of incidence is going to be the angle of the tangent line above the horizontal line containing the x intercept. So the angle is (in degrees)

    90 - arctan[(f'(x) - xint) / x].

    Also, if you assumed that the tangent is the x axis on a new plane the slope of the reflection line would be the negative slope of the incidence line. To find the actual slop of the reflection line you could add the slope of the tangent line to the negative slope of the incidence line.


    slope of reflection = -tan(angle of incidence) + f ''(x)

    It doesn't seem to work though.
  4. Oct 13, 2006 #3


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    The angles for a line parallel to the axis of the parabola is relatively simple. The angles for other lines are extremely difficult to calculate. You want parallel to the axis anyway since those are the only rays that pass through the focus.

    To avoid vertical lines (which have infinite slope) assume your parabola has axis on the positive x-axis, vertex at (0,0). Say, [itex]y= \sqrt{x}[itex]. Imagine a beam of light coming along the line y= c. It will intersect the parabola at (c2, c). The derivative of [itex]y= \sqrt{x}= x^{\frac{1}{2}}[/itex] is [itex]\frac{1}{2}x^{-\frac{1}{2}}[/itex] which, at (c2,c), is [itex]\frac{1}{2c}[/itex]. The angle of incidence is the arctan of that. The angle of reflection will be the same. If you draw that line of reflection up through the point of reflection, you will see that angle that line makes with the line of tangency above that point is the same as below ("vertical angles" from geometry) and so is the same as the angle of incidence. But that means that the angle that line makes with the x-axis is twice the angle of incidence. Since
    [tex]tan(2\theta)= \frac{2 tan(\theta)}{1- tan^2(\theta)}[/tex]
    the tangent of that angle is
    [tex]\frac{\frac{1}{c}}{1- \frac{1}{4c^2}}= \frac{4c}{4c^2- 1}[/tex]
    Thus, the equation of that line is
    [tex]y= \frac{4c}{4c^2- 1}(x- c^2)+ c[/tex]
    That will cross the x-axis (the axis of the parabola) when y= 0:
    [tex]\frac{4c}{4c^2- 1}(x- c^2)+ c= 0[/tex]
    [tex]\frac{4c}{4c^2- 1}(x- c^2)= -c[/tex]
    [tex]x- c^2= -c\left(\frac{4c^2- 1}{4c}\right)= -c^2+ \frac{1}{4}[/tex]
    so that
    [tex]x= \frac{1}{4}[/tex]
    independent of c.
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