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Reflecting Surface

  1. Jun 6, 2006 #1

    mrjeffy321

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    In the question, I am given a diagram of a curved surface which is symetic around the X axis which will reflect all incoming parallel light rays to the origin.
    A point on the curve (point P(x, y) is chooses, which happens to be the point the light reflects off the curve/mirror) and a tangent line is drawn as well as the reflected light rays path towards the origin. Angle measures are given for the angle between the tangent line and the incoming light ray (theta), between the tangent line and the reflect path of the light (also theta), and the angle measure between the X axis and the reflected path of the light (phi). We are also told that phi = 2*theta so that we can use an appropriate trigonometric identity.
    The question asks to determine a differential equation which describes the shape of the curve C.

    I know that the slope of the tangent line is the instantaneous slop of the curve at point P, also the derivative of the curve equation (dx/dy = y’). I also know that the slop of the line equals the tangent of theta.

    For reasons I am not totally sure about,
    y/x = tan (2*theta)
    So then using that “appropriate trigonometric identity”, we can re-write this as,
    y/x = 2tan(theta) / 1 – tan^2(theta)
    then we can substitute y’ in for tan(theta) to get,
    y/x = 2*(y’) / 1 – (y’)^2

    At this point, I am lost, I do not know how to proceed from here.
    I do know that the answer to the question is suppose to be:
    dy/dx = (-x + sqrt(x^2 + y^2)) / y
     
  2. jcsd
  3. Jun 6, 2006 #2
    At first glance, it looks like they are saying,

    y/x = tan (phi)

    and then just substituting in phi = 2*theta
     
  4. Jun 6, 2006 #3

    mrjeffy321

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    The reason I substituted,
    y/x = 2tan(theta) / 1 – tan^2(theta)
    for,
    y/x = tan (2*theta)
    was because that is what the teacher did in class while semi-solving the problem on the board.
     
    Last edited: Jun 6, 2006
  5. Jun 7, 2006 #4

    mrjeffy321

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    I figured it out (and by "I", of course I mean "someone else").

    I was left with this when last I stopped,
    y/x = 2(y') / 1 - (y')^2
    and the goal was to find an equation for y' (y' = ....)

    By multiplying both sides by x, then both by 1 - (y')^2, you are left with,
    y(1-(y')^2) = 2(y')x
    then distribute the y on the left into the parenthesis,
    y - y(y')^2 = 2(y')x
    then rearrange it to look like this,
    -y(y')^2 - 2x(y') + y = 0
    then treat this as a quadratic equation and solve for (y').

    In this end, the answer should be,
    y' = dy/dx = (-x + sqrt(x^2 + y^2)) / y

    It is always important to keep the goal in mind while solving this stuff, I quickly forgot the goal and was lost in all the symbol rearranging.
     
    Last edited: Jun 7, 2006
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