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I Proof of image formation property of spherical mirrors

  1. May 25, 2016 #1
    Hi. I'm trying to proof the image formation property of a concave spherical mirror. I know you can do this easily with a particular choice of rays (namely one that hits the vertex and one that passes through the center of the sphere) but I would like to show that a generic ray yields the same result because

    1) Any two non-parallel rays will cross paths sonner or later so to use only two rays gives you a result but not a satisfactory proof.

    2) The rays chosen for this easy proof pass through special points of the problem's geometry and could be a particular result.

    I know there is a purely geometrical approach to this general proof I'm seeking but since I want to keep consistency with the work I've already done, I'm doing it with Analytical Geometry.

    The Mirror is given by the equation

    [itex]y=R-R\sqrt{1-\left(\frac{\displaystyle x}{\displaystyle R}\right)^2}[/itex],

    where [itex]R[/itex] is the radius of the mirror and its axis is parallel to the y-axis. The object in front of the mirror has coordinates [itex](x_{1},y_{1})[/itex].

    The ray reflected at the mirror's vertex has a line equation given by

    [itex]y=-\frac{\displaystyle y_{1}}{\displaystyle x_{1}}\,x[/itex]

    and the one passing through the mirror's center has a line equation

    [itex]y=R+\frac{\displaystyle y_{1}-R}{\displaystyle x_{1}}\,x{\displaystyle }[/itex].

    These rays cross paths at [itex]x_{2}=\frac{\displaystyle R}{\displaystyle R-2y_{1}}\,x_{1}[/itex] and showing that

    [itex]y_{2}=-\frac{\displaystyle R}{\displaystyle R-2y_{1}}\,y_{1}[/itex],

    we obtain the known amplification factor [itex]x_{2}=-\frac{\displaystyle y_{2}}{\displaystyle y_{1}}\,x_{1}[/itex].

    Now... let's consider a ray that hits the mirror at a point [itex](x_{0},y_{0})[/itex]. When this ray is reflected it will have a line equation

    [itex]y=y_{0}+\tan\left(2\arctan\left(-\frac{\displaystyle R}{\displaystyle x_{0}}\sqrt{1-\left(\frac{\displaystyle x_{0}}{\displaystyle R}\right)^2}\right)-\arctan\left(\frac{\displaystyle y_{1}-y_{0}}{\displaystyle x_{1}-x_{0}}\right)\right)(x-x_{0})[/itex].

    Under the paraxial approximation (which allows us to consider [itex]x_{0}\ll R[/itex]) and using the property of arctan addition we can rewrite it as

    [itex]y=y_{0}+\tan\left(\arctan\left(\frac{\displaystyle 2x_{0}R}{\displaystyle R^2-x^2_{0}}\right)-\arctan\left(\frac{\displaystyle y_{1}-y_{0}}{\displaystyle x_{1}-x_{0}}\right)\right)(x-x_{0})[/itex].

    I know we can simplify it a bit more by using the [itex]\tan(a+b)[/itex] formula but the way I kept it above is easier to contemplate. I have tested all these rays with mathematica and they converge to the same point so I know everything up to here is ok. My problem is:

    If I equal this last ray to one of the others (the one that reflects at the mirror's vertex, for instance) I'll obtain the x at which they cross paths as a function of [itex]x_{0}[/itex] and [itex]y_{0}[/itex]. Since all the rays converge to the same point (given the paraxial approximation) I would not expect them to depend upon the point at which they touch the mirror.

    What am I missing here? There is some simplification I did not do? Could you lend me a hand?

    Thank you very much.
  2. jcsd
  3. May 26, 2016 #2
    I haven't attempted to follow all of your equations, but since no one has responded yet I'll ask one question - you derived a general equation and then applied the paraxial approximation. So why do you still have trig functions in that equation? For instance, if x0 ≪ R then R2 - x02 ≈ R2 so the argument of the first arctan can be written 2x0/R which << 1 in the paraxial approximation. Etc.

    What I'm guessing is that if you rewrite the equation fully making use of the paraxial approximation then either x0 and y0 won't appear, or if they do they would be with other expressions that they are much less than.
    Last edited: May 26, 2016
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