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## Main Question or Discussion Point

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.

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.