I'm trying to find the focal point and center of curvature of a concave mirror. Just using the radius for the center doesn't seem to work. I know C = 2f, but I'm not sure how to find f or C, given the radius of a perfect circle. is r = C and I'm just to drawing it right?

The object should be at the same place (but inverted) if it's placed at the center of curvature right?

my drawing

When I try to use an optics simulator the rays seem to bounce off something behind the mirror.

optics simulator


closed as off-topic by ACuriousMind, JamalS, Kyle Kanos, David Z Jan 16 '15 at 21:06

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  • $\begingroup$ A circular mirror is an approximation of a parabolic mirror. This approximation gets worse the further you are from the main/symmetry axis. So you would get better results if you would make the object a lot smaller, such that the parallel beam stays closer to the axis, or use a parabolic mirror. $\endgroup$ – fibonatic Jan 16 '15 at 16:35

You are correct in the way that you use the center and radius. You are also mostly correct that $C=2f$, however this is only true when the size of the mirror is small, compared to the radius of the mirror. The relationship $C=2f$ holds best for small angular diameters.

What you are noticing in your optical simulation is that when the angular diameter of the mirror is too large, the light rays that reflect off the edges of the mirror do not reflect into the focal point. As a result, there is not a single point where all light rays parallel to the principle axis focus anymore. This inconvenience is known as spherical aberration and is part of the reason why better optical systems use parabolic reflectors instead of spherical ones. However, your simulation is probably written such that all parallel rays are programmed to reflect into the focal point, and so the simulation is forced to be unphysical in order for that to happen. Take it is a lesson that simulations are great at modeling physical realities, but are only as good as their program.


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