Mirrors 1.2 2005/10/26 11:59:10 GMT-5 2005/10/28 13:52:53.151 GMT-5 Paul Padley padley@rice.edu Paul Padley padley@rice.edu geometric optics Mirrors We come up with the mirror equation.

Notice that in Mirrors we get a virtual image to the right of the surface. Thus for mirrors we say that s i is positive to the left of the mirror. This allows to retain correspondence between s i being negative and an image being virtual. Again we use the small angle approximation. By inspection of the figure we see that 2 θ i = α + β and θ i = α + γ . Now we multiply the second equation by two and subtract the first equation from it and we get: 2 α + 2 γ − α − β = 0 or α − β = − 2 γ . Using the small angle approximation we see that this is h s o + h s i = − 2 h r where I have used the fact that s i is negative to the right of the mirror. So I can write the mirror equation as 1 s o + 1 s i = − 2 R or 1 s o + 1 s i = 1 f where for a mirror 1 / f = − 2 / R