Question

in the derivation of convex mirror formula why pc=2f

Answer 1

see diagram.

Let a ray parallel to the principal axis be incident on the mirror at M. It makes an angle Theta with the normal at M.  The reflected ray also makes angle theta with the normal at the mirror.  Extend the rays backwards. The ray meets the principal axis on the other side at F. The normal meets the principal axis at center of the curvature of mirror at C.  let PC = R.  P is the geometrical center of mirror.

Now, angle MCP is theta, as CP is parallel to incident ray.
Now angle FMC is theta as it is equal to angle theta on the other side of intersection at M.

Now angle MFP is 2 * theta as it is the external angle of Δ MFC and it is sum of angles FMC and FCM. 

tan (2 theta) = P'M / P'F,     
   as the angle theta is small,  2 theta is nearly = P'M / P'F
tan (theta) = P'M / P'C ,    
    as the angle theta is small, theta is nearly = P'M/ P'C

So  P'M / P'F = 2 P'M/P'C

So  P'C = 2 P'F

Now since P'P is very small , compared to other distances,
   P'C is nearly = PC   and P'F is nearly = PF.

So    PC = 2 PF          R = 2 f

We call F as focal point, as rays parallel to principal axis seem to diverge from this point, after reflection from the mirror.  f is the focal length.