Derive the formula:1/v+1/u=1/f

The  formula is a relation between the object distance u ,
inmage distance v and the focal length from the pole of the concave mirror. The formula
is valid for the images in convex mirror and even for the images in
lens.

We consider the image formed by aconcave mirror whose
focal length is f and whose  radius of curvature is r =
2f.

Let P be the pole of the concave mirror. Let P, F , C
be the pole, focucal point , and centre of curvature along principal axis . So, PC = 2PF
, as r = 2f.

Let AB  be a vertically standinding object
beyond  C on the principal axis. 

Then the ray starting
from B parallel to  principal axis incident on the mirror at D reflects through the
focus F. Let the reflected ray be CFB' .

The another ray
starting from B through the centre C  incident on the mirror at E retraces its path by
reflection being normal to the mirror.

Now BE and DF
produced meet at B'.

Now drop the perpendicular from B' to
PC to meet at A'.

Drop the perpendicular from D to  PC to
meet at G.

Now PF = f , the focal length.  PA = u object
distance  from the mirror. PA' = v the image distance.

Now
consider the similar triangles ABC and A'BC.

AB/AB' = AC/
A'C =( PU-PC)(PC-PA') = (u-2f)/((2f-v).....(1)

Consider the
similar triangles DFG and A'B'F.

DG/A'B'  = PF/PA'
PF/(PA'-PF)= f/(v-f)... (2)

DG = AB. So (2) could be
rewritten as:

AB/A'B' + f/v
....................(3).

From (2) and (3), LHS being same ,
we can equate right sides.

(u-2f)/(2f-v) =
f/(v-f).

(u-2f)(v-f) =
(2f-v)f.

uv-2fv -fu +2f^2 = 2f^2
-fv

uv = fu +fv

Dvide by
uvf;

1/f = 1/v+1/u.