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

neela | High School Teacher | (Level 3) Valedictorian

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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.