Question

Prove 1/v plus1/u=1/f

Answer 1

$hey \: mate$



•here is your answer...,




•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..







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.by Ishita..
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