Question

Write the derivation of mirror formula. 

(OR) 

Derive \(\frac{1}{f}=\frac{1}{u}\,+\,\frac{1}{v} . \).

(OR) 

A student wants to find the image distance for a given object distance of a mirror. Then derive a formula for the mirror.

Answer 1

Derivation of mirror formula :

In the figure P = pole, C = centre of curvature and F= focus of the concave miror. 

Object AB is placed beyond C. Image AB’ is formed in between F and C. 

From the diagram triangles A’B’C and ABC are similar triangles. = AB/A'B' = BC/B'C ...........(1)

Triangles A'B'F and DNF are similar triangles.

DN/A'B' =FN/ B'F or AB = DN

AB/A'B' = FN/ B'F .............(2)

From (1) and  (2)

 BC/B'C = FN/B'F

BC/B'C = FP/B'F ..............(3)

Hera FP = FN as AD is very closer to principal axis.

and also PF = focal length = f

PB' = image distance (v)

PB = object distance (u)

PC = radius of curvature =2f

BC = PB - PC = u - 2f

B'C = PC - PB' = 2f - v

B'F = PB'- PF = v - f

Then (3) BC /B'C = FP / B'F

(PB - PC)/(PC - PB') = (FP)/(PB' - PF )

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

By using sign convention 

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

2f² - vf = -uf + uv + 2f²  -2vf

vf + uf = uv

Dividing both sides with uvf then 

Hence , 1/f=1/v - 1/u ............(4)

The relation between focal length (f) object distance (u) and the image distance (v) is called Mirror Formula ,Which is 1/f = 1/v - 1/u .