Question

Derive the thin lens formula.

Answer 1

Consider an object placed in front of a concave lens of focal length " f " on the principle axis of the lens. Concave lens forms a virtual and erect image at a distance of " q " from the optical centre of the lens as shown in the diagram below.

consider similar triangles $\Delta$CO'P and $\Delta$II'P

$\frac{O{O}^{\prime }}{I{I}^{\prime }}$ = $\frac{O^{\prime }P}{I{I}^{\prime }P}$

$\frac{O{O}^{\prime }}{I{I}^{\prime }}=\frac{p}{q}$ ...................( 1 )

similarly in similar triangle $\Delta$EFP and $\Delta$II'F

$\frac{EP}{I{I}^{\prime }}$ = $\frac{PF}{I^{\prime }F}$

$\frac{O{O}^{\prime }}{I{I}^{\prime }}=\frac{PF}{PF-P{I}^{\prime }}$

$\frac{O{O}^{\prime }}{I{I}^{\prime }}-\frac{f}{f-q}$ ........................( 2 )

equation ( 1 ) and ( 2 )

$\frac{p}{q}-\frac{f}{f-q}$

p ( f - q ) = fq

pf - pq = fq

Dividing both sides by "pqf"

$\frac{1}{q}-\frac{1}{f}=\frac{1}{p}$

$-\frac{1}{q}+\frac{1}{f}=\frac{1}{p}$

For a concave lens:

$\frac{1}{f}-\frac{1}{p}+\frac{1}{q}$