Question
Derive the lens formula $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$ for a thin concave lens, using the necessary ray diagram.

Answer

The formation of image by a concave lens 'L' is shown in fig. AB is object and A' B' is the image. Triangles ABO and A' B' O are similar

$\frac{\text{AB}}{\text{A}'\text{B}}=\frac{\text{OB}}{\text{OB}'}\dots(\text{i})$

Also triangles NOF and A' B' F are similar

$\frac{\text{NO}}{\text{A}'\text{B}}=\frac{\text{OP}}{\text{FB}'}$

But NO = AB

$\frac{\text{AB}}{\text{A}'\text{B}'}=\frac{\text{OB}}{\text{FB}'}\dots(\text{i})$

Comparing equation (i) and (ii)

$\frac{​\text{OB}​}{\text{OB}'}=\frac{\text{OF}}{\text{FB}'}$

$\Rightarrow\frac{\text{OB}}{\text{OB}'}=\frac{\text{OF}}{\text{OF}-\text{OB}'}$

Using sign conventions of coordinate geometry

OB = -u, OB' = -v, OF = -f

$\frac{-\text{u}}{-\text{v}}=\frac{-\text{f}}{-\text{f}+\text{v}}$

$\Rightarrow\text{uf}-\text{uv}=\text{vf}$

$\Rightarrow\text{uv}=\text{uf}-\text{uf}$

Dividing throughout by uvf, we get

$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$

This is the required lens formula.

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