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.
$\frac{-\text{u}}{-\text{v}}=\frac{-\text{f}}{-\text{f}+\text{v}}$ $\Rightarrow\text{uf}-\text{uv}=\text{vf}$ $\Rightarrow\text{uv}=\text{uf}-\text{uf}$ 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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