Question
Using lens maker’s formula, derive the thin lens formula $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$ for a biconvex lens.
✓
Answer
Ace to lens maker's formula,
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=(\text{n}_{21}-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)\ ...(1)$
When object is at placed at infinity,
$\text{u}=\infty$
Image is obtained at focus,
$\text{v}=\text{f}$
Using these values in Eq (1),
$\frac{1}{\text{f}}-\frac{1}{\infty}={(\text{n}_{21}-1)}\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$
$\Rightarrow\frac{1}{\text{f}}=(\text{n}_{21}-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)\ ...(2)$
$\therefore$ By Eq (1) and (2),
$\Rightarrow\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=(\text{n}_{21}-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)\ ...(1)$
When object is at placed at infinity,
$\text{u}=\infty$
Image is obtained at focus,
$\text{v}=\text{f}$
Using these values in Eq (1),
$\frac{1}{\text{f}}-\frac{1}{\infty}={(\text{n}_{21}-1)}\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$
$\Rightarrow\frac{1}{\text{f}}=(\text{n}_{21}-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)\ ...(2)$
$\therefore$ By Eq (1) and (2),
$\Rightarrow\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
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