Question

If an object is placed at $A(OA>f)$; where $f$ is the local length of the lens, the image is formed at $B$. A perpendicular is erected at $O$ and $C$ is choosen on it such that the $\angle BCA$ is a right angle. Then the value of $f$ will be:

• A. $\frac{AB}{O{C}^2}$
• B. $\frac{\left(AC\right)\left(BC\right)}{OC}$
• C. $\frac{O{C}^2}{AB}$
• D. $\frac{\left(OC\right)\left(AB\right)}{AC+BC}$

Answer 1

The correct option is C $\frac{O{C}^2}{AB}$


From lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{f}=\frac{u-v}{vu}$
$\Rightarrow f=\frac{vu}{u-v}$
$(\because u=-OA,v=OB)$
Or, $f=\frac{OB\times (-OA)}{-OA-OB}$
Or, $f=\frac{OA\times OB}{OA+OB}$
Or, $f=\frac{OA\times OB}{AB}.....\left(1\right)$
From geometry,
$B{C}^2=O{B}^2+O{C}^2.....\left(i\right)$
$A{C}^2=O{A}^2+O{C}^2......\left(ii\right)$
On adding $\left(i\right)$ and $\left(ii\right)$
$B{C}^2+A{C}^2=O{B}^2+O{A}^2+2O{C}^2$
$\Rightarrow A{B}^2=O{B}^2+O{A}^2+2O{C}^2$
$\because [A{B}^2=(OB+OA{)}^2]$
$\Rightarrow O{B}^2+O{A}^2+2\left(OB\right)\left(OA\right)=O{B}^2+O{A}^2+2O{C}^2$
$\Rightarrow OB\times OA=O{C}^2................\left(2\right)$
On putting $\left(2\right)$ in $\left(1\right)$,
$\Rightarrow f=\frac{O{C}^2}{AB}$