Browse Questions

# An object is placed at $A\;(OA > f)$ Here, f is the focal length of the lens. The image is formed at B, A perpendicular is erected at 0 and C is chooses such that $\angle BCA =90^{\circ}$. Let $OA=a,OB=b$ and $OC=c$ then f is :

$(a)\;\frac{c^2}{a^2+b^2+c^2} \\ (b)\;\frac{c}{a+b} \\ (c)\;\frac{c^2}{a+b} \\ (d)\;None$

From $\large\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ we have
$\large\frac{1}{b}+\frac{1}{a}=\frac{1}{f}$
or $f= \large\frac{ab}{a+b}$----(i)
Further $AC^2+BC^2=AB^2$
or $(a^2+c^2) +(b^2+c^2)=(a+b)^2$
or $a^2+b^2+2c^2=a^2+b^2+2ab$
$\therefore ab=c^2$
Substituting this in equ (i) we gey,
$f= \large\frac{c^2}{a+b}$
Hence c is the correct answer.