Question

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 O and C is chosen such that $\angle BCA={90}^o$. Let OA = a, OB = b and OC = c. Then the value of f is

• A. $\frac{(a+b{)}^3}{c^2}$
• B. $\frac{(a+b)c}{(a+c)}$
• C. $\frac{c^2}{a+b}$
• D. $\frac{a^2}{a+b+c}$

Answer 1

The correct option is C $\frac{c^2}{a+b}$

For a lens,

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

$⟹\frac{1}{b}+\frac{1}{a}=\frac{1}{f}$

$⟹f=\frac{ab}{a+b}$

From pythagoras theorem,

$A{B}^2=A{C}^2+B{C}^2$

$⟹(a+b{)}^2=(a^2+c^2)+(b^2+c^2)$

$⟹c^2=ab$

$⟹f=\frac{c^2}{a+b}$