Question

If an object is placed at A (OA > f); Where f is the focal length of the convex lens the image is found to be formed at B. A perpendicular is erected at O and C is chosen on it such that the angle $\angle$BAC is a right angle. Then the value of f will be

• A. AB/OC${}^2$
• B. (AC)(BC)/OC
• C. (OC)(AB)/AC+BC
• D. OC${}^2$/AB

Answer 1

Answer: (D)

OC${}^2$/AB

Given $OA(=u)>f$ and $u=-ive,f=+ive\left(convexlens\right)$ ,

From lens equation:

$v=\frac{uf}{u+f}$

$v$ will be +ive i.e. the image will be formed on other side of the lens.

In right angled triangle $ACB$

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

But in right angled $△AOC$

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

and in right angled $△BOC$

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

and $AB=OA+OB$

Hence $(OA+OB{)}^2=O{A}^2+O{C}^2+O{B}^2+O{C}^2$

or $O{A}^2+O{B}^2+2OA\times OC=O{A}^2+2O{C}^2+O{B}^2$

or $2uv=2O{C}^2$ (as $OA=u,OB=v$)

or $uv=O{C}^2$

Now again by lens equation, (for $u=-ive$)

$f=\frac{uv}{u+v}=\frac{O{C}^2}{OA+OB}=\frac{O{C}^2}{AB}$