Question

The distance between two point objects P and Q is 32 cm. A convex lens of focal length 15 cm is placed between them so that the images of both the objects are formed at the same place. The distance of P from the lens could be

• A. 12 cm (or) 20 cm
• B. 16 cm (or) 7 cm
• C. 18 cm (or) 12 cm
• D. 20 cm (or) 7 cm

Answer 1

The correct option is A

12 cm (or) 20 cm

Since images are formed at the same place, So one image must be real and other must be virtual
$|x|+|y|=32$

For P

$\frac{1}{y}-\frac{1}{-x}=\frac{1}{5}$

$\Rightarrow \frac{1}{-y}+\frac{1}{x}=\frac{1}{15}$

For Q

$\frac{1}{-y}-\frac{1}{-(32-x)}=\frac{1}{15}$

$\Rightarrow \frac{1}{-y}+\frac{1}{32-x}=\frac{1}{5}$

Adding (1) and (2), we get

$\frac{1}{x}+\frac{1}{32-x}=\frac{2}{15}$

$\frac{32}{x(32-x)}=\frac{2}{15}$

$\Rightarrow 32x-x^2=240$

$\Rightarrow x^2-32x+240=0$

$\Rightarrow x^2-20x-12x+240=0$

$\Rightarrow x(x-20)-12(x-20)=0$

$\Rightarrow (x-12)(x-20)=0$

$\Rightarrow x=12cm,20cm$