Question

An object is placed at a distance of 12 cm from a convex lens on its principal axis and a virtual image of certain size is formed. If the object is moved 8 cm away from the lens, a real image of the same size as that of the virtual image is formed. The focal length of the lens is

• A. 15 cm
• B. 16 cm
• C. 18 cm
• D. 19 cm

Answer 1

Answer: (B)

$\frac{1}{v_{1}}-\frac{1}{-12}=\frac{1}{f} $
$\Rightarrow \frac{1}{v_{1}}=\frac{1}{f}-\frac{1}{12}=\frac{12-f}{12 f}$
$v_{1}=\frac{12 f}{12-f}$
$m_{1}=\frac{I}{O}=\frac{v_{1}}{u}$
$=\frac{12 f}{(12-f)(-12)}-\frac{f}{(12-f)}$
$\frac{1}{v_{2}}-\frac{1}{-20}=\frac{1}{f} $
$\Rightarrow \frac{1}{v_{2}}=\frac{1}{f}-\frac{1}{20}=\frac{20-f}{20 f}$
$v_{2}=\frac{20 f}{20-f}$
$m_{2}=\frac{I}{O}=\frac{v_{2}}{u}$
$=\frac{20 f}{(20-f)(-20)}=-\frac{f}{(20-f)}$
$m_{1}=-\mu_{2}$
$-\frac{f}{12-f}=\frac{f}{20-f}$
$ \Rightarrow -20+f=12-f$
$f=16 \,c m$