Question

An object is kept at a distance of $16 \,cm$ from a thin lens and the image formed is real. If the object is kept at a distance of $6\, cm$ from the same lens the image formed is virtual. If the size of the images formed are equal, the focal length of the lens will be

• A. 15 cm
• B. 17 cm
• C. 21 cm
• D. 11 cm

Answer 1

Answer: (D)

As the lens can form both real and virtual images,
it is a convex lens and $u=-v e$
$\therefore$ we get $\frac{1}{v_{1}}+\frac{1}{16}=\frac{1}{f} $
$\Rightarrow v_{1}=\frac{16 f}{16-f}$
And $\frac{1}{v_{2}}+\frac{1}{6}=\frac{1}{f}$
and $v_{2}=\frac{6 f}{6-f}$
As the images are of the same size
$\frac{v_{1}}{u_{1}}=-\frac{v_{2}}{u_{2}} $
Or $\frac{\frac{16 f}{16-f}}{16}=-\frac{\frac{6 f}{6-f}}{6} $
Or $-(6-f)=16-f$
Or $f=11 \,cm$