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, then the image formed is virtual. If the size of the image formed are equal, then the focal length of the lens will be

• A. 19 cm
• B. 17 cm
• C. 21 cm
• D. 11 cm

Answer 1

The correct option is D 11 cm

Since, both real and virtual image are formed by lens, it must be a convex lens.

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

$⟹\frac{u}{v}+1=\frac{u}{f}$

$⟹\frac{1}{m}+1=\frac{u}{f}$

$⟹m=\frac{f}{f+u}$

When, $u=-16cm$, magnification $m=-1$ since image formed is real

• $⟹-1=\frac{f}{f-16}$
  

When, $u=-6cm$, magnification $m=1$ since image formed is virtual

$⟹1=\frac{f}{f-6}$

Hence, $-\frac{f}{f-16}=\frac{f}{f-6}$

$⟹-(f-6)=f-16$

$⟹2f=22$

$⟹f=11cm$

Answer-(D)