Q. An object placed 10 $cm$ in front of a lens has an image $20\ cm$ behind the lens. What is the power of the lens (in $dioptres$)
A) 1.5
B) 3.0
C) $- 15.0$
D) +15.0
Explanation
(d) $u = - 10\ cm,\ v = 20\ cm$
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{20} - \left( - \frac{1}{10} \right) = \frac{3}{20} \Rightarrow f = \frac{20}{3}cm$
Now $P = \frac{100}{f} = \frac{100}{20/3} = + 15\ D$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.