Q. Two lenses are placed in contact with each other and the focal length of combination is $80\ cm$. If the focal length of one is $20\ cm$, then the power of the other will be
A) 1.66 $D$
B) 4.00 $D$
C) $- 1.00\ D$
D) $- 3.75\ D$
Explanation
(d) $\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \Rightarrow \frac{1}{80} = \frac{1}{20} + \frac{1}{f_{2}} \Rightarrow f_{2} = - \frac{80}{3}cm$
$\therefore\ $Power of second lens
$P_{2} = \frac{100}{f_{2}} = \frac{100}{- 80/3} = - 3.75\ D$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.