Q. A converging lens is to project the image of a lamp 4 times the size of the lamp on a wall at a distance of 10 m from the lamp. The focal length of the lens is
A) 1.6 m
B) 2.67 m
C) 4.4 m
D) -1.6 m
Explanation
(a) Magnification $m = \frac{v}{u}$
$- 4 = \frac{v}{u}$
$Or\ \ \ v = \ - 4\ u$
$Now,\ v - u = 10$
$Or\ \ - 5u = 10$
$Or\ \ \ u = \ - 2\ m$
$v = 8\ m$
$Again,\ \ \frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$Or\ \ \ \frac{1}{f} = \frac{1}{8} - \frac{1}{- 2}$
$Or\ \ \frac{1}{f} = \frac{1}{8} + \frac{1}{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ = \frac{1 + 4}{8} = \frac{5}{8}$
$Or\ f = \frac{8}{5}m = 1.6\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.