Q. The size of the image of an object, which is at infinity, as formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 20 cm is placed between the convex lens and the image at a distance of 26 cm from the convex lens, calculate the new size of the image
A) 1.25 cm
B) 2.5 cm
C) 1.05 cm
D) 2 cm
Explanation
(b) Image formed by convex lens at $I_{1}$ will act as a virtual object for concave lens. For concave lens
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$or\ \ \frac{1}{v} - \frac{1}{4} = \frac{1}{- 20}$
Or $v = 5$cm
Magnification for concave lens
$m = \frac{v}{u} = \frac{5}{4} = 1.25$
As size of the image at $I_{1}$ is 2 cm. Therefore, size of image at $I_{2}$ will be $2 \times 1.25 = 2.5$ cm.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.