Q. A convex lens forms an image of an object placed 20 cm away from it at a distance of 20 cm on the other side of the lens. If the object is moved 5 cm towards the lens, the image will move
A) 5 cm towards the lens
B) 5 cm away from the lens
C) 10 cm towards the lens
D) 10 cm away from the lens
Explanation
(d) Clearly, $2f = 20\ $cm or $f = 10\ cm$
Now, $u = - 15$ cm, $v = ?$
$F = 10\ cm$
$\frac{1}{v} - \frac{1}{- 15} = \frac{1}{10}$
Or $\ \frac{1}{v} + \frac{1}{15} = \frac{1}{10}\ $or$\ \ \frac{1}{v} = \frac{1}{10} - \frac{1}{15}$
Or $\ \frac{1}{v} = \frac{3 - 2}{30} = \frac{1}{30}$ or$\ v = 30\ cm$
The charge in image distance is $(30 - 20)$cm $ie$, 10 cm
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.