Q. An object is placed 30 cm to the left of a diverging lens whose focal length is of magnitude 20 cm. Which one of the following correctly states the nature and position of the virtual image formed? Nature of image Distance from lens
A) Inverted, enlarged 60 cm to the right
B) Erect, diminished 12 cm to the left
C) Inverted, enlarged 60 cm to the left
D) Erect, diminished 12 cm to the right
Explanation
(b) When an object is placed between $2f$ and $f$(focal length) of the diverging lens, the image is virtual, erect and diminished as shown in the graph. To calculate the distance of the image from the lens, we apply
[image:9_image94.jpeg]
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Longrightarrow \frac{1}{- 20} = \frac{1}{v} - \frac{1}{30}$
$\Rightarrow v = - \frac{(20)(30)}{20 + 30}$
$= - 12$ cm (to the left to the diverging lens.)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.