Question

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 right
D	erect, diminished	12 cm to the left
E	inverted enlarged	12 cm to the left

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (B)

When an object is placed between $2 f$ 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

$ \frac{1}{f}=\frac{1}{v}-\frac{1}{u} $
$ \Rightarrow \frac{1}{-20}=\frac{1}{v}+\frac{1}{30} (as \mu=-30 \,\,cm )$
$ \Rightarrow v=-\frac{(20)(30)}{20+30}$
Thus, distance $=-12 \,cm$ ( to the left of the diverging lens)