Question

lf the focal length of the lens is $20\, cm$. what is the distance of the image from the lens in the following figure?

• A. 5.5 cm
• B. 7.5 cm
• C. 12.0 cm
• D. 20.0 cm

Answer 1

Answer: (B)

7.5 cm

Answer 2

According to figure, the point on the right side of the lens of which rays converge will behave as virtual object at the lens.
$\therefore u = + 12\, cm, f = 20\, cm$
By lens formula
$\frac{1}{f}= \frac{1}{v}- \frac{1}{u}$
$\therefore \frac{1}{20}= \frac{1}{v}- \frac{1}{12}$
or $\frac{1}{v}=\frac{1}{20} + \frac{1}{12} = \frac{3+5}{60} = \frac{8}{60}$
or $u = \frac{60}{8} = 7.5\, cm$
So, image will be formed on same side of the virtual object at a distance of 7.5 cm from the lens.