Ray Optics and Optical Instruments - Result Question 33
33. A biconvex lens has a radius of curvature of magnitude $20 cm$. Which one of the following options best describe the image formed of an object of height $2 cm$ placed $30 cm$ from the lens?
(a) Virtual, upright, height $=1 cm$
(b) Virtual, upright, height $=0.5 cm$
[2011]
(c) Real, inverted, height $=4 cm$
(d) Real, inverted, height $=1 cm$
Show Answer
Answer:
Correct Answer: 33. (c)
Solution:
- (c) $R=20 cm$
$h_0=2$
$u=-30 cm$
We have, $\frac{1}{f}=(\mu-1)(\frac{1}{R_1}-\frac{1}{R_2})$
$=(\frac{3}{2}-1)[\frac{1}{20}-(-\frac{1}{20})]$
$\Rightarrow \frac{1}{f}=(\frac{3}{2}-1) \times \frac{2}{20}$
$\therefore f=20 cm$
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \Rightarrow \frac{1}{20}=\frac{1}{v}+\frac{1}{30}$
$\frac{1}{v}=\frac{1}{20}-\frac{1}{30}=\frac{10}{600}$
$v=60 cm$
$m=\frac{h_i}{h_0}=\frac{v}{u}$
$\Rightarrow h_i=\frac{v}{u} \times h_0=\frac{60}{30} \times 2=-4 cm$
So, image is inverted.
