Ray Optics And Optical Instruments Ques 31
31. 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?
[2011]
(a) Virtual, upright, height $=1$ $ cm$
(b) Virtual, upright, height $=0.5$ $ cm$
(c) Real, inverted, height $=4 $ $cm$
(d) Real, inverted, height $=1 $ $ cm$
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Answer:
Correct Answer: 31.(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$
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u} \Rightarrow \frac{1}{20}=\frac{1}{v}+\frac{1}{30}$
$\frac{1}{v}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}$
$v=60$ $ 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.
BETA
