Optics Ques 207
- Consider a concave mirror and a convex lens (refractive index $=1.5$ ) of focal length $10 cm$ each, separated by a distance of $50 $ $cm$ in air (refractive index =1) as shown in the figure. An object is placed at a distance of $15 $ $cm$ from the mirror. Its erect image formed by this combination has magnification $M _1$. When the set-up is kept in a medium of refractive index $\frac{7}{6}$, the magnification becomes $M _2$. The magnitude $\frac{M _2}{M _1}$ is
(2015 Adv.)
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Answer:
Correct Answer: 207.$(7)$
Solution:
- Case I
Reflection from mirror
$ \begin{aligned} \frac{1}{f} & =\frac{1}{v}+\frac{1}{u} \Rightarrow \frac{1}{-10}=\frac{1}{v}+\frac{1}{-15} \\ \Rightarrow \quad v & =-30 \end{aligned} $
For lens
$ \left|M _1\right|=\left|\frac{v _1}{u _1}\right|\left|\frac{v _2}{u _2}\right|=(\frac{30}{15}) \quad (\frac{20}{20})=2 \times 1=2 \quad $ (in air)
Case II
For mirror, there is no change.
For lens, $\quad \frac{1}{f _{\text {air }}}=(\frac{3 / 2}{1}-1 )\quad (\frac{1}{R _1}-\frac{1}{R _2})$
$ \frac{1}{f _{\text {medium }}}=(\frac{3 / 2}{7 / 6}-1 )\quad (\frac{1}{R _1}-\frac{1}{R _2}) $
with
$ f _{\text {air }}=10 $ $cm $
We get
$ \frac{1}{f _{\text {medium }}}=\frac{4}{70} cm^{-1} $
$ \begin{aligned} & \frac{1}{v}-\frac{1}{-20}=\frac{4}{70} \\ & \frac{1}{v}+\frac{1}{20}=(\frac{2}{7}) \quad (\frac{2}{10})=\frac{4}{70} \\ & \frac{1}{v}=\frac{4}{70}-\frac{1}{20} \quad \Rightarrow \quad v=140 \\ &\left|M _2\right|=\left|\frac{v _1}{u _1}\right|\left|\frac{v _2}{u _2}\right|=(\frac{30}{15}) \quad( \frac{140}{20}) \\ &=(2) (\frac{140}{20})=14 \\ & \frac{|M _2|}{|M _1|}=\frac{14}{2}=7 \end{aligned} $
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