Optics 7 Question 32
33. 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 $M _1$.
(2015 Adv.) 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
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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
(in air)
$$ \left|M _1\right|=\left|\frac{v _1}{u _1}\right|\left|\frac{v _2}{u _2}\right|=\frac{30}{15} \quad \frac{20}{20}=2 \times 1=2 $$
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
$$ \frac{1}{f _{\text {medium }}}=\frac{4}{70} cm^{-1} $$
$$ \begin{aligned} & \frac{1}{v}-\frac{1}{-20}=\frac{4}{70} \\ & \frac{1}{v}+\frac{1}{20}=\frac{2}{7} \quad \frac{2}{10}=\frac{4}{70} \\ & \frac{1}{v}=\frac{4}{70}-\frac{1}{20} \quad \Rightarrow \quad v=140 \\ &\left|M _2\right|=\left|\frac{v _1}{u _1}\right|\left|\frac{v _2}{u _2}\right|=\frac{30}{15} \quad \frac{140}{20} \\ &=(2) \frac{140}{20}=14 \\ & \frac{M _2}{M _1}=\frac{14}{2}=7 \end{aligned} $$
