Q113 / 150 · MHT CET 2024 (09 May Shift 2) · Physics
Concave and convex lenses are placed touching each other. The ratio of magnitudes of their power is $2: 3$. The focal length of the system is 30 cm . The focal lengths of individual lens are
- A$5 \mathrm{~cm},-10 \mathrm{~cm}$
- Correct answerB$10 \mathrm{~cm},-15 \mathrm{~cm}$
- C$10 \mathrm{~cm},-20 \mathrm{~cm}$
- D$20 \mathrm{~cm},-30 \mathrm{~cm}$
Solution
Power $(P)=\frac{1}{f}...(i)$
Given, $\frac{\mathrm{P}_{\text {concave }}}{\mathrm{P}_{\text {convex }}}=\frac{2}{3}...(ii)$
Let the focal length of the convex lens be $\mathrm{f}_{\text {convex }}=\mathrm{f}$
From (i) and (ii),
$\mathrm{f}_{\text {concave }}=-\frac{3}{2} \mathrm{f} \cdot(-\mathrm{ve}$ as lens is concave $)$
Using the equivalent focal length for two lenses in contact, we get
$\begin{aligned} \frac{1}{f_{e q}} & =\frac{1}{f}+\frac{1}{-\frac{3}{2} f} \\ \Rightarrow \quad \frac{1}{30} & =\frac{1}{f}-\frac{2}{3 f} \\ & =\frac{3 f-2 f}{3 f^2}\end{aligned}$
$\begin{aligned} & \Rightarrow \quad \frac{1}{30}=\frac{1}{3 \mathrm{f}} \\ & \therefore \quad, \mathrm{f}=10 \mathrm{~cm} \\ & \therefore \quad \mathrm{f}_{\text {convex }}=10 \mathrm{~cm} \text { and } \mathrm{f}_{\text {concave }}=\frac{-3}{2} \times 10=-15 \mathrm{~cm}\end{aligned}$
Hence, the correct answer is option: 'B'.