Q132 / 150 · MHT CET 2021 (21 Sep Shift 1) · Physics
A particle executes linear S.H.M. along the principal axis of a convex lens of focal length $8 \mathrm{~cm}$. The mean position of oscillation is at $14 \mathrm{~cm}$ from the lens with amplitude $1 \mathrm{~cm}$. The amplitude of oscillating image of the particle is nearly
- A$3 \mathrm{~cm}$
- B$5 \mathrm{~cm}$
- Correct answerC$2 \mathrm{~cm}$
- D$4 \mathrm{~cm}$
Solution
$\mathrm{f}=8 \mathrm{~cm}$, when the particle is at mean position, $\mathrm{u}=-14 \mathrm{~cm}$
$$
\begin{aligned}
& \frac{1}{\mathrm{~V}}=\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{u}}=\frac{1}{8}-\frac{1}{14}=\frac{3}{56} \\
& \therefore \mathrm{V}=\frac{56}{3} \approx 19 \mathrm{~cm}
\end{aligned}
$$
When the particle is at one of the extreme positions its distance from the lens is $14+1=15 \mathrm{~cm}$
$$
\therefore \mathrm{u}=-15 \mathrm{~cm}
$$
Again, $\frac{1}{\mathrm{~V}}=\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{u}}=\frac{1}{8}-\frac{1}{15}=\frac{7}{120}$
$$
\therefore \mathrm{v}=\frac{120}{7} \approx 17 \mathrm{~cm}
$$
Amplitude of the image $=19-17=2 \mathrm{~cm}$
Hence, the correct answer is option: 'C'.