Question

A particle executes linear S.H.M. along the principal axis of a convex lens of focal length 8 cm. The mean position of oscillation is at 14 cm from the lens with amplitude 1 cm. The amplitude of the oscillating image of the particle is nearly

• A. 3 cm
• B. 4 cm
• C. 2 cm
• D. 1 cm

Hint:

For small oscillations, image amplitude is magnification times object amplitude.

Answer 1

The Correct Option is C

Step 1: Lens formula.
For a convex lens,
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Step 2: Mean position of the object.
Given:
\[ f = 8\,\text{cm}, \quad u = 14\,\text{cm} \]
\[ \frac{1}{8} = \frac{1}{v} + \frac{1}{14} \]
\[ \frac{1}{v} = \frac{1}{8} - \frac{1}{14} = \frac{3}{56} \Rightarrow v = \frac{56}{3} \approx 18.7\,\text{cm} \]
Step 3: Linear magnification.
\[ m = \frac{v}{u} = \frac{18.7}{14} \approx 1.33 \]
Step 4: Amplitude of image.
Amplitude of object \(= 1\,\text{cm}\). Hence,
\[ \text{Amplitude of image} = m \times 1 \approx 1.33 \approx 2\,\text{cm} \]
Step 5: Conclusion.
The amplitude of oscillating image is nearly 2 cm.