Question

A particle executes linear S.H.M. The mean position of oscillation is at 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 oscillating image of the particle is nearly

• A. 3 cm
• B. 5 cm
• C. 2 cm
• D. 4 cm

Hint:

For small oscillations about the mean position, the amplitude of the image can be found using the derivative \(\frac{dv}{du} = \frac{f^2}{(u+f)^2}\). This gives a good approximation when the object amplitude is small compared to the object distance.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A particle oscillates in SHM about a mean position that lies on the principal axis of a convex lens (f = +8 cm). Mean object distance = 14 cm from lens (so u₀ = -14 cm using Cartesian sign convention). Amplitude of object oscillation = 1 cm. We need the amplitude of the image oscillation.

Step 2: Key Formula or Approach:
Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\). For a small change in u, the change in v is approximately \(\Delta v = \left|\frac{dv}{du}\right| \Delta u\). The amplitude of image = \(\left|\frac{dv}{du}\right| \times\) (amplitude of object).

Step 3: Detailed Explanation:
From lens formula, \(v = \frac{uf}{u+f}\). Differentiate with respect to u: \[ \frac{dv}{du} = \frac{f^2}{(u+f)^2}. \] At the mean position, \(u = -14\ \text{cm}\), \(f = 8\ \text{cm}\), so \(u+f = -6\ \text{cm}\). Then \[ \frac{dv}{du} = \frac{64}{36} = \frac{16}{9} \approx 1.7778. \] Amplitude of image = \(\left|\frac{dv}{du}\right| \times 1\ \text{cm} \approx 1.78\ \text{cm}\), which is nearly 2 cm. (Exact calculation of extreme positions: at \(u = -13\ \text{cm}\), \(v = 104/5 = 20.8\ \text{cm}\); at \(u = -15\ \text{cm}\), \(v = 120/7 \approx 17.1429\ \text{cm}\); half the difference = \((20.8 - 17.1429)/2 = 1.8286\ \text{cm} \approx 1.83\ \text{cm}\). The closest option is 2 cm.)

Step 4: Final Answer:
Option (C) 2 cm.