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

Answer: (c). u = -14 cm: 1/v = 1/8 - 1/14 = 6/112 ⇒ v ≈ 18.67 cm For small motion along the axis, longitudinal magnification = (v/u)² = ((18.67)/(14))² ≈ 1.78 Image amplitude ≈ 1.78 × 1 ≈ 2 cm So the answer is (c) 2 cm.

A particle executes linear S.H.M. The mean position of oscillation is… | MHT CET 2021 (21 Sep, Shift 1)

Q81MHT CET 2021 · 21 SEP · SHIFT 1Difficult

Spherical Surfaces & Thin Lenses

A particle executes linear S.H.M. The mean position of oscillation is at the principal axis of a convex lens of focal length $8\text{ cm}$ . 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

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Answer: (c)

$u = -14$ cm: $\frac{1}{v} = \frac{1}{8} - \frac{1}{14} = \frac{6}{112} \Rightarrow v \approx 18.67$ cm
For small motion along the axis, longitudinal magnification $= \left(\frac{v}{u}\right)^2 = \left(\frac{18.67}{14}\right)^2 \approx 1.78$
Image amplitude $\approx 1.78 \times 1 \approx 2$ cm
So the answer is (c) 2 cm.

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Spherical Surfaces & Thin Lenses question from MHT CET 2021 (21 Sep, Shift 1)

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