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Question
A particle executes a simple harmonic motion of amplitude 1.0 cm along the principal axis of a convex lens of focal length 12 cm. The mean position of oscillation is at 20 cm from the lens. Find the amplitude of oscillation of the image of the particle.
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Solution
When the particle is at point B,
\[\frac{1}{v_B} = \frac{1}{f} + \frac{1}{u}\]
\[\frac{1}{v_B} = \frac{1}{12} - \frac{1}{19}\]
\[\Rightarrow v_B = \frac{12 \times 19}{7}\]
\[ \Rightarrow v_B = 32 . 57 cm\]
When particle is at point A,
\[\frac{1}{v_A} = \frac{1}{f} + \frac{1}{u}\]
\[\frac{1}{v_A} = \frac{1}{12} - \frac{1}{21}\]
\[\Rightarrow v_A = \frac{12 \times 21}{9}\]
\[ \Rightarrow v_A = 28 \text{ cm }\]
\[\text{ Amplitude of image } = \frac{v_A - v_B}{2}\] \[= \frac{4 . 5}{2} = 2 . 2 \text{ cm }\]
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