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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.
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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