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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jun 2008 11:04:08 IST
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a particle executes a simple harmonic motion of amplitude 1 cm horizontally along the principle axisof a convex lens of focal length12 cm. the mean position of occilationis at 20 cm from the lens. find the amplitude of occilationof the image of the particle.
my ans:2.39cm approx
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Given f=12cm
Object is at a distance of 19 cm from the lens and 21 cm in extreme.Now using lens formula u may calculate image distance for u=-19cm and u=-21cm.
Clearly the image would oscillate between this two values. v=228/7=32.57cm and 252/9 =28cm
Thus amplitude of vibration of image is
(32.57-28)/2= 4.57/2=2.285 cm
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