A particle executes a simple harmonic motion of amplitude 1.0cm along the principal axis of a convex lens of focal length 12cm. The mean position of oscillation is at 20cm from the lens. Find the amplitude of oscillation of the image of the particle.
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When the object is at 19cm from the lens, let the image will be at,$ v_1.$
$\Rightarrow\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}_1}-\frac{1}{-19}=\frac{1}{12}$
$\Rightarrow\text{v}_1=32.57\text{cm}$
Again, when the object is at 21cm from the lens, let the image will be at, $v_2.$
$\Rightarrow\frac{1}{\text{v}_2}-\frac{1}{\text{u}_2}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}_2}-\frac{1}{21}=\frac{1}{12}$
$\Rightarrow\text{v}_2=28\text{cm}$
$\therefore$ Amplitude of vibration of the image is A $=\frac{\text{A}'\text{B}'}{2}=\frac{\text{v}_1-\text{v}_2}{2}$
$\Rightarrow\text{A}=\frac{32.57}{2}=2.285\text{cm}$
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