A particle executes linear S.H.M. along the p | vedclass

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(C) Given: Focal length $f = 8 \,cm$. The mean position of the particle is $u_0 = -14 \,cm$. The amplitude of the particle is $A_p = 1 \,cm$.First, fi

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(C) Given: Focal length $f = 8 \,cm$. The mean position of the particle is $u_0 = -14 \,cm$. The amplitude of the particle is $A_p = 1 \,cm$.First, find the position of the image $v_0$ when the particle is at the mean position:$\frac{1}{v_0} = \frac{1}{f} + \frac{1}{u_0} = \frac{1}{8} - \frac{1}{14} = \frac{7-4}{56} = \frac{3}{56}$$v_0 = \frac{56}{3} \approx 18.67 \,cm$.Next, find the position of the image $v_1$ when the particle is at the extreme position $u_1 = -14 - 1 = -15 \,cm$:$\frac{1}{v_1} = \frac{1}{f} + \frac{1}{u_1} = \frac{1}{8} - \frac{1}{15} = \frac{15-8}{120} = \frac{7}{120}$$v_1 = \frac{120}{7} \approx 17.14 \,cm$.The amplitude of the oscillating image is the difference between the image positions:$A_i = |v_0 - v_1| = |18.67 - 17.14| = 1.53 \,cm$.Rounding to the nearest integer, the amplitude is approximately $2 \,cm$.

$A$ particle executes linear $S.H.M.$ along 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 the oscillating image of the particle is nearly: (in $\,cm$)

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