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.

Answer 1

When the particle is at point B,

$\frac{1}{v_{\mathrm{B}}}=\frac{1}{f}+\frac{1}{u}$
$\frac{1}{v_B}=\frac{1}{12}-\frac{1}{19}$
$$\begin{gathered} \Rightarrow v_{\mathrm{B}}=\frac{12\times 19}{7} \\ \Rightarrow v_{\mathrm{B}}=32.57\mathrm{cm} \end{gathered}$$

When particle is at point A,

$\frac{1}{v_{\mathrm{A}}}=\frac{1}{f}+\frac{1}{u}$
$\frac{1}{v_{\mathrm{A}}}=\frac{1}{12}-\frac{1}{21}$
$$\begin{gathered} \Rightarrow v_{\mathrm{A}}=\frac{12\times 21}{9} \\ \Rightarrow v_{\mathrm{A}}=28\mathrm{cm} \end{gathered}$$

$\mathrm{Amplitude}\mathrm{of}\mathrm{image}=\frac{v_{\mathrm{A}}-v_{\mathrm{B}}}{2}$
$=\frac{4.5}{2}=2.2\mathrm{cm}$