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 particles at A,

$\frac{1}{v_A}=\frac{1}{f}+\frac{1}{u}$

$=\frac{1}{12}-\frac{1}{19}$

$\Rightarrow v_A=\frac{12\times 19}{7}$

Similarly $v_B=\frac{12\times 12}{9}$

Amplitude of image $=\frac{v_A-v_B}{2}$

$=\frac{4.5}{2}=2.2\text{cm}$