Question

A particle executes SHM of amplitude 1 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.

• A. 2 cm
• B. 2.6 cm
• C. 1 cm
• D. 2.3 cm

Answer 1

The correct option is D 2.3 cm

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v_1}=\frac{1}{12}-\frac{1}{21}=\frac{9}{21\times 12}$
$or{v}_1=28cm$
$\frac{1}{v_2}=\frac{1}{12}-\frac{1}{19}=\frac{7}{19\times 12}$
$or{v}_2=\frac{19\times 12}{7}=\frac{228}{7}=32.6cm$
$\Delta x=32.6-28=4.6cm$

amplitude = $\frac{\Delta x}{2}$
$=2.3cm$