Question

A point object is placed on the axis of a thin convex lens of focal length 0.05 m at a distance of 0.2 m from the lens and its image is formed on the axis. If the object is now made to oscillate along the axis with a small amplitude of A cm, then what is the amplitude of oscillation of the image? [you may assume,$\frac{1}{1+x}\approx 1-x,wherex<<1]$

• A. $\frac{4A}{9}\times 10^{-2}m$
• B. $\frac{5A}{9}\times 10^{-2}m$
• C. $\frac{A}{3}\times 10^{-2}m$
• D. $\frac{A}{9}\times 10^{-2}m$

Answer 1

Answer: (D)

$u=-0.2,f=0.05;\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{100}{5}-\frac{10}{2}=20-5=15or,v=\frac{1}{15}m$
$\frac{dv}{v^2}=-\frac{du}{u^2}\therefore dv=A.\frac{v^2}{u^2}$
$A_{image}=A\times \frac{1\times 25}{225\times 1}cm=\frac{A}{9}cmor,A_{image}=\frac{A}{9}\times 10^{-2}m$