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,$\frac{1}{1+x}\approx1-x, where x < < 1]$

• A. $\frac{4A}{9}\times10^{-2}m$
• B. $\frac{5A}{9}\times10^{-2}m$
• C. $\frac{A}{3}\times10^{-2}m$
• D. $\frac{A}{9}\times10^{-2}m$

Answer 1

Answer: (D)

$u=-0.2, f=0.05;\frac{1}{v} =\frac{1}{f}+\frac{1}{u} =\frac{100}{5}-\frac{10}{2} =20-5 =15 or, v=\frac{1}{15}m$
$\frac{dv}{v^{2}}=-\frac{du}{u^{2}} \therefore dv=A.\frac{v^{2}}{u^{2}}$
$A_{image} =A\times\frac{1\times25}{225\times1} cm =\frac{A}{9}cm or, A_{image} =\frac{A}{9}\times10^{-2}m$