Question

A thin convex lens of focal length ' $f'$ is put on a plane mirror as shown in the figure. When an object is kept at a distance ' $a$ ' from the lens-mirror combination, its image is formed at a distance $\frac{a}{3}$ in front of the combination. The value of ' $a$ ' is

• A. f
• B. $2f$
• C. $3f$
• D. $\frac{3}{2}$f

Answer 1

Answer: (B)

The combination acts like a silvered lens
$\frac{1}{f_{ eq }}=\frac{2}{f_{e}}+\frac{1}{f_{ m }}=\frac{2}{f}+\frac{1}{\infty}$
$f_{ eq }=\frac{f}{2}$ (Combination behaves as a concave mirror as object and image lie on the same side.)
Applying Mirror formula, we get
$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$
$\Rightarrow \frac{1}{-a}+\frac{1}{-a / 3}=\frac{1}{-f / 2}$
$\Rightarrow \frac{4}{a}=\frac{2}{f}$
$\Rightarrow a=2f$