Question 231 of 295

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 $\dfrac{a}{3}$ in front of the combination. The value of $a$ is:

A$2f$B$\dfrac{3}{2}f$C$3f$D$f$

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Answer: (A) $2f$

Equivalent power of combination is

${P}_{eff}=2{P}_{lens}+{P}_{mirror}$ ...(i)

$∵{P}_{mirror}=0$ for plane mirror and ${P}_{lens}=\dfrac{1}{f}$

Also combination behaves as a mirror. So. ${P}_{eff}=\dfrac{1}{{f}_{mirror}}$

In equation (i)

$-\dfrac{1}{{f}_{mirror}}=\dfrac{2}{f}\Rightarrow {f}_{mirror}=-\dfrac{f}{2}$

By mirror formula

$\dfrac{1}{{f}_{mirror}}=\dfrac{1}{u}+\dfrac{1}{v}$

Here $u=-a,v=-\dfrac{a}{3},{f}_{mirror}=-\dfrac{f}{2}$

$\Rightarrow -\dfrac{2}{f}=\dfrac{1}{-a}+\dfrac{3}{-a}=\dfrac{-4}{a}$

$\Rightarrow \dfrac{-2}{f}=\dfrac{-4}{a}\Rightarrow a=2f$