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

Answer 1

The correct option is A $2f$

The first image will be formed from refraction from lens, its position can be determined by lens formulas

The lens formulas is given as $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$...(i)

where $u=-a,f=+f$, substituting values in equation (i)

$\frac{1}{v}-\frac{1}{-a}=\frac{1}{f}\Rightarrow v=\frac{af}{a-f}$

The image formed will be left side of lens provide $a>f$

The mirror will form the image at right side of the lens with distance equal to

$\frac{af}{a-f}$

Now the third image will be formed by lens again for which the coordinate system will be reversed of earlier as the rays are traveling from left to right:

$u=\frac{af}{a-f}$, $f=+f$ and $v=\frac{a}{3}$

applying lens formulas

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$...(i)

$\frac{3}{a}-\frac{a-f}{af}=\frac{1}{f}$...(ii) multiplying and dividing by f

$\frac{3f}{af}-\frac{a-f}{af}=\frac{a}{af}$

$3f-a+f=a\Rightarrow a=2f$

Hence, correct answer is option A