If h_{1} and h_{2} are the sizes of the images formed in two conjugate positions, then, the size of object is given as h = √h_{1}h_{2} = √9×1 = 3cm.

The size of the image in the two conjugate positions is

In the first case, if the image is formed, the magnification is given as

In the first case, if the image is formed, the magnification is given as \(\frac{v}{u} = \frac{9}{3}\)Such that, v = 3u

Also, v = 20+u, since “v” and “u” interchange the conjugate positions.

∴ 3u = 20 + u

Or 2u = 20

Or u = 10

And, v = 20+ u = 20+10 = 30 cm

Now, focal length is calculated as

\(\frac1f=\frac1v+\frac1u\)

= \(\frac{1}{30} + \frac{1}{10}\)

= \(\frac{10+30}{300}\)

= \(\frac{40}{300} = 7.5\)

**Thus, correct answer is option (d)**