Question

A real image of an object is formed at a distance of $20\, cm$ from a lens. On putting another lens in contact with it, the image is shifted $10 \,cm$ towards the combination. The power of the lens is

• A. $2\,D$
• B. $5\,D$
• C. $6\,D$
• D. $10\,D$

Answer 1

Answer: (B)

As image formed is real, therefore lens must be convex, $v = 20 \,cm$. Let $f_1$, be focal length for this lens.
$\therefore \frac{1}{f_1} = \frac{1}{v}-\frac{1}{u} = \frac{1}{20} -\frac{1}{u}$
After placing it in contact with another lens, the image shifted to $10 \,cm$ towards the combination
i.e., $v = (20 - 10) \,cm = 10 \,cm$
$\therefore \frac{1}{10}-\frac{1}{u} = \frac{1}{f_{1}}+\frac{1}{f_{2}} = \left(\frac{1}{20}-\frac{1}{u}\right)+\frac{1}{f_{2} }$
$\Rightarrow f_{2} = 20 \,cm $
$ \therefore P =\frac{100}{20} \,m^{-1} = 5 \,D$