Question

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

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

Answer 1

The correct option is B $5D$

As image formed is real, therefore lens must be convex, $v=20cm$.
Let $f_1$, be the focal length for this lens and $f_2$ be the focal length of the other lens put in contact later.
for the first lens,

$\therefore \frac{1}{f_1}=\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{20}-\frac{1}{u_1}$-----------------(1)

After placing it in contact with another lens, the image shifted to $10cm$ towards the combination

i.e., $v_2=(20-10)cm=10cm$
Let $f_{net}$ be the effective focal length of the combination, which can be given as:

$\therefore \frac{1}{f_{net}}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{10}-\frac{1}{u_1}$
$\Rightarrow \frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{10}-\frac{1}{u_1}$--------------(2)
From equation (1) and (2):

$\frac{1}{10}-\frac{1}{u}=\frac{1}{f_1}+\frac{1}{f_2}=[\frac{1}{20}-\frac{1}{u}]+\frac{1}{f_2}$
$\Rightarrow \frac{1}{10}-\frac{1}{u}=\frac{1}{20}-\frac{1}{u}+\frac{1}{f_2}$

$\Rightarrow \frac{1}{10}=\frac{1}{20}+\frac{1}{f_2}$

$\Rightarrow \frac{1}{10}-\frac{1}{20}=\frac{1}{f_2}$
$\Rightarrow \frac{1}{f_2}=\frac{1}{20}$

Hence, $f_2=20cm$

$\therefore P=\frac{100}{20}{m}^{-1}=5D$
So, the correct option is $\left(B\right)$