Question

Two object are placed on the principal axis of a thin convergnig lens. One is $10\text{cm}$ from the lens and the other is on the other side of the lens at a distance of $40\text{cm}$ from the lens. If he image of both objects coincide, the focal length of the lens (in $\text{cm}$) is

• A. $20$
• B. $16$
• C. $15$
• D. $14$

Answer 1

The correct option is B $16$

In the first case, the image is virtual when $u=-10\text{cm}$
$\frac{1}{f}=\frac{1}{v_1}-\frac{1}{u}\Rightarrow \frac{1}{f}=\frac{1}{v_1}+\frac{1}{10}\Rightarrow \frac{1}{f}-\frac{1}{10}=\frac{1}{v_1}$
$v_1=\frac{10f}{10-f}\text{(virtual)}\to \left(1\right)$
In the second case, the image is real when $u=-40\text{cm}$
$v_2=\frac{40-f}{40-f}\text{(real)}\to \left(2\right)$
scince the image is at the same plane.
$v_2=-v_1$
$\frac{40f}{40-f}=-\left(\frac{10f}{10-f}\right)\Rightarrow (40-4f)=(-40+f)$
$5f=80$
$f=\frac{80}{5}$
$f=16\text{cm}$