Question

The distance between two point sources of light is $24\text{cm}$. Where would you place a converging lens of focal length $9\text{cm}$, so that the images of both the sources are formed at the same point.

• A. $3\text{cm}$ from source of light
• B. $5\text{cm}$ from source of light
• C. $6\text{cm}$ from source of light
• D. $8\text{cm}$ from source of light

Answer 1

The correct option is C $6\text{cm}$ from source of light

For source $1$, $\left({\text{S}}_1\right)$:

Applying lens formula, and substituting the values from the image, we have

$\frac{1}{v_1}-\frac{1}{-x}=\frac{1}{9}$

$\therefore \frac{1}{v_1}=\frac{1}{9}-\frac{1}{x}...\left(1\right)$

For source $2$, $\left({\text{S}}_2\right)$:

$\frac{1}{v_2}-\frac{1}{-(24-x)}=\frac{1}{9}$

$\therefore \frac{1}{v_2}=\frac{1}{9}-\frac{1}{24-x}...\left(2\right)$

Since, sign convention for ${\text{S}}_1$ and ${\text{S}}_2$ are just opposite. Hence,

$v_1=-v_2$

$or$, $\frac{1}{v_1}=-\frac{1}{v_2}...\left(3\right)$

From equation $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$

$\frac{1}{9}-\frac{1}{x}=\frac{1}{24-x}-\frac{1}{9}$

Solving this equation we get, $x=6\text{cm}$.

Therefore, the lens should be kept at a distance of $6\text{cm}$ from either of the object.