Question

Where should an object be placed from a converging lens of the focal length of $20\mathrm{cm}$, so as to obtain a real image of magnification $2$?

Answer 1

Step 1: Given information

The focal length of a converging lens, $\mathrm{f}=+20\mathrm{cm}$.

The magnification is positive ($+$) for all virtual images and negative ($-$) for all real images and if the image is inverted and real the magnification will be negative.

Hence, magnification of the image, $\mathrm{m}=-2$.

Step 2: To find

The distance of the object is placed from a converging lens.

Step 2: Finding the distance of the object is placed from a converging lens

We know the formula,

Magnification,

$\mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}$

∴ $\frac{\mathrm{v}}{\mathrm{u}}=-2$

Or,

$\mathrm{v}=-2\mathrm{u}$

Now,

By using the lens formula,

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}–\frac{1}{\mathrm{u}}$

By substituting the values,

$\frac{1}{20}=-\frac{1}{2u}-\frac{1}{\mathrm{u}}$

$\frac{1}{20}=-\frac{3}{2}\mathrm{u}$

$\mathrm{u}=-30\mathrm{cm}$

Therefore, the object should be placed at a distance of $30\mathrm{cm}$ from the lens.