Question

The optic axis of a thin equiconvex lens is the x – axis. The co-ordinates of a point object and its image are ($-40\text{cm}$, $1\text{cm}$) and ($50\text{cm}$, $-2\text{cm}$) respectively. The lens is located at x= -n cm. Find n

Answer 1

Linear magnification,$m=\frac{-2\text{cm}}{1\text{cm}}=-2$
i.e., image is real and inverted.
$\left|v\right|=2\left|u\right|$
Let$\left|u\right|=x$
then $\left|v\right|=2x$
Now,
$\left|u\right|+\left|v\right|=50-(-40)=90\therefore x_0+2x_0=90$
or $x_0=30\text{cm}$
Hence, distance of the object from the lens is $30\text{cm}$ and of image is $60\text{cm}$. Therefore, the lens should be located at $x=-10\text{cm}$ as shown in the figure.

Why this question? For lenses, the magnification is given by $m=\frac{v}{u}$, where $v$ and $u$ are the image distance and object distance respectively taken according to the sign convention.