Question

Use the mirror equation to deduce that an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

Answer 1

For a concave mirror, the focal length (f) is negative.

∴ f < 0

When the object is placed on the left side of the mirror, the object distance (u) is negative.

∴ u < 0

It is placed between the focus (f) and the pole.

∴ f > u > 0

${\displaystyle \frac{1}{\text{f}}<\frac{1}{\text{u}}<0}$

${\displaystyle \frac{1}{\text{f}}-\frac{1}{\text{u}}<0}$

For image distance v, we have the mirror formula:

${\displaystyle \frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}}$

${\displaystyle \frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}}$

∴ ${\displaystyle \frac{1}{\text{v}}<0}$

∴ v > 0

The image is formed on the right side of the mirror. Hence, it is a virtual image.

For u < 0 and v > 0, we can write:

${\displaystyle \frac{1}{\text{u}}>\frac{1}{\text{v}}}$

v > u

Magnification, m = ${\displaystyle \frac{\text{v}}{\text{u}}>1}$

Hence, the formed image is enlarged.