Question

An object is placed at a distance of 10cm from a diverging mirror of a focal length of 15 cm. Find the position of the image formed and also calculate the magnification of the image.

Answer 1

Step 1: Given:

object distance, $u=-10cm$

the focal length of the mirror, $f=15cm$

mirror type = diverging.

Step 2: Finding the image distance using the mirror formula:

We know the formula

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

Therefore, $\frac{1}{v}=\frac{1}{f}–\frac{1}{u}$

Substituting the known values in the above equation we get,

$$\begin{gathered} \frac{1}{v}=\frac{1}{15}–\left(\frac{-1}{10}\right) \\ \frac{1}{v}=\frac{1}{15}+\frac{1}{10} \\ \frac{1}{v}=\frac{(2+3)}{30} \end{gathered}$$

Hence $v=\frac{30}{5}=6cm$

The nature of the image formed:

1. When the v is positive, the image is formed behind the mirror, and in front of the mirror when the v is negative.
2. Our resultant value indicates that the image is produced behind the mirror at a distance of 6cm from the pole of the mirror in this scenario.
3. When v is positive, the resulting image is virtual.
4. Hence, the image is formed behind the mirror and it is virtual.