Question

An arrow $2.5cm$ high is placed at a distance of $25cm$ from a diverging mirror of a focal length of $20cm$. Find the nature, position, and size of the image formed.

Answer 1

Step 1: Given data

Height of the object: $h_1=2.5cm$

Distance of object from mirror: $u=-25cm$

Focal length of the mirror: $f=20cm$

Step 2: Formula used

The focal length of the mirror is given by the formula $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$, where $v$ is the distance between mirror and image and $u$ is the distance between mirror and object. This is known as the mirror formula.

The magnification is given by the formula $m=\frac{h_2}{h_1}=-\frac{v}{u}$, where $h_2$ is the height of image, $h_1$ is height of object, $v$ is distance of image from mirror and $u$ is distance of object from mirror.

Step 3: Find the distance between image and object using mirror formula.

$$\begin{gathered} \frac{1}{20}=\frac{1}{v}-\frac{1}{25} \\ \Rightarrow \frac{1}{v}=\frac{1}{20}+\frac{1}{25} \\ \Rightarrow \frac{1}{v}=\frac{9}{100} \\ \Rightarrow v=11.11cm \end{gathered}$$

Step 4: Find the height of the image

$$\begin{gathered} m=\frac{h_2}{h_1}=\frac{-v}{u} \\ \Rightarrow \frac{h_2}{2.5}=\frac{-11.11}{-25} \\ \Rightarrow h_2=+1.11cm \end{gathered}$$

Hence, the image will be formed at $11.11cm$ behind the mirror, which will be virtual, erect, and $1.11cm$ in size.