Question

Size of the image of an object by a mirror having a focal length of $20cm$ is observed to be reduced to ${\frac{1}{3}}^{rd}$ of its size. At what distance the object has been placed from the mirror? What is the nature of the image and the mirror?

Answer 1

Step 1: Given data

1. Focal length of the mirror $f=20cm$
2. The image is reduced to ${\frac{1}{3}}^{rd}$ of its size

Step 2: Formula used

1. Magnification $m=-\frac{v}{u}$
2. $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

Where $f$ is the focal length, $v$ is the size of the image and $u$ is the size of the object.

Step 3: Find the distance the object has been placed from the mirror

The image is reduced to ${\frac{1}{3}}^{rd}$ of its size

So $m=\frac{1}{3}$

Let $v$ is the size of the image and $u$ is the size of the object

$$\begin{gathered} \therefore m=-\frac{v}{u} \\ \Rightarrow \frac{1}{3}=\frac{-v}{u} \\ \Rightarrow u=-3v \end{gathered}$$

Focal length of the mirror $f=20cm$

$$\begin{gathered} \therefore \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \\ \Rightarrow \frac{1}{20}=\frac{1}{v}+\frac{1}{-3v} \\ \Rightarrow v=\frac{40}{3}cm \end{gathered}$$

$$\begin{gathered} \therefore u=-3v \\ =-40cm \end{gathered}$$

So, the object is placed at a distance of $80cm$ in front of the convex mirror.

Step 4: Find the nature of the image

The nature of the mirror is convex and the nature of the image is virtually erect and diminished.

Hence the object is placed at a distance of $80cm$ in front of the convex mirror and the nature of the image is virtual erect and diminished.