Question

An object is placed 20 cm in front of a convex mirror with a focal length of 10 cm. Find the position and the nature of the image.

Answer 1

Step 1: Mirror formula,

Write the mirror formula
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
here, $f,vandu$are the focal length of the mirror, the position of the image formed, and the position of the object from the mirror respectively.

Step 2: Calculate the distance of the image,

It is given that $f=10cm$ and $u=-20cm$. Substitute the values in the above expression
$$\begin{gathered} \frac{1}{10}=\frac{1}{v}+\frac{1}{-20} \\ \frac{1}{v}=\frac{1}{10}+\frac{1}{20} \\ \frac{1}{v}=\frac{3}{20} \\ \Rightarrow v=\frac{20}{3}cm \end{gathered}$$
The position of the image formed is $\frac{20}{3}cm$ to the right side of a convex mirror.
Step 3. Determine the nature of the image,
Write the expression of magnification produced by the mirror
$m=-\frac{v}{u}$
Substitute the values of $vandu$ in the above expression
$m=-\frac{\left(20/3\right)}{\left(-20\right)}=\frac{1}{3}=0.33$

The magnification is less than one which implies that the size of the image is less than the size of the object. The value is positive, which shows that the image is erected in nature.

Therefore, the nature of the image formed is virtual, erect, and diminished