Question

A square wire of side 3cm is placed 25 cm away from a concave mirror of focal length 10 cm. What is the area enclosed by the image of the wire? The center of the square lies on the principal axis and perpendicular to the axis.

Answer 1

Step 1: Mirror formula and given data:

We the mirror formula
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
here, $f,uandv$ are the focal length of the given mirror, distance of the object from the pole of mirror and the distance of the image formed from the pole of the mirror respectively.

Given :
Focal length of the mirror, $f=10cm$
position of the wire, $u=25cm$
Side of the wire= $h_{object}=3cm$

Step 2: Calculate the distance of the image,
Substitute the values in mirror formula. (Negative sign is used because it is considered that the focal length and the object is on left side of the mirror)
$$\begin{gathered} \frac{1}{-10}=\frac{1}{v}+\frac{1}{-25} \\ \frac{1}{v}=\frac{1}{25}-\frac{1}{10} \\ \frac{1}{v}=-\frac{3}{50} \\ \Rightarrow v=\frac{-50}{3}cm \end{gathered}$$
Therefore, the position of image is $\frac{50}{3}cm$ to the left side of the mirror.

Step 3: Calculate the magnification,

The expression for magnification caused by the mirror is given as
$m=-\frac{v}{u}=\frac{h_{image}}{h_{object}}$
Substitute the value$h_{object}=3cm,u=-25cmandv=-50/3cm$ in the above expression
$$\begin{gathered} \frac{h_{image}}{3}=-\frac{\left[\frac{-50}{3}\right]}{-25} \\ {h}_{image}=-2cm \end{gathered}$$
Hence, the image height would be 2 cm.
The negative sign shows that the image would be inverted in nature.

Now, the height of the image formed by the mirror is 2 cm, therefore, the area enclosed by the image would be ${\left(2cm\right)}^2=4c{m}^2$