Question

A square of side 3 cm is placed at a distance of 25 cm from a concave mirror of a focal length of 10 cm. The center of the square is at the axis of the mirror and the plane is normal to the axis. The area enclosed by the image of the square is

• A. 4cm²
• B. 6cm²
• C. 16cm²
• D. 36cm²

Answer 1

The correct option is (D) 4 cm²

Given,

Side of the square = 3 cm

Object distance u = 25 cm

The focal length of the mirror f = 10 cm

We know,

From mirror formula

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

Putting all the values

$\frac{1}{v}=\frac{1}{-10}+\frac{1}{25}$

After solving

$v=\frac{-50}{3}$

Now,

Magnification is given as

$m=\frac{h_i}{h_o}=\frac{-v}{u}$

Or, $\frac{h_i}{3}=\frac{\left(\frac{-50}{3}\right)}{-25}$

After solving

$h_i=-2cm$

So if the side of the image is -2 cm then its area will be

$Area={(-2)}^2=4$ ${cm}^2$

So the area enclosed by the image is $4{cm}^2$