Question

A square wire of side 3.0cm is placed 25.0cm away from a concave mirror of focal length 10cm. Find the area enclosed by the image of the wire in $c{m^2}$ . The centre of the wire is on the axis of the mirror, with the all sides normal to the axis.

Answer 1

Hint: Here we will be applying the mirror equation. To find the area of the image we must know the size of the image formed at the surface of the mirror. Mirror equation can be used to find the distance of the image formed from the mirror. The obtained image distance can be further used to find the magnification.

Formula used:

1. Focal length =$\dfrac{{\text{1}}}{{\text{f}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{u}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{v}}}$
2. Linear magnification ${\text{m}} = - \dfrac{{\text{v}}}{{\text{u}}} = \dfrac{{{h_i}}}{{{h_o}}}$
   Where,
   f = the focal length
   v = image distance
   u = object distance
   ${h_i}$ = size of the image formed
   ${h_o}$ = size of the object.

Complete step by step answer:
The given data from the problem is as follows,
Let us consider the Height of the object (that is of the square wire) =3cm
Object distance u= -25cm
Focal length of concave mirror =-10cm. (Using Cartesian sign convention)
To find the area of an image formed at the mirror we need to first find out the image distance as well as size of the image.
From the rearrangement of mirror equation, we get,
⇒$\dfrac{{\text{1}}}{{\text{f}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{v}}} + \dfrac{{\text{1}}}{{\text{u}}}$
⇒$\dfrac{{\text{1}}}{v}{\text{ = }}\dfrac{{\text{1}}}{{\text{f}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{u}}}$
⇒$\dfrac{{\text{1}}}{v}{\text{ = }}\dfrac{{\text{1}}}{{ - 10}} - \dfrac{{\text{1}}}{{25}}$
⇒$\dfrac{1}{v} = \dfrac{{ - 3}}{{50}}$
$\therefore v = \dfrac{{ - 50}}{3}cm$ --------------- (1)
We know that, magnification of mirror can be find out as $m = \dfrac{{{h_i}}}{{{h_o}}} = \dfrac{{ - v}}{u}$
By putting (1) in the above equation we get,
$m = \dfrac{{ - v}}{u} = \dfrac{{ - \left( {\dfrac{{ - 50}}{3}} \right)}}{{ - 25}} = - \dfrac{2}{3}$
By calculating the size of the image i.e.
⇒${h_i} = m \times {h_o} = \dfrac{{ - 2}}{3} \times 3 = - 2cm$
Thus area of the image of square wire,
⇒$A = l \times b = {\left( { - 2cm} \right)^2} = 4c{m^2}$
The area enclosed by the image of wire= $4c{m^2}$

Additional information:
Cartesian sign convention for spherical mirror:
All distances are measured from the pole of the mirror and along the principal axis.
The distances measured in the direction of the incident light are taken as positive while those measured in the direction opposite to the incident light are taken as negative.
Heights or distances measured upward and perpendicular to the principal axis are considered positive while those measured downwards are considered negative.
The linear magnification produced by a spherical mirror is defined as the ratio of the height of the image to the height of the object. Denoted by ‘m’.
Linear magnification: ${\text{m}} = - \dfrac{{\text{v}}}{{\text{u}}} = \dfrac{{{h_i}}}{{{h_o}}}$
The magnification will be positive if the image is erect. Magnification will be negative if the image is inverted with respect to the object.

Note: Focal length is defined as the distance between the pole P of the concave mirror and the focus F.
Linear magnification is produced by the spherical mirror which is the ratio of the size of the image formed at the mirror to that of the size of the object.