91Ó°ÊÓ

The focal length of the mirror, $f=12\text{cm}$

The object’s distance from the mirror, $p=+18\text{cm}$

A mirror is Concave.

A concave mirror is a diverging mirror with its reflective surface bugling opposite of the light source.

The focal length is positive if the mirror is a concave mirror. The focal length is negative if the mirror is a convex mirror. The image distance is positive if the image is a real image and is on the mirror side of the object.

Magnification refers to the ratio of image length to object length measured in planes that are perpendicular to the optical axis.

Formulae:

The radius of curvature of a mirror is,

$r=2f$ ….. (i)

The mirror equation is,

$\frac{1}{f}=\frac{1}{i}+\frac{1}{p}$ ….. (ii)

Where, $f$is the focal length, $p$is the object distance from the mirror, $i$is the image distance.

The lateral magnification of an object,

$m=\frac{h_i}{h_o}=\frac{−i}{p}$ ….. (iii)

Where, $p$is the object distance from the mirror, $i$is the image distance, $h_i$is the height of the image, and $h_o$is the height of the object.

Since the mirror is concave, the focal length must be positive, i.e., $f=+12\begin{array}{rcl}& & \text{cm}\end{array}$.

Thus, the radius of curvature of the mirror can be given using equation (i) as follows:

$\begin{array}{rcl}r& =& 2f\\ & =& 2×12\text{cm}\\ & =& +24\text{cm}\end{array}$

Hence, the radius of curvature is $\begin{array}{rcl}& & +24\text{cm}\end{array}$.

Now, the image distance can be calculated by rearranging equation (ii) as follows:

$\begin{array}{rcl}\frac{1}{i}& =& \frac{1}{f}−\frac{1}{p}\\ & =& \frac{p−f}{pf}\end{array}$

$\begin{array}{rcl}i& =& \frac{pf}{(p−f)}\\ & =& \frac{18\text{cm}×12\text{cm}}{(18−12)\text{cm}}\\ & =& +36\text{cm}\end{array}$

Hence, the image distance is $+36\text{cm}$.

The lateral magnification of the mirror can be given using equation (iii) as follows:

$\begin{array}{rcl}m& =& \frac{−i}{p}\\ & =& −\frac{36\text{cm}}{18\text{cm}}\\ & =& −2.0\end{array}$

Hence, the lateral magnification is $-2.0$.

From the calculations based on part (b), it is found that the image distance is positive in value. Thus, for an image distance to be positive, the image can be concluded to be real.

Hence, the image formed by the mirror is real $\left(R\right)$.

The lateral magnification of the mirror is given to be negative value. Again, we know that the lateral magnification can be given as:

$\begin{array}{rcl}m& =& \frac{h_i}{h_o}\\ & =& \frac{−i}{p}\\ & =& −2.0\end{array}$

Thus, the image height needs to be negative which is possible only in an inverted image case.

Hence, the image is inverted $\left(I\right)$.

For spherical mirrors, real images form on the side of the mirror where the object is located and virtual images form on the opposite side. Since the image is real, it is formed on the same side as the object.

Hence, the image is on the same side as the object O.