91Ó°ÊÓ

The ratio of the height of the image to the height of the object is given as the lateral magnification of the lens or the mirror. Now, the magnification of the mirror or the lens can also be given as the negative value of the ratio of the image distance to the object distance from the mirror.

Formulae:

The mirror equation is,

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

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

The lateral magnification of an object is,

$m=\frac{-i}{p}$….. (ii)

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

Draw the graph as below.

From the given graph, at $p=5\text{cm}$, the lateral magnification is,

$m=2$

The image distance of the spherical mirror can be calculated using equation (i) as follows:

$\begin{array}{rcl}i& =& -mp\\ & =& -2×5\text{cm}\\ & =& -10\text{cm}\end{array}$

Now, the focal distance of the mirror can be calculated using the given data in equation (ii) as follows:

$\begin{array}{rcl}\frac{1}{f}& =& \frac{1}{5\text{cm}}+\frac{1}{10\text{cm}}\\ \frac{1}{f}& =& \frac{1}{10\text{cm}}\\ f& =& 10\text{cm}\end{array}$

Now, for the given object distance,$p_1=14\text{cm}$, the image distance from the mirror can be given using equation (ii) as follows:

$\begin{array}{rcl}\frac{1}{10\text{cm}}& =& \frac{1}{14\text{cm}}+\frac{1}{i_1}\\ \frac{1}{i_1}& =& \frac{1}{10\text{cm}}-\frac{1}{14\text{cm}}\\ \frac{1}{i_1}& =& \frac{4\text{cm}}{140{\text{cm}}^2}\end{array}$

$\begin{array}{rcl}{i}_1& =& \frac{140{\text{cm}}^2}{4\text{cm}}\\ & =& \frac{70}{2}\text{cm}\\ & =& 35\text{cm}\end{array}$

Now, the lateral magnification of the mirror can be given using equation (i) as follows:

$\begin{array}{rcl}m& =& -\frac{35\text{cm}}{14\text{cm}}\\ & =& -2.5\end{array}$

Hence, the magnification for $p=14.0\text{cm}$is $m=-2.5$.