Question

An object is moved along the central axis of a spherical mirror while the lateral magnification m of it is measured. Figure 34-35 gives m versus object distance p for the range${\mathit{p}}_{\mathbf{a}}\mathbf{=}\mathbf{2}\mathbf{}\mathbf{\text{cm}}$ and ${\mathit{p}}_{\mathbf{b}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{cm}}$. What is m for $\mathit{p}\mathbf{=}\mathbf{14}\mathbf{}\mathbf{\text{cm}}$?

Short answer

The lateral magnification for $p=14.0\text{cm}$is $m=-2.5$.

Step-by-step solution

Step 1: The given data:

• Horizontal range of the object distances in the graph,$p_a=2.0\text{cm}$to$p_b=8.0\text{cm}$.

• The object distance from the mirror, $p=14\text{cm}$

Determining the concept of lateral magnification:

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.

Determining the magnification for  :

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:

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

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 $m=-2.5$.