An object is placed against the center of a spherical mirror, and then moved$\mathbf{70}\mathbf{\text{cm}}$from it along the central axis as theimage distance $\mathit{i}$ is measured. Figure 34-36 gives$\mathbf{i}$versus object distance$\mathit{p}$out to${\mathbf{p}}_{\mathbf{s}}\mathbf{=}\mathbf{40}\mathbf{\text{cm}}$. What is$\mathit{i}$for $\mathbf{p}\mathbf{=}\mathbf{70}\mathbf{\text{cm}}$?

• Object distance from the mirror,$p=70\text{cm}$.
• Horizontal scale of the object versus image distance graph, $p_s=40\text{cm}$.

The focal length, the image distance, and the object distance of the mirror are related by the mirror equation to each other. Thus, for any given object distance, the image distance is given as per the equation. Thus, for the common mirror, the image distance is solely dependent on the position of the object. The focal length of the mirror is the distance from a mirror at which parallel light rays will meet.

Formulae:

The mirror equation is,

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

Where, localid="1663047904621" $f$is the focal length, $p$is the object distance from the mirror, $i$ is the image distance.

From the given graph:

At $p=20cm$, the image distance is $i=\infty$.

Thus, the value of the focal length can be given using equation (i) as follows:

$$\begin{gathered} \frac{1}{f}=\frac{1}{20cm}+\frac{1}{\infty } \\ f=20cm \end{gathered}$$

Now use the same equation (i), the new image distance for the given object distance, i.e., for$p=70\text{cm}$can be given as:

$$\begin{gathered} \frac{1}{20\text{cm}}=\frac{1}{70\text{cm}}+\frac{1}{i} \\ \frac{1}{i}=\frac{1}{20\text{cm}}-\frac{1}{70\text{cm}} \\ \frac{1}{i}=\frac{50\text{cm}}{1400{\text{cm}}^2} \end{gathered}$$

$$\begin{gathered} i=\frac{1400{\text{cm}}^2}{50\text{cm}} \\ i=28\text{cm} \end{gathered}$$

Hence, the image distance for $p=70\text{cm}$is $2\text{8 cm}$.