Question

An object is placed against the center of a spherical mirror and then moved 70 cm from it along the central axis as the image distance i is measured. Figure 34-48 gives i versus object distance p out to ${\mathbf{p}}_{\mathbf{s}}\mathbf{}\mathbf{=}\mathbf{40}\mathbf{}\mathbf{cm}$. What is the image distance when the object is 70 cm from the mirror?

Short answer

The image distance is $i_{new}=-21\text{cm}$.

Step-by-step solution

Step 1: Given data

• Object distance,$p_{given}=70\text{cm}$ .

• Horizontal Scale, $p_s=40\mathrm{cm}$.

Determining the concept

The focal length of the lens is constant for every situation. Using this concept, find the focal length from the graph. Using the value in the equation and the given object distance, find the new image distance.

Formulae are as follows:

$\frac{1}{f}=\left(\frac{1}{p}+\frac{1}{i}\right)$

Here, p is the pole, f is the focal length, and i is the image distance.

Determining the image distance

From the graph, take one point Q having co-ordinate $Q\left(15,-10\right).$

So,

$\left(\frac{1}{p}+\frac{1}{i}\right)=\frac{1}{f}$

$\begin{array}{rcl}\left(\frac{1}{15}+\frac{1}{-10}\right)& =& \frac{1}{f}\\ & =& \frac{1}{p_{given}}+\frac{1}{i_{new}}\\ & & \end{array}$

Rearrange for the i_new;

$\begin{array}{rcl}\frac{1}{i_{new}}& =& \frac{-10+15}{-10\times 15}-\frac{1}{70}\\ & =& -\frac{5}{150}-\frac{1}{70}\\ & & \end{array}$

Solve for i_new;

$\begin{array}{rcl}\frac{1}{i_{new}}& =& -\frac{100}{2100}\\ & =& -21\text{cm}\\ & & \end{array}$

Therefore, the image distance is$i_{new}=-21\text{cm}$ .