Question

An object is placed against the center of a thin lens and then moved away from it along the central axis as the image distance is measured. Figure 34-41 gives i versus object distance p out to ${\mathbf{p}}_{\mathbf{s}}\mathbf{=}\mathbf{60}\mathbf{}\mathbf{cm}$. What is the image distancewhen $\mathbf{p}\mathbf{=}\mathbf{100}\mathbf{}\mathbf{cm}$?

Short answer

The image distance when $\mathrm{p}=100\mathrm{cm}$is$+43\mathrm{cm}$.

Step-by-step solution

Listing the given quantities

Horizontal scale ${\mathrm{p}}_{\mathrm{s}}=60\mathrm{cm}$

$\mathrm{p}=100\mathrm{cm}$

Understanding the concepts of lens equation

We can find the focal length by using the lens equation and the data given in the graph. Using this focal length, we can find the image distance for the given object distance.

Formula:

$\frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{i}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{p}}$

Calculations of the image distance when p=100 cm

Using the lens equation,

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{i}}+\frac{1}{\mathrm{p}}$

From the graph, at $\mathrm{p}=30$, $\mathrm{i}=\infty$

Therefore,

$\begin{array}{rcl}\frac{1}{\mathrm{f}}& =& \frac{1}{\infty }+\frac{1}{30}\\ & =& \frac{1}{30}\\ & & \end{array}$

Now, we have to find the image distance at $\mathrm{p}=100\mathrm{cm}$

$$\begin{gathered} \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{i}}+\frac{1}{\mathrm{p}} \\ \mathrm{i}=\frac{1}{\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{p}}} \\ =\frac{1}{\frac{1}{30}-\frac{1}{100}} \\ =42.8 \\ \approx +43\mathrm{cm} \end{gathered}$$

The image distance when $\mathrm{p}=100\mathrm{cm}$is $+43\mathrm{cm}$.