Question

An object is moved along the central axis of a thin lens while the lateral magnification m is measured. Figure 34-43 gives m versus object distance p out to p_s. What is the magnification of the object when the object is p=14 cmfrom the lens?

Short answer

Magnification of the object when it is at 14 cm from the lens is m= -2.5.

Step-by-step solution

Listing the given quantities

• Horizontal scale, p_s = 8 cm

• Object distance, p = 14 cm.

Understanding the concepts of lens equation and magnification

By using the thin lens equation and the formula for magnification, we can find the magnification when object is at 14.0 cm from the lens.

Formula:

Magnification,$\mathrm{m}=-\frac{\mathrm{i}}{\mathrm{p}}$

Thin lens equation, $\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{i}}$

Here f is the focal length, i is the image distance, p is the object distance, m is the magnification.

Calculations of the magnification of the object

Since the magnification is given by,

$$\begin{gathered} \mathrm{m}=-\frac{\mathrm{i}}{\mathrm{p}} \\ \mathrm{i}=-\mathrm{m}\times \mathrm{p} \end{gathered}$$

From the graph, we can say that, at p = 5 cm, m = 2 cm

Therefore,

$$\begin{gathered} \mathrm{i}=-\left(2\right)\times \left(5\right) \\ =-10.0\mathrm{cm} \end{gathered}$$

Now, we have to find the focal length.

Thin lens equation is given as

$$\begin{gathered} \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{i}} \\ =\frac{1}{5}+\frac{1}{-10} \end{gathered}$$

Hence,

f = 10.0 cm

Now, we have to find the image distance, with the same focal length and when object is at a distance p = 14cm.

Thus,

$$\begin{gathered} \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{i}} \\ \frac{1}{10}=\frac{1}{14}+\frac{1}{\mathrm{i}} \\ \frac{1}{\mathrm{i}}=\frac{1}{10}-\frac{1}{14} \\ \frac{1}{\mathrm{i}}=0.0286 \\ \mathrm{i}=35\mathrm{cm} \end{gathered}$$

Now, using this image distance and given object distance, the magnification is

$$\begin{gathered} \mathrm{m}=-\frac{\mathrm{i}}{\mathrm{p}} \\ =-\frac{35}{14} \\ =-2.5 \end{gathered}$$

Therefore the magnification of the object when it is at 14 cm from the lens is m = -2.5.