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?

• Horizontal scale, p_s = 8 cm

• Object distance, p = 14 cm.

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.

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.