Question

Figure 34-40 gives the lateral magnification of an object versus the object distancefrom a lens asthe object is moved along the central axis of the lens through a range of values for p out to ${\mathbf{p}}_{\mathbf{s}}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$. What is the magnification of the objectwhen the object is $\mathbf{35}$cmfrom the lens?

Short answer

Magnification of the object when the object is at from the lens is $+30$.

Step-by-step solution

Listing the given quantities 

• Horizontal scale${\mathrm{p}}_{\mathrm{s}}=20.0\mathrm{cm}$

• Object distance $\mathrm{p}=35\mathrm{cm}$

Understanding the concepts of magnification

By using the formula for magnification and the given graph, we can find the value of the image distance. Using this image distance in the lens equation, we can find the focal length of the lens.

We have the new object distance and the focal length. From this, we can find the new image distance. Once again using the formula for magnification, we can find the required magnification.

Formula:

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

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

We have,

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

From the graph, at $\mathrm{p}=15\mathrm{cm}$, when $\mathrm{m}=0.5\mathrm{cm}$

Therefore, image distance,

$$\begin{gathered} \mathrm{i}=-0.5\times 15 \\ =-7.5\mathrm{cm} \end{gathered}$$

The lens equation is,

$$\begin{gathered} \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{i}}+\frac{1}{\mathrm{p}} \\ \frac{1}{\mathrm{f}}=\frac{1}{-7.5}+\frac{1}{15} \\ \mathrm{f}=\frac{1}{\frac{1}{-7.5}+\frac{1}{15}} \\ =-15\mathrm{cm} \end{gathered}$$

Now, we have to find the image distance, when$\mathrm{p}=35\mathrm{cm}$.

$$\begin{gathered} \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{i}}+\frac{1}{\mathrm{p}} \\ \frac{1}{\mathrm{i}}=\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{p}} \\ \mathrm{i}=\frac{1}{\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{p}}} \\ =\frac{1}{\frac{1}{-15}-\frac{1}{35}} \\ =-10.5\mathrm{cm} \end{gathered}$$

Now using the magnification expression,

$$\begin{gathered} \mathrm{m}=-\frac{\mathrm{i}}{\mathrm{p}} \\ =-\frac{\left(-10.5\right)}{35} \\ =+0.30 \end{gathered}$$

Therefore the magnification of the object when the object is at $35\mathrm{cm}$from the lens is $+0.30$.