91Ó°ÊÓ

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

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

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.

We have,

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

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

Therefore, image distance,

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

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 $+0.30$.