91Ó°ÊÓ

The lens is converging

Focal length,$\mathrm{f}=35.0\mathrm{cm}$

Object distance, $\mathrm{p}=+25$

By using the thin lens equation and the formula for magnification, we can find all the required quantities.

Formula:

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

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

Since the lens is converging, the focal length value should be positive, i.e.

$\mathrm{f}=+35\mathrm{cm}$

Thin lens equation is

$\begin{array}{c}\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{i}}\\ \frac{1}{35}=\frac{1}{25}+\frac{1}{\mathrm{i}}\\ \frac{1}{\mathrm{i}}=\frac{1}{35}−\frac{1}{25}\\ \frac{1}{\mathrm{i}}=−0.01143\end{array}$

$\begin{array}{c}\mathrm{i}=-87.5\mathrm{cm}\\ ≈-88\mathrm{cm}\end{array}$

Magnification is,

$\begin{array}{c}\mathrm{m}=−\frac{\mathrm{i}}{\mathrm{p}}\\ =−\left(\frac{−88}{25}\right)\\ =3.52\\ ≈3.5\end{array}$

Lateral magnification $\mathrm{m}=+3.5$

As the image distance $\mathrm{i}$is negative, the image is virtual $\left(\mathrm{V}\right)$.

As the magnification is positive, the image is non-inverted $\left(\mathrm{NI}\right)$ .

For thin lens, the real images form on the opposite side as the object and virtual images form on the same side as the object.

since the image is non-inverted, it forms on the same side of the object.