91Ó°ÊÓ

The lens is diverging

Focal length,$\mathrm{f}=−12.0\mathrm{cm}$

Object distance, $\mathrm{p}=+8$

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 diverging, the focal length value should be negative, i.e.

$\mathrm{f}=−12\mathrm{cm}$

Thin lens equation is

$\begin{array}{c}\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{i}}\\ \frac{1}{(−12)}=\frac{1}{8}+\frac{1}{\mathrm{i}}\\ \frac{1}{\mathrm{i}}=\frac{1}{(−12)}−\frac{1}{8}\\ \frac{1}{\mathrm{i}}=−0.2083\end{array}$

$\mathrm{i}=−4.8\mathrm{cm}$

Image distance $\mathrm{i}=−4.8\mathrm{cm}$

Magnification is,

$\begin{array}{c}\mathrm{m}=−\frac{\mathrm{i}}{\mathrm{p}}\\ =−\left(\frac{−4.8}{8}\right)\\ =+0.60\end{array}$

Lateral magnification $\mathrm{m}=+0.60$

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 $\left(NI\right)$.

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.