Question

When a diverging lens is held 13.0 cm above a line of print, as in Figure 26.29, the image is 5.0 cm beneath the lens. (a) Is the image real or virtual? (b) What is the focal length of the lens?

Answer 1

Given

$$ \begin{array}{l} o=13.0 \mathrm{~cm} \\ i=5.0 \mathrm{~cm} \end{array} $$

(a)

The image formed is virtual

(b)

The focal length is

$$ \begin{array}{l} f=\frac{u v}{u+v} \\ =\frac{(13.0 \mathrm{~cm}) \times(-5.0 \mathrm{~cm})}{(13.0 \mathrm{~cm})+(-5.0 \mathrm{~cm})} \\ \hline f=-8.125 \mathrm{~cm} \end{array} $$