Tutorial Exercise

A $1\mathrm{.}00-cm-$high object is placed $3\mathrm{.}38$ cm to the left of a converging lens of focal length $8\mathrm{.}93$ cm $.$ A diverging lens of focal length $-16\mathrm{.}00$ cm is $6\mathrm{.}00$ cm to the ric image. Is the image inverted or upright? Is the image real or virtual?

Step $1$

The $1\mathrm{.}00-cm-$high object serves as a real object, located at a distance $p_1=3\mathrm{.}38cm$ in front of the converging lens $L_1,$ as shown in the diagram.

Figure

Let $f_1$ be the focal length of the converging lens. The thin lens equation gives the image distance for this lens as

$q_1=\frac{p_1{f}_1}{p_1-f_1}$

$=-(c{m}^2)$

$=-,cm$