Answer
(a)Diagram below;
(b)$s'=-6.60$ $cm$,$y'=0.240$ $cm$, The image formed is virtual
Work Step by Step
Given that $R= -22.0$ $cm$
$f=\frac{R}{2}=-11.0$ $cm$
We use the following formula:
$\frac{1}{f}=\frac{1}{s}+\frac{1}{s′}$
Rearranging gives:
$s′=\frac{f\times{s}}{s−f}$
$s′=\frac{(-11)×16.5}{16.5−(-11.0)}$
$s′=-6.60$ $cm$ is the position of the image
$m=-\frac{s′}{s}=−\frac{-6.60}{16.5}=0.4$
$m=\frac{y′}{y}$
$y′=m×y=(0.4)×0.600=0.240$ $cm$
As the image distance is negative and the image height is positive, we can conclude that the image formed is virtual.
