Answer
(a) Diagram below
(b) s'=33.0 cm; y'=-1.2 cm; The image formed is real.
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:
$s'=\frac{fs}{s-f}$
$s'=\frac{11\times16.5}{16.5-11.0}$
$s'= 33.0$ $cm$ is the position of the image
$m=-\frac{s'}{s}=-\frac{33}{16.5}=-2$
$m=\frac{y'}{y}$
$y'=m\times{y}=(-2)\times{0.600}=-1.2$ $cm$
As the image distance is positive and the image height is negative, we can conclude that the image formed is real.
