Answer
(a) $29cm$ and $9.8cm$
(b) $29cm$ real and inverted; $9.8cm$ virtual and upright
Work Step by Step
(a) We can find the required two locations as follows:
$\frac{1}{d_i}+\frac{1}{d_{\circ}}=\frac{1}{f}$
$\implies \frac{1}{\pm 2d_i}+\frac{1}{d_{\circ}}=\frac{1}{19.5}$
This simplifies to:
$d_{\circ}=19.5(1\pm0.5)$
$\implies d_{\circ}=29cm$ and $d_{\circ}=9.8cm$
(b) Case 1: $\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_{\circ}}$
$\implies \frac{1}{d_i}=\frac{1}{19.5cm}-\frac{1}{29.25cm}$
$\implies d_i=58.5cm$
and $m=-\frac{d_i}{d_{\circ}}$
We plug in the known values to obtain:
$m=-\frac{58.5cm}{29.25cm}=-2$
Thus, for $29.25cm$ object distance, the image formed is real and inverted.
(b) Case 2: $\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_{\circ}}$
$\implies \frac{1}{d_i}=\frac{1}{19.5cm}-\frac{1}{9.75cm}$
$\implies d_i=-19.5cm$
and $m=-\frac{d_i}{d_{\circ}}$
$\implies m=-\frac{-19.5cm}{9.75cm}=2$
Thus, for $9.75cm$ object distance, the image is virtual and upright.
