Answer
(a) $12cm$ to the left
(b) $-0.12$
Work Step by Step
(a) We can determine the required image location as follows:
$d_{i1}=\frac{f_1d_{\circ 1}}{d_{\circ 1}-f_1}$
We plug in the known values to obtain:
$d_{i1}=\frac{(-6.00cm)(18.0cm)}{18.0cm-(-6.00cm)}$
$\implies d_{i1}=-\frac{108cm^2}{24cm}=-4.50cm$
The object distance for the converging lens is $d_{\circ 2}=20.0cm+4.50cm=24.50cm$
Now the image formed by the second lens is
$d_{i2}=\frac{f_2d_{\circ 2}}{d_{\circ 2}-f_2}$
We plug in the known values to obtain:
$d_{i2}=\frac{(24.50cm)(8.00cm)}{(24.50cm-8.00cm)}$
$\implies d_{i2}=12cm$ to the left of the converging lens
(b) We can determine the required magnification as follows:
$m=\frac{d_{i1}d_{i2}}{d_{\circ 1}d_{\circ 2}}$
We plug in the known values to obtain:
$m=\frac{(-4.50cm)(12.0cm)}{(18.0cm)(24.5cm)}$
$\implies m=-0.12$
