Answer
(a) 12.0 cm to the right of the second lens.
(b) $-6.0$
Work Step by Step
(a) We can find the required location of the image as follows:
$d_{i1}=\frac{(d_{\circ 1})(f_1)}{d_{\circ 1}-f_1}$
We plug in the known values to obtain:
$d_{i1}=\frac{(12.0cm)(8.0cm)}{12.0cm-8.0cm}$
$\implies d_{i1}=24cm$
The object distance is $d_{\circ 2}=20.0cm-24.0cm=-4.0cm$
Now the image formed by the second lens is located at
$d_{i2}=\frac{(d_{\circ 2})(f_2)}{d_{\circ 2}-f_2}$
We plug in the known values to obtain:
$d_{i2}=\frac{(-4.0cm)(-6.0cm)}{-4.0cm+6.00cm}$
$\implies d_{i2}=12.0cm$ to the right of the second lens.
(b) The required magnification can be determined as follows:
$m=(-\frac{d_{i1}}{d_{\circ 1}})-(\frac{d_{i2}}{d_{\circ 2}})$
$\implies m=(\frac{d_{i1} d_{i2}}{d_{\circ}1 d_{\circ 2}})$
We plug in the known values to obtain:
$m=\frac{(24.0cm)(12.0cm)}{(12.0cm)(-4.0cm)}=-6.0$