Answer
(a) $5mm$
(b) $0.15$
Work Step by Step
(a) We can find the required location of the image as follows:
The location of the image formed by the first lens is given as
$d_{i1}=\frac{d_{\circ 1}}{f_1-d_{\circ 1}}$
We plug in the known values to obtain:
$d_{i1}=\frac{(-12cm)(24cm)}{24cm+12cm}$
$\implies d_{i1}=-8.0cm$
Thus, the object distance for the second lens is $d_{\circ 2}=8.0cm+6.00cm=14.0cm$
Now, the image distance formed by the second lens is given as
$d_{i2}=\frac{(-12cm)(14.0cm)}{(14+12)cm}$
$\implies d_{i2}=-6.46cm$
Now the require image location is
$6.4cm-6.00cm=0.46cm=4.6mm\approx 5mm$
(b) We can determine the require magnification as
$m=(-\frac{d_{i 1}}{d_{\circ 1} })(-\frac{d_{i2}}{d_{\circ 2}})$
We plug in the known values to obtain:
$m=(\frac{-8.00cm}{24cm})(\frac{-6.46cm}{14.0cm})$
$\implies m=0.15 $
