(a)
Hint: Use the lens equation.

Step 1: Find the final image formed when the rays are incident on the convex lens first.
The focal length of the convex lens, $$f${}_1$​= 30 cm
The focal length of the concave lens, $$f${}_2$​ = −20 cm
Distance between the two lenses, d = 8.0 cm
When the parallel beam of light is incident on the convex lens first:
Object distance, ${\mathrm{u}}_1=\infty$
Image distance, ${\mathrm{v}}_1=?$
According to the lens formula,

$\frac{1}{{\mathrm{v}}_1}-\frac{1}{{\mathrm{u}}_1}=\frac{1}{{\mathrm{f}}_1}$

$\frac{1}{{\mathrm{v}}_1}=\frac{1}{30}-\frac{1}{\infty }=\frac{1}{30}$
${\mathrm{v}}_1=30<mspace\; width="1em"></mspace>\mathrm{cm}$

$\mathrm{For}<mspace\; width="1em"></mspace>\mathrm{concave}<mspace\; width="1em"></mspace>\mathrm{lens},$
$\mathrm{Object}<mspace\; width="1em"></mspace>\mathrm{distance},<mspace\; width="1em"></mspace>{\mathrm{u}}_2=(30-\mathrm{d})=30-8=22<mspace\; width="1em"></mspace>\mathrm{cm}$
$\mathrm{lmage}<mspace\; width="1em"></mspace>\mathrm{distance},<mspace\; width="1em"></mspace>{\mathrm{v}}_2=?$
$\frac{1}{{\mathrm{v}}_2}=\frac{1}{22}-\frac{1}{20}=\frac{10-11}{220}=\frac{-1}{220}$
$\therefore <mspace\; width="1em"></mspace>{\mathrm{v}}_2=-220<mspace\; width="1em"></mspace>\mathrm{cm}$

 The parallel incident beam appears to diverge from a point that is$$$\left(220-\frac{\mathrm{d}}{2}=220-4=216<mspace\; width="1em"></mspace>\mathrm{cm}\right)$ from the centre of the combination of the two lenses.

Step 1: Find the final image formed when the rays are incident on the concave lens first.
 When the parallel beam of light is incident on the concave lens first:

Object distance, ${\mathrm{u}}_2=\infty$

Image distance, ${\mathrm{v}}_2=?$
Using the lens formula,

$\frac{1}{{\mathrm{v}}_2}-\frac{1}{{\mathrm{u}}_2}=\frac{1}{{\mathrm{f}}_2}$
$\frac{1}{{\mathrm{v}}_2}=\frac{1}{{\mathrm{f}}_2}+\frac{1}{{\mathrm{u}}_2}$

$\frac{1}{{\mathrm{v}}_2}=\frac{1}{-20}+\frac{1}{-\infty }=-\frac{1}{20}$
$\therefore <mspace\; width="1em"></mspace>{\mathrm{v}}_2=-20<mspace\; width="1em"></mspace>\mathrm{cm}$

For the convex lens:

Object distance, ${\mathrm{u}}_1=-\left(20+\mathrm{d}\right)=-\left(20+8\right)=-28<mspace\; width="1em"></mspace>\mathrm{cm}$

Image distance, ${\mathrm{v}}_1=?$
Applying the lens formula:

$$$\frac{1}{{\mathrm{v}}_1}-\frac{1}{{\mathrm{u}}_1}=\frac{1}{{\mathrm{f}}_1}$
$\frac{1}{{\mathrm{v}}_1}=\frac{1}{30}+\frac{1}{-28}=\frac{14-15}{420}=\frac{-1}{420}$
$\therefore <mspace\; width="1em"></mspace>{\mathrm{v}}_2=-420<mspace\; width="1em"></mspace>\mathrm{cm}<mspace\; width="1em"></mspace>$

Hence, the parallel incident beam appears to diverge from (420-4=416 cm). The diversion happens from the left of the centre of the combination of the two lenses.

Step 3: Compare the effective focal length in both cases.
The answer does depend on the side of the combination at which the parallel beam of light is incident. The notion of effective focal length does not seem to be useful for this combination.

(b)

Hint: Use the lens formula.

Step 1: Find the magnification produced by the convex lens.
For the convex lens:
Object distance, u${}_1$​=-40 cm

Image distance, ${\mathrm{v}}_1=?$

According to the lens formula:

$\frac{1}{{\mathrm{v}}_1}-\frac{1}{{\mathrm{u}}_1}=\frac{1}{{\mathrm{f}}_1}<mspace\; width="1em"></mspace>$
$\frac{1}{{\mathrm{v}}_1}=\frac{1}{30}+\frac{1}{-40}=\frac{4-3}{120}=\frac{1}{120}$
$\therefore <mspace\; width="1em"></mspace>{\mathrm{v}}_1=120<mspace\; width="1em"></mspace>\mathrm{cm}<mspace\; width="1em"></mspace>$
$\mathrm{Magnification},<mspace\; width="1em"></mspace>{\mathrm{m}}_1=\frac{{\mathrm{v}}_1}{{\mathrm{u}}_1}=\frac{120}{-40}=-3$

Hence, the magnification due to the convex lens is 3.

Step 2: Find the magnification produced by the concave lens.
For the concave lens:

Object distance, ${\mathrm{u}}_1=+120-8=112<mspace\; width="1em"></mspace>\mathrm{cm}$

Image distance, ${\mathrm{v}}_2=?$
According to the lens formula:

$\frac{1}{{\mathrm{v}}_2}-\frac{1}{{\mathrm{u}}_2}=\frac{1}{{\mathrm{f}}_2}$
$\frac{1}{{\mathrm{v}}_2}=\frac{1}{-20}+\frac{1}{112}=\frac{-112+20}{2240}=\frac{-92}{2240}$
$\therefore {\mathrm{v}}_2<mspace\; width="1em"></mspace>=\frac{-2240}{92}<mspace\; width="1em"></mspace>\mathrm{cm}<mspace\; width="1em"></mspace>$
$\mathrm{Magnification},<mspace\; width="1em"></mspace>{\mathrm{m}}_2=\frac{{\mathrm{v}}_2}{{\mathrm{u}}_2}<mspace\; width="1em"></mspace>=\frac{-2240}{92}\times \frac{1}{112}=\frac{-20}{92}$

Step 3:  Find the magnification produced by the combination.
The magnification produced by the combination:

$\mathrm{m}={\mathrm{m}}_1{\mathrm{m}}_2$
$\mathrm{m}=-3\times -\frac{20}{92}=\frac{60}{92}=0.652$

The magnification is given by:

$\mathrm{m}=\frac{\mathrm{height}<mspace\; width="1em"></mspace>\mathrm{of}<mspace\; width="1em"></mspace>\mathrm{the}<mspace\; width="1em"></mspace>\mathrm{image}}{\mathrm{height}<mspace\; width="1em"></mspace>\mathrm{of}<mspace\; width="1em"></mspace>\mathrm{the}<mspace\; width="1em"></mspace>\mathrm{object}}=\frac{\mathrm{h}\text{'}}{\mathrm{h}}=0.652$
$\mathrm{h}\text{'}=0.652<mspace\; width="1em"></mspace>\times \mathrm{h}<mspace\; width="1em"></mspace>$
$\mathrm{h}=0.652\times 1.5=0.98<mspace\; width="1em"></mspace>\mathrm{cm}<mspace\; width="1em"></mspace>$

Hence, the height of the image is 0.98 cm.