(a) According to the question, the focal length of the convex lens, $\mathrm{f}_1=30 \mathrm{~cm}$
the focal length of the convex lens, $\mathrm{f}_2=-20 \mathrm{~cm}$
separation between two lenses $=8 \mathrm{~cm}$.
If a parallel beam of light is incident from the left on the convex lens, $\mathrm{f}_1=30 \mathrm{~cm}, \mathrm{u}_1=-\propto$
As, $\frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}_1}=\frac{1}{\mathrm{f}_1} \Rightarrow \frac{1}{\mathrm{v}_1}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{u}_1}$ $=\frac{1}{30}+\frac{1}{(-\infty)}=\frac{1}{30} \quad \therefore v_1=30 \mathrm{~cm}$.
This image acts as a virtual object for the second lens. $f_2=-20 \mathrm{~cm}, \mathrm{u}_2=+(30-8)=+22 \mathrm{~cm}$.
Using $\frac{1}{\mathrm{v}_2}-\frac{1}{\mathrm{u}_2}=\frac{1}{\mathrm{f}_2}$
$$
\frac{1}{\mathrm{v}_2}=\frac{1}{\mathrm{f}_2}+\frac{1}{\mathrm{u}_2}=\frac{1}{-20}+\frac{1}{22}=\frac{-11+10}{220}
$$
$$
\Rightarrow \frac{1}{v_2}=-\frac{1}{220} \quad \therefore v_2=-220 \mathrm{~cm} \text {. }
$$
Therefore the parallel beam of light appears to damage from a point $(220-4)=216 \mathrm{~cm}$ from the centre of the two - lens system.
(ii) Let the parallel beam of light be incident from the left on the concave lens
then, $f_1=-20 \mathrm{~cm}$. $u_1=-\infty$.
$$
\begin{gathered}
\text { As, } \frac{1}{v_1}=\frac{1}{f_1}+\frac{1}{u_1} \\
\Rightarrow \frac{1}{v_1}=\frac{1}{-20} \\
\therefore v_1=-20 \mathrm{~cm}
\end{gathered}
$$
This image acts as real image for the second convex
lens: $\mathrm{f}_2=+30 \mathrm{~cm}$.
$$
\begin{aligned}
&\mathrm{u}_2=-(20+8)=-28 \mathrm{~cm} . \\
&\therefore \frac{1}{\mathrm{v}_2}=\frac{1}{\mathrm{f}_2}+\frac{1}{\mathrm{u}_2} \\
&\quad=\frac{1}{30}+\frac{1}{-28}=\frac{14-15}{420}=-\frac{1}{420}
\end{aligned}
$$
The parallel beam of light appears to diverge from a point $416 \mathrm{~cm}$ on the left of the centre of the two lenses.
The answer depends on which side of the lens, the object is placed. The notion of the effective focal length is not meaningful.
(b) Here object distance, $\mathrm{u}_1=-40 \mathrm{~cm}$ $\mathrm{f}_1=30 \mathrm{~cm}$.
As, $\frac{1}{\mathrm{v}_1}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{u}_1}=\frac{1}{30}+\frac{1}{-40}$ $=\frac{4-3}{120}=\frac{1}{120} \therefore \mathrm{v}_1=120 \mathrm{~cm}$.
$\therefore$ Magnification produced by the convex lens, $\mathrm{m}_1=\frac{120}{-40}=-3 \cdot \mathrm{m}_1=|3|$
The image formed by the convex lens acts as virtual object for the concave lens.
$\mathrm{u}_2=+(120-8)=112 \mathrm{~cm}, \mathrm{f}_2=-20 \mathrm{~cm}$
As, $\frac{1}{\mathrm{v}_2}=\frac{1}{\mathrm{f}_2}+\frac{1}{\mathrm{u}_2}=\frac{1}{-20}+\frac{1}{112}$
$=\frac{-112+20}{20 \times 112}=\frac{-92}{20 \times 112} \quad \therefore v_2=\frac{-112 \times 20}{92}$
$\therefore$ Magnification produced by the concave lens, $\mathrm{m}_2=\frac{\mathrm{v}_2}{\mathrm{u}_2}=\frac{112 \times 20}{92 \times 112}=\frac{20}{92}$
$\therefore$ Total magnification produced by the combination, $\mathrm{m}=\mathrm{m}_1 \times \mathrm{m}_2=3 \times \frac{20}{92}=0.652$
Size of image $=\mathrm{m} \times$ size of object
$$
=0.652 \times 1.5=0.98 \mathrm{~cm}
$$