(a) (i) Ray diagram of compound microscope:

In a compound microscope there are two lenses - objective $\left(\mathrm{O}\right)$ and Eyepiece $\left(\mathrm{E}\right)$.

Object $\mathit{P}\mathit{Q}$ is placed in front of the objective at a distance more than the focal length of the objective.

An inverted, magnified, real image ${\mathit{P}}_1{\mathit{Q}}_1$ is formed in front of eyepiece within the optical centre and the focus of the eyepiece. This acts as the object of the eyepiece. An (erect with respect to ${\mathit{P}}_1{\mathit{Q}}_1$, inverted with respect to $\mathit{P}\mathit{Q}$ ), magnified, virtual image ${\mathit{P}}_2{\mathit{Q}}_2$ is formed at a distance $\mathrm{D}$ (minimum distance of distinct vision) from the eyepiece.

Magnification:

For objective:

Object distance $=\mathit{u}$
Image distance $=\mathit{v}$

Applying the lens formula,
$\frac{1}{\mathit{v}}-\frac{1}{-\mathit{u}}=\frac{1}{{\mathit{f}}_0}$
Or, $1+\frac{\mathit{v}}{\mathit{u}}=\frac{\mathit{v}}{{\mathit{f}}_0}$
Or, $\frac{\mathit{v}}{\mathit{u}}=\frac{\mathit{v}}{{\mathit{f}}_0}-1$
For eyepiece:


Magnification of the combination of objective and eyepiece $=\mathit{m}={\mathit{m}}_{\mathit{O}}\times {\mathit{m}}_{\mathit{e}}$
$\mathit{m}=\frac{\mathit{v}}{\mathit{u}}\times \left(1+\frac{\mathit{D}}{\mathit{f}\mathit{e}}\right)$

${\mathrm{P}}_1{\mathrm{Q}}_1$ image is formed very close to the eyepiece, hence $\mathit{v}$ can be approximated as the distance between the two lenses i.e., the length of the tube(L).
$\therefore \mathit{m}=\left(\frac{\mathit{L}}{{\mathit{f}}_0}-1\right)\left(1+\frac{\mathit{D}}{{\mathit{f}}_{\mathit{e}}}\right)$
Since, ${\mathit{f}}_{\mathit{e}}\ll \mathit{D}$ and ${\mathit{f}}_0\ll \mathit{L}$, hence the above expression may be approximated as,
$\mathit{m}=\frac{\mathit{L}}{{\mathit{f}}_0}\times \frac{\mathit{D}}{{\mathit{f}}_{\mathit{e}}}$
(ii) Given, $\mathit{u}=1.5\mathrm{cm},{\mathit{f}}_0=1.25\mathrm{cm},{\mathit{f}}_{\mathit{e}}=5\mathrm{cm},\mathit{D}$ $=25\mathrm{cm}$
Here, all alphabets are in their usual meanings Applying lens formula for objective lens,
$\frac{1}{\mathit{v}}-\frac{1}{\mathit{u}}=\frac{1}{{\mathit{f}}_0}$

$\therefore \mathit{v}=7.5\mathrm{cm}$
Magnification $=\mathit{m}=\frac{\mathit{v}}{\mathit{u}}\times \left(1+\frac{\mathit{D}}{{\mathit{f}}_{\mathit{e}}}\right)$

$\therefore \mathit{m}=-30$
OR
(b) (i) Astronomical telescope in normal adjustment:
In an astronomical telescope there are two lenses - objective $\left(\mathrm{O}\right)$ and Eyepiece $\left(\mathrm{E}\right)$.
The two lenses are so placed during focussing that the foci of the lenses meet at a point.
Objective is directed towards the object at infinity.
Parallel rays coming from the object meet at the focus of the objective and forms an inverted, real image ${\mathit{P}}_1{\mathit{Q}}_1$ in front of eyepiece. This point is the focus of eyepiece too.
This acts as the object of the eyepiece. An (inverted with respect to ${\mathit{P}}_1{\mathit{Q}}_1$, erect with respect to original object), highly magnified, real image is formed at infinity.
Magnification $=\mathit{m}$

Or, $\mathit{m}=$
Or, $\mathit{m}=\frac{\mathit{\beta }}{\mathit{\alpha }}$


[ $\mathit{\alpha }$ and $\mathit{\beta }$ being very small, $\tan \mathit{\alpha }=\mathit{\alpha }$ and $\tan \mathit{\beta }=\mathit{\beta }$ ]
Or, $\mathit{m}=\frac{\frac{{\mathit{Q}}_1{\mathit{P}}_1}{{\mathit{Q}}_1\mathit{E}}}{\frac{{\mathit{Q}}_1{\mathit{P}}_1}{{\mathit{Q}}_1\mathit{O}}}$
Or, $\mathit{m}=\frac{{\mathit{Q}}_1\mathit{O}}{{\mathit{Q}}_1\mathit{E}}$
∴ $\mathit{m}=\frac{{\mathit{f}}_{\mathit{o}}}{{\mathit{f}}_{\mathit{e}}}$


$\therefore {\mathit{f}}_{\mathit{o}}=2.9{\mathit{f}}_{\mathit{e}}$


$\therefore {\mathit{f}}_{\mathit{e}}=\frac{150}{3.9}$
$=38.46\mathrm{cm}$
${\mathit{f}}_{\mathit{O}}=2.9{\mathit{f}}_{\mathit{c}}$
$=2.9\times 38.46$
$=111.54\mathrm{cm}$
$$