(a)
Hint: L=f₀ + f_e

Step: Find the separation between the lens in normal adjustment.

For normal adjustment, the separation between the objective lens and the eyepiece = ${\mathrm{f}}_0+{\mathrm{f}}_{\mathrm{e}}=140 +5 =145 \mathrm{cm}$

(b)

Hint: Angle subtended by the tower and its image is same.

Step 1: Find the angle subtended by the tower at the telescope.
Height of the tower, h${}_{\mathrm{i}}$ =100 m
Object distance, u = 3Km = 3000 m
The angle subtended by the tower at the telescope:

Step 2: Find the height of the image.
If h₂ is the height of the image of the tower formed by the objective lens,

the angle subtended by the image produced by the objective lens: $\mathrm{\theta }=\frac{{\mathrm{h}}_2}{{\mathrm{f}}_0}=\frac{{\mathrm{h}}_2}{140} \mathrm{rad}$

$\Rightarrow \frac{1}{30}=\frac{{\mathrm{h}}_2}{140}$
$\Rightarrow {\mathrm{h}}_2=\frac{140}{30}=4.7 \mathrm{cm}$

Therefore, the objective lens forms a 4.7 cm tall image of the tower.

(c)

Hint: \(m=(1+\frac{d}{f_{e}})\)

Step 1:  Find the magnification of the eyepiece.
The magnification of the eyepiece:

$\mathrm{m}=1+\frac{\mathrm{d}}{{\mathrm{f}}_{\mathrm{e}}}=1+\frac{25}{5}=1+5=6$

Step 2: Find the height of the final image.
Height of the final image = mh₂ = 6 x 4.7 = 28.2 cm