An astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance

$f_o$ = 40cm ; $f_e$ = 4cm ; u₀ = 200cm

$\frac{1}{{\mathrm{f}}_0}=\frac{1}{{\mathrm{v}}_0}-\frac{1}{{\mathrm{u}}_0} ; {\mathrm{V}}_0=50 \mathrm{cm} ; \mathrm{L}={\mathrm{V}}_0+{\mathrm{f}}_{\mathrm{e}}$

= 50 + 4 = 54cm

when image is formed at least distance of distinct vision

$$\begin{gathered} \frac{1}{{\mathrm{f}}_{\mathrm{e}}}=\frac{1}{{\mathrm{v}}_{\mathrm{e}}}-\frac{1}{{\mathrm{u}}_{\mathrm{e}}} ; {\mathrm{L}}_{\mathrm{D}}={\mathrm{V}}_0+{\mathrm{u}}_{\mathrm{e}} {\mathrm{V}}_{\mathrm{e}}=-25 \mathrm{cm} \\ \frac{1}{{\mathrm{f}}_{\mathrm{e}}}=\frac{1}{{\mathrm{v}}_{\mathrm{e}}}-\frac{1}{{\mathrm{u}}_{\mathrm{e}}} ; {\mathrm{u}}_{\mathrm{e}}=\frac{100}{29} L_D=50+\frac{100}{29}=53.44 \mathrm{cm} \end{gathered}$$