Explanation

(b) For the relaxed eye, magnifying power is
$M = - \frac{v_{0}}{u_{0}}\frac{D}{f_{e}}$
$\therefore - 45 = - \frac{v_{0}}{u_{0}} \times \frac{25}{5},\ \frac{v_{0}}{u_{0}} = 9$
For objective lens, image is real
$\therefore v_{0} = + v_{0},u_{0} = - \frac{v_{0}}{9}$
Given, $f_{0} = 1\ $cm
Form $\frac{1}{v_{0}} - \frac{1}{u_{0}} = \frac{1}{f_{0}}$
$\frac{1}{v_{0}} + \frac{9}{v_{0}} = \frac{1}{1};{\ v}_{0} =$10 cm
Length of the tube $= v_{0} + f_{e} = 10 + 5 = 15\ $cm