Here we want the distance between given objective and eye lens for the required magnification of 30. Let the final image is formed at least distance of distinct vision for eyepiece. $m_e$ = $\left(1+\frac{D}{f_e}\right)$ = $\frac{D}{u_e}$ or $m_e$ = 1 + $\frac{25}{5}$ = $\frac{25}{u_e}$ or $m_e$ = 6 and $u_e$ = $\frac{25}{6}$ cm Now magnification by objective lens m = $m_0 \times m_e$ or $m_0$ = $\frac{m}{m_e}$ = $\frac{30}{6}$ = 5 Also $m_0$ = $\frac{v_0}{u_0}$ ⇒ $m_0$ = $\frac{v_0}{u_0}$ = $-5$ (Since real image is formed by objective.) So , the relation $v_0$ - 5 $u_0$ Now lens formula for objective lens $\frac{1}{v_0} - \frac{1}{u_0}$ = $\frac{1}{f_0}$ or $\frac{1}{-5u_0} - \frac{1}{u_0}$ = $\frac{1}{1.25}$ $\frac{1+5}{5u_0}$ = $\frac{1}{1.25}$ or $5u_0$ = 7.5 or $u_0$ = - 1.5 cm Also $v_o$ = – 5 $u_o$ = 7.5 cm So, required distance between objective and eyepiece L = $v_0$ + $\left| u_e \right|$ = 7.5 + $\frac{25}{6}$ = 11.67 cm