The image is formed at least distance of distinct vision for sharp focus. The separation between two lenses will be $v_o + \lvert u_e \rvert$ Let us find first $v_o$ the image distance for objective lens. $\frac{1}{v_0} - \frac{1}{u_0}$ = $\frac{1}{f_0}$ or $\frac{1}{v_0}$ - $\frac{1}{-9}$ = $\frac{1}{8}$ or $\frac{1}{v_0}$ = $\frac{1}{8} - \frac{1}{9}$ = $\frac{1}{72}$ ⇒ $v_0$ = 72 mm = 7.2 cm Also we can find object distance for eyepiece $u_e$, as we know $v_e$ = D = 25 cm = 250 mm $\frac{1}{v_e} - \frac{1}{u_e}$ = $\frac{1}{f_e}$ ⇒ $\frac{1}{-250} - \frac{1}{u_e}$ = $\frac{1}{25} - \frac{1}{u_e}$ = $\frac{1}{25} + \frac{1}{250}$ = $\frac{10+1}{250}$ $u_e$ = - $\frac{250}{11}$ mm = - 22.7 mm = - 2.27 cm Separation between lenses L = $v_o + \lvert u_e \rvert$ = 72 + 22.7 or L = 94.7 mm = 9.47 cm Magnifying power m = $-\frac{v_0}{u_0} \left[1 + \frac{D}{f_e}\right]$ m = - $\frac{72}{-9} \left[1 + \frac{250}{25}\right]$ = 8 [11] = 88