(a) We want the final image at least distance of distinct vision. Let the object in front of objective is at distance $u_0$. Let us first find the $u_e$, the object distance for eyepiece. Here $v_e$ = - 25 , $f_e$ = 6.25 , $u_e$ = ? $\frac{1}{v_e} - \frac{1}{u_e}$ = $\frac{1}{f_e}$ $\frac{1}{-25} - \frac{1}{u_e}$ = $\frac{1}{6.25}$ or $-\frac{1}{u_e}$ = $\frac{1}{6.25} + \frac{1}{25}$ = $\frac{4+1}{25}$ $u_e$ = – 5 cm So, image distance of objective lens $v_o$ = 15 – $u_e$ = 15 – 5 = 10 cm Now we can get required position of object in point of objective. $\frac{1}{v_0} - \frac{1}{u_0}$ = $\frac{1}{f_0}$ or $\frac{1}{+10} - \frac{1}{u_e}$ = $\frac{1}{2}$ $\frac{1}{u_0}$ = $\frac{1}{10} - \frac{1}{2}$ = $\frac{1-5}{10}$ ⇒ $u_0$ = - 2.5 cm So, the object should be 2.5 cm in front of objective lens. Magnifying power (most strained eye) m = - $\frac{v_0}{u_0}\left[1+\frac{D}{f_e}\right]$ or m = - $\frac{10}{-2.5}\left[1+\frac{25}{6.25}\right]$ = 4 [5] = 20 (b) We want the final image at infinity. Let us again assume the object in front of objective at distance $u_o$. Since $v_e$ = ∞ ∴ $\frac{1}{v_e} - \frac{1}{u_e}$ = $\frac{1}{f_e}$ The object distance $u_e$ for the eyepiece should be equal to $f_e$ = 6.25 cm to obtain final image at ∞. So, image distance of objective lens $v_o$ = 15 – $f_e$ = 15 – 6.25 = 8.75 cm Now, lens formula, $\frac{1}{v_0} - \frac{1}{u_0}$ = $\frac{1}{f_0}$ ⇒ $\frac{1}{8.75} - \frac{1}{u_0}$ = $\frac{1}{2}$ or $\frac{1}{u_0}$ = $\frac{1}{8.75} - \frac{1}{2}$ = $\frac{2-8.75}{17.5}$ ⇒ $u_0$ = - $\frac{17.5}{6.75}$ cm = - 2.59 cm Magnifying power (most relaxed eye) m = - $\frac{v_0}{u_0}\left[\frac{D}{f_e}\right]$ ⇒ m = - $\frac{8.75}{-2.59}\left[\frac{25}{6.25}\right]$ = 13.5