The focal length of the objective and the eyepiece of a telescope are $$50\,cm$$  and $$5\,cm$$  respectively. If the telescope is focused for distinct vision on a scale distant $$2\,m$$  from its objective, then its magnifying power will be :

Solution :
$$\eqalign{ & {\text{Given: }}{f_0} = 50\,cm,{f_e} = 5\,cm \cr & d = 25\,cm,{u_0} = - 200\,cm \cr & {\text{Magnification }}M = ? \cr & {\text{As }}\frac{1}{{{v_0}}} - \frac{1}{{{u_0}}} = \frac{1}{{{f_0}}} \cr & \Rightarrow \frac{1}{{{v_0}}} = \frac{1}{{{f_0}}} + \frac{1}{{{u_0}}} = \frac{1}{{50}} - \frac{1}{{200}} = \frac{{4 - 1}}{{200}} = \frac{3}{{200}} \cr & {\text{or }}{v_0} = \frac{{200}}{3}\,cm \cr & {\text{Now }}{v_e} = d = - 25\,cm \cr & {\text{From,}}\,\,\frac{1}{{{v_e}}} - \frac{1}{{{u_e}}} = \frac{1}{{{f_e}}} \cr & - \frac{1}{{{u_e}}} = \frac{1}{{{f_e}}} - \frac{1}{{{v_e}}} = \frac{1}{5} + \frac{1}{{25}} = \frac{6}{{25}} \cr & {\text{or, }}{v_e} = \frac{{ - 25}}{6}\,cm \cr & {\text{Magnification }}M = {M_0} \times {M_e} \cr & = \frac{{{v_0}}}{{{u_0}}} \times \frac{{{v_e}}}{{{u_e}}} = \frac{{\frac{{ - 200}}{3}}}{{200}} \times \frac{{ - 25}}{{\frac{{ - 25}}{6}}} = - \frac{1}{3} \times 6 = - 2 \cr} $$