c
અહીં  ,\( f_0 = 2.0 \,cm,   f_e = 6.25\, cm, u_0 = ?\)

\({v_e}\, = \,\, - 25\,\,cm\,\,\,\,\,\,\because \,\,\,\frac{1}{{{v_e}}}\,\, - \,\,\frac{1}{{{u_e}}}\,\, = \,\,\,\frac{1}{{{{{f}}_e}}}\,\)

\(\therefore \,\,\,\frac{1}{{{u_e}}}\,\, = \,\,\frac{1}{{{v_e}}}\,\, - \,\,\frac{1}{{{{{f}}_e}}}\,\, = \,\,\frac{1}{{ - 25}}\,\, - \,\,\frac{1}{{6.25}}\,\, = \,\,\frac{{ - 1 - 4}}{{25}}\,\, = \,\,\frac{{ - 5}}{{25}}\,\,\,\)

\(\Rightarrow \,\,\,{u_e} = \,\, - 5\,\,cm\)

જેમ આઇપીસી અને ઓબ્જેક્ટિવ વચ્ચે અંતર  \( = {\text{ 15  cm}}{\text{.}}\)

\({v_0} = \,\,15 - 5\,\, = \,\,10\,\,cm\)

\(\because \,\,\,\frac{1}{{{v_0}}}\,\, - \,\,\,\frac{1}{{{u_0}}}\,\, = \,\,\frac{1}{{{{{f}}_0}}}\,\,\,\,\,\)

\(\therefore \,\,\frac{1}{{{u_0}}}\,\, = \,\,\frac{1}{{{v_0}}}\,\, - \,\,\frac{1}{{{{{f}}_0}}}\,\, = \,\,\frac{1}{{10}} - \frac{1}{2}\,\, = \,\,\,\frac{{1 - 5}}{{10}}\,\,\,\,\, \)

\(\Rightarrow \,\,\,{u_0} = \frac{{ - 10}}{4}\,\, = \,\, - 2.5\,\,cm\)

મેગ્નિફાઇંગ પાવર  \(\, = \,\,\frac{{{v_0}}}{{|{u_0}|}}\,\,\left[ {1 + \frac{D}{{{{{f}}_e}}}} \right]\,\, = \,\,\frac{{10}}{{2.5}}\,\,\left[ {1 + \frac{{25}}{{6.25}}} \right]\,\, = \,\,20\)