$\frac{1}{{{f_L}}} = \frac{1}{{{f_m}}} - \frac{2}{{{f_l}}}$
${f_m} \to \infty $
$\frac{1}{{{f_l}}} = \left( {\mu - 1} \right)\left( {\frac{1}{R} - \frac{1}{\infty }} \right)$
$\frac{1}{{{f_l}}} = \frac{{\mu - 1}}{R}$
$\frac{1}{{{f_l}}} = \frac{R}{{\mu - 1}}$
${f_m} = \frac{{ - fl}}{2} = \frac{{ - R}}{{2\left( {\mu - 1} \right)}}$
Given that $\frac{R}{{2\left( {\mu - 1} \right)}} = 28$ ...........$(1)$
$\frac{1}{{{f_l}}} = \left( {\mu - 1} \right)\left( {\frac{1}{\infty } - \frac{1}{{ - R}}} \right)$
$\frac{1}{{{f_l}}} = \frac{{\mu - 1}}{R}$
${f_m} = \frac{{ - R}}{2}$
$\frac{1}{{{f_F}}} = \frac{1}{{{f_m}}} - \frac{2}{{{f_l}}}$
$\frac{1}{{{f_m}}} = \frac{2}{R} - \frac{{2\left( {\mu - 1} \right)}}{R}$
${f_F} = 10\,\,cm$
$\frac{2}{R} + \frac{{2\left( {\mu - 1} \right)}}{R} = \frac{1}{{10}}$
$\frac{{2\mu }}{R} = \frac{1}{{10}}$ ..........$(2)$
$R = 20\mu $
$\frac{{20\mu }}{{2\left( {2\mu - 1} \right)}} = 28$
$\mu = 2.8\left( {\mu - 1} \right)$
$\mu = 2.8\mu - 2.8$
$\mu \frac{{2.8}}{{1.8}} = \frac{{14}}{9}$
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