The focal length of objective and eye piece of a microscope are 1 cm and 5 cm respectively. If the magnifying power for relaxed eye is 45, then length of the tube is

Magnifying power for relaxed eye is

$\begin{array}{l}\mathrm{M}=-\frac{{\mathrm{v}}_{\mathrm{o}}}{{\mathrm{u}}_{\mathrm{o}}}\frac{\mathrm{D}}{{\mathrm{f}}_{\mathrm{e}}}\\ \mathrm{Here}\mathrm{M}=-45,{\mathrm{f}}_{\mathrm{o}}=1\mathrm{cm}\\ {\mathrm{f}}_{\mathrm{e}}=5\mathrm{cm},\mathrm{D}=25\mathrm{cm}\\ \therefore -45=\frac{{\mathrm{v}}_{\mathrm{o}}}{{\mathrm{u}}_{\mathrm{o}}}.\frac{25}{5};\mathrm{or}\frac{{\mathrm{v}}_{\mathrm{o}}}{{\mathrm{u}}_{\mathrm{o}}}=9;\\ \mathrm{For}\mathrm{objective}\mathrm{image}\mathrm{is}\mathrm{real}.\\ \\ {\mathrm{u}}_{\mathrm{o}}=-\frac{{\mathrm{v}}_{\mathrm{o}}}{9};{\mathrm{f}}_{\mathrm{o}}=+1\mathrm{cm}\\ \mathrm{Substituting}\mathrm{in}\frac{1}{{\mathrm{v}}_{\mathrm{o}}}-\frac{1}{{\mathrm{u}}_{\mathrm{o}}}=\frac{1}{{\mathrm{f}}_{\mathrm{o}}},\\ \mathrm{we}\mathrm{get},\frac{1}{{\mathrm{v}}_{\mathrm{o}}}+\frac{9}{{\mathrm{v}}_{\mathrm{o}}}=1\mathrm{or}{\mathrm{v}}_{\mathrm{o}}=10\mathrm{cm}\\ \therefore \mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{tube}\\ \mathrm{L}={\mathrm{v}}_{\mathrm{o}}+{\mathrm{f}}_{\mathrm{e}}=10+5=15\mathrm{cm}\end{array}$