• # question_answer The focal lengths of the objective and eye-piece of a microscope are 1 cm and 5 cm respectively. If the magnifying power for the normal adjustment is -45, the length of the tube is (Least distance of clear vision = 25 cm) A)  23 cm                        B)  25 cmC)    15 cm             D)    13 cm

$m=\frac{{{v}_{0}}}{{{u}_{0}}}\frac{D}{{{f}_{e}}}=-45\Rightarrow \frac{{{v}_{0}}}{{{u}_{0}}}=-9$ Also $\frac{1}{{{V}_{0}}}-\frac{1}{{{u}_{0}}}=\frac{1}{{{f}_{0}}}\Rightarrow {{V}_{0}}=10\,cm.$ $\therefore$Length of tube $={{V}_{0}}+{{f}_{e}}=15\,cm.$ Hence, the correction option is (c).