The focal lengths of the objective and eye-piece of a microscope are 1 cm & 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)

15 cm

For a microscope,

$m =\frac{v_0}{u_0}\frac{D}{f_e}=-45$

$⇒\frac{v_0}{u_0}=-9$

also $\frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0}$

$⇒ v_0 = 10 cm$

∴ Length of tube = $v_0 + f_e = 15 cm$