The length of the compound microscope is 15 cm . The magnifying power for relaxed eye is 25 . If the focal length of eye lens is 6 cm then the object distance for objective lens will be

Given parameters: microscope length $L = 15\text{ cm}$, magnifying power $M = 25$, eyepiece focal length $f_e = 6\text{ cm}$, and standard least distance of distinct vision $D = 25\text{ cm}$.

For a relaxed eye with final image at infinity, the objective image forms at the eyepiece focal point. Thus, the objective image distance is determined by $v_o = L - f_e = 15 - 6 = 9\text{ cm}$.

The magnifying power formula yields $M = \left( \frac{v_o}{u_o} \right) \left( \frac{D}{f_e} \right)$. Substituting known values gives $25 = \left( \frac{9}{u_o} \right) \left( \frac{25}{6} \right)$.

Solving for the object distance $u_o$ results in $u_o = \frac{9 \times 25}{25 \times 6} = \frac{9}{6} = 1.5\text{ cm}$.

Final object distance: $1.5\text{ cm}$