In a compound microscope, the focal length of two lenses are 1.5cm and 6.25cm. An object is placed at 2cm from the objective and the final image is formed at 25cm from the eye lens. The distance between the two lenses is __ (in cm)

For objective                                                     For eyepiece

$\begin{array}{l}\mu =-2cm\text{ f=1.5cm v=-25cm f=6.25cm}\\ \frac{1}{v}-\frac{1}{4}=\frac{1}{f}\frac{1}{v}-\frac{1}{4}=\frac{1}{f}\\ \frac{1}{v}=\frac{1}{1.5}=\frac{1}{2}\frac{1}{4}=\frac{1}{v}-\frac{1}{f}=\frac{1}{-25}+\frac{1}{-6.25}\\ v=6cm\text{ 4=-5cm}\end{array}$

Distance b/n lenses = 6+5=11cm