A compound microscope consists of an objective lens of focal length $1cm$ and an eye piece of focal length $5cm$ with a separation of $10cm$ . The distance between an object and the objective lens,at which the strain on the eye is minimum is  $\frac{n}{40}$. The value of n is ____

Strain is minimum this implies final image is at far point and intermediate image is at focal length of eye piece . That means intermediate image distance from objective is (10-5=5cm).

$\begin{array}{l}Lens formula for objective lens,\\ \text{}\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\text{}\\ \Rightarrow \frac{1}{+1}=\frac{1}{+5}-\frac{1}{u}\text{}\\ \Rightarrow u=-\frac{5}{4}\text{}\\ Object dis\tan ce=\frac{50}{40}cm\end{array}$