Pls solve this question and explain the answer in detail

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$$\begin{gathered} for compound microscope, focal length of objective lens is less than the eyepiece lens. \\ magnification =M = -\frac{L}{f_0}\times \frac{25}{f_e} \\ M = 30 \\ L = \frac{M\times f_0{f}_e}{25}=\frac{30\times 1.25\times 5}{25}=7.5 cm \\ where L is the dis\tan ce between the two lenses. \end{gathered}$$

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