Question

56.The length of the tube of a compound microscope 15cm.The focal length of objective and eye lenses are 1cm and 5cm respectively.The magnifying power of microscope for relaxed vision is --------

Answer 1

$$\begin{gathered} \mathbf{Magnification}\mathbf{}\mathbf{of}\mathbf{}\mathbf{compound}\mathbf{}\mathbf{microscope}\mathbf{:} \\ \mathrm{M}=\frac{\mathrm{L}\mathrm{d}}{{\mathrm{f}}_{\mathrm{o}}{\mathrm{f}}_{\mathrm{e}}} \\ \mathrm{Here}, \\ \mathrm{L}=\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{tube} \\ \mathrm{d}=\mathrm{Least}\mathrm{distance}\mathrm{of}\mathrm{distinct}\mathrm{vision} \\ {\mathrm{f}}_{\mathrm{o}}=\mathrm{Focal}\mathrm{length}\mathrm{of}\mathrm{objective} \\ {\mathrm{f}}_{\mathrm{e}}=\mathrm{Focal}\mathrm{length}\mathrm{of}\mathrm{eyepiece} \\ \mathrm{Hence}, \\ \mathrm{M}=\frac{15\times 25}{1\times 5} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{75} \end{gathered}$$