(a) (i) Draw a ray diagram to show the working of a compound microscope. Obtain the expression for the total magnification for the final image to be…

Linear magnification due to the objective is \[\begin{aligned} & \tan \beta=\left(\frac{\mathrm{h}}{\mathrm{f}_{0}}\right)=\left(\frac{\mathrm{h}^{\prime}}{\mathrm{L}}\right) \\ & \mathrm{m}_{0}=\frac{\mathrm{h}^{\prime}}{\mathrm{h}}=\frac{\mathrm{L}}{\mathrm{f}_{0}} \end{aligned} \] Here L is the distance between the second focal point of the objective and the first focal point of the eyepiece. Linear magnification due to eyepiece is \[\mathrm{m}_{\mathrm{e}}=\left(1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right) \] Thus, the total magnification is, \(\displaystyle \mathrm{m}=\mathrm{m}_{0} \times \mathrm{m}_{\mathrm{e}}\) \[\mathrm{m}=\frac{\mathrm{L}}{\mathrm{f}_{0}}\left(1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right) \] \[\mathrm{m}=\frac{v_{0}}{u_{0}}\left(1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right) \]