(a) The instrument is called compound microscope because the focal length of objective lens is smaller than the focal length of eyepiece. (b) Power of objective $= P_o = 100 \, \mathrm{D}$ $\therefore \quad f_o = \frac{1}{100} \,\mathrm{m} = 1 \,\mathrm{cm}$ $\text{Power of eyepiece} = P_E = 40 \, \mathrm{D}$ $\therefore \quad f_E = \frac{1}{40} \,\mathrm{m} = 2.5 \,\mathrm{cm}$ Power of eyepiece $= P_E = 40 \, \mathrm{D}$ $\therefore \quad f_E = \frac{1}{40} \,\mathrm{m} = 2.5 \,\mathrm{cm}$ Length of tube $= L = 20 \, \mathrm{cm}$ $\mathrm{D} =$ least distance of distinct vision $= 25 \, \mathrm{cm}$Angular Magnification $= \frac{L}{f_o} \times \frac{D}{f_R}$$\therefore$ Angular Magnification $= \frac{20}{1} \times \frac{25}{2.5} = 200$