In an astronomical telescope in normal adjustment a straight black line of length L is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is I. The magnification of the telescope is

The situation is shown in the figure.Let f0 and fe be the focal lengths of the objective and eyepiece respectively.For normal adjustment distance of the objective from the eyepiece (tube length) f0 + feTreating the line on the objective as the object and eyepiece as the lens.∴  u=-fo+fe and f=fe As 1v-1u=1f∴  1v-1-fo+fe=1fe1v=1fe-1fo+fe=fo+fe-fefefo+fe=fofefo+fe or   v=fefo+fefo Thus, IL=v|u|=fefo+fefofo+fe=fefo or fofe=LI∴The magnification of the telescope in normal adjustment ism=fofe=LI                    . . . .(i)