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

Question

Infinity Learn · Accepted Answer

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)

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

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