In an astronomical telescope in normal adjust | vedclass

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(A) Let $f_{o}$ and $f_{e}$ be the focal lengths of the objective and eyepiece respectively.In normal adjustment,the distance between the objective an

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$L/l$

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(A) Let $f_{o}$ and $f_{e}$ be the focal lengths of the objective and eyepiece respectively.In normal adjustment,the distance between the objective and the eyepiece is $f_{o} + f_{e}$.The line of length $L$ on the objective lens acts as an object for the eyepiece.The distance of this object from the eyepiece is $u = -(f_{o} + f_{e})$.Using the lens formula for the eyepiece: $1/v - 1/u = 1/f_{e}$.$1/v = 1/f_{e} + 1/u = 1/f_{e} - 1/(f_{o} + f_{e}) = f_{o} / [f_{e}(f_{o} + f_{e})]$.So,$v = [f_{e}(f_{o} + f_{e})] / f_{o}$.The magnification $m$ produced by the eyepiece for this object is $m = v/u = [f_{e}(f_{o} + f_{e}) / f_{o}] / [-(f_{o} + f_{e})] = -f_{e}/f_{o}$.The ratio of the image length $l$ to the object length $L$ is $|m| = l/L = f_{e}/f_{o}$.The magnification of the astronomical telescope in normal adjustment is $M = f_{o}/f_{e}$.Therefore,$M = L/l$.

In an astronomical telescope in normal adjustment,a straight black line of length $L$ is drawn on the objective lens. The eyepiece forms a real image of this line. The length of this image is $l$. The magnification of the telescope is

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