Four lenses of focal length + 15, cm, + 20, c | vedclass

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(A) The magnification $M$ of an astronomical telescope in normal adjustment is given by the formula $M = -\frac{f_o}{f_e}$,where $f_o$ is the focal le

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$+ 15$

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(A) The magnification $M$ of an astronomical telescope in normal adjustment is given by the formula $M = -\frac{f_o}{f_e}$,where $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eye-piece.To obtain the largest magnification $M$,the ratio $\frac{f_o}{f_e}$ must be maximized.This implies that for a given objective lens,the focal length of the eye-piece $f_e$ should be the smallest available value.Among the given focal lengths $(+ 15\, cm, + 20\, cm, + 150\, cm, + 250\, cm)$,the smallest value is $+ 15\, cm$.Therefore,the focal length of the eye-piece should be $+ 15\, cm$.

Four lenses of focal length $+ 15\, cm, + 20\, cm, + 150\, cm$ and $+ 250\, cm$ are available for making an astronomical telescope. To produce the largest magnification,the focal length of the eye-piece should be.....$cm$

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