In an astronomical telescope in normal adjust | vedclass

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(A) In normal adjustment,the distance between the objective lens and the eyepiece is $f_0 + f_e$.The line of length $L$ is at the objective lens. The

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$\frac{L}{I}$

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(A) In normal adjustment,the distance between the objective lens and the eyepiece is $f_0 + f_e$.The line of length $L$ is at the objective lens. The distance of this line from the eyepiece is $u = -(f_0 + f_e)$.The magnification $m$ produced by the eyepiece is given by the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f_e}$.Magnification $m = \frac{v}{u} = \frac{f_e}{f_e + u}$.Substituting $u = -(f_0 + f_e)$,we get $m = \frac{f_e}{f_e - (f_0 + f_e)} = \frac{f_e}{-f_0} = -\frac{f_e}{f_0}$.Since the magnification is also the ratio of the image length $I$ to the object length $L$ (with a negative sign for inversion),we have $\frac{-I}{L} = -\frac{f_e}{f_0}$.Therefore,$\frac{I}{L} = \frac{f_e}{f_0}$.The magnifying power of the telescope is $M = \frac{f_0}{f_e} = \frac{L}{I}$.

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

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