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,so i

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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,so its distance from the eyepiece is $u = -(f_0 + f_e)$.Using the magnification formula for a lens,$m = \frac{I}{L} = \frac{f_e}{f_e + u}$.Substituting $u = -(f_0 + f_e)$:$\frac{I}{L} = \frac{f_e}{f_e - (f_0 + f_e)} = \frac{f_e}{-f_0}$.Since the magnifying power of the telescope in normal adjustment is $M.P. = -\frac{f_0}{f_e}$,we have $\frac{I}{L} = \frac{1}{M.P.}$.Therefore,$M.P. = \frac{L}{I}$.

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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