In an astronomical telescope in normal adjust | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{L}{I}$

Vedclass · Answer

(A) In normal adjustment,the distance between the objective lens and the eyepiece is $f_0 + f_e$. Since the line of length $L$ is on the objective lens,its distance from the eyepiece is $u = -(f_0 + f_e)$. The magnification $m$ produced by a lens is given by $m = \frac{f}{f + u}$. For the eyepiece,the magnification of the image of length $I$ is $\frac{I}{L} = \frac{f_e}{f_e + u}$. Substituting $u = -(f_0 + f_e)$,we get: $\frac{I}{L} = \frac{f_e}{f_e - (f_0 + f_e)} = \frac{f_e}{-f_0}$. Since the magnification of the telescope is $M = -\frac{f_0}{f_e}$,we have $\frac{I}{L} = \frac{1}{M}$. Therefore,the magnification $M = \frac{L}{I}$.

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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