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 $f_0 + f_e$. Since the line of length $L$ is on the objective len

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$L/l$

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(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,it acts as an object for the eyepiece. The distance of this object from the eyepiece is $u = -(f_0 + f_e)$. The magnification of the eyepiece is given by $m_e = f_e / (f_e + u)$. Substituting the value of $u$: $m_e = f_e / (f_e - (f_0 + f_e)) = f_e / (-f_0) = -f_e / f_0$. The magnitude of magnification is $|m_e| = l / L = f_e / f_0$. Since the magnification of the telescope is $M = f_0 / f_e$,we have $M = L / l$.

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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