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(C) Given: Focal length of objective $f_o = 40\, cm$,focal length of eyepiece $f_e = 4\, cm$,and object distance $u_o = -200\, cm$.For the objective l

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$54$

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(C) Given: Focal length of objective $f_o = 40\, cm$,focal length of eyepiece $f_e = 4\, cm$,and object distance $u_o = -200\, cm$.For the objective lens,using the lens formula $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$:$\frac{1}{v_o} - \frac{1}{-200} = \frac{1}{40}$$\frac{1}{v_o} = \frac{1}{40} - \frac{1}{200} = \frac{5-1}{200} = \frac{4}{200} = \frac{1}{50}$Thus,$v_o = 50\, cm$.For the final image to be formed at infinity (normal adjustment),the image formed by the objective must lie at the principal focus of the eyepiece. Therefore,the distance between the lenses is $l = v_o + f_e$.$l = 50\, cm + 4\, cm = 54\, cm$.

An astronomical telescope has an objective and an eyepiece of focal lengths $40\, cm$ and $4\, cm$ respectively. To view an object $200\, cm$ away from the objective,the lenses must be separated by a distance of.....$cm$.

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