The magnifying power of an astronomical teles | vedclass

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(C) For an astronomical telescope in normal adjustment (relaxed vision),the magnifying power is given by $m = \frac{f_o}{f_e} = 16$. From this,we get

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$32\, cm, 2\, cm$

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(C) For an astronomical telescope in normal adjustment (relaxed vision),the magnifying power is given by $m = \frac{f_o}{f_e} = 16$. From this,we get $f_o = 16 f_e$ ..........$(i)$ The length of the telescope tube in normal adjustment is the sum of the focal lengths of the objective and the eye lens: $L = f_o + f_e = 34\, cm$ ..........$(ii)$ Substituting equation $(i)$ into equation $(ii)$: $16 f_e + f_e = 34$ $17 f_e = 34$ $f_e = 2\, cm$ Now,substituting $f_e = 2\, cm$ back into equation $(i)$: $f_o = 16 	imes 2 = 32\, cm$ Therefore,the focal lengths are $f_o = 32\, cm$ and $f_e = 2\, cm$.

The magnifying power of an astronomical telescope for relaxed vision is $16$. On adjusting,the distance between the objective and eye lens is $34\, cm$. Then the focal length of the objective and eye lens will be respectively:

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