Answer :

Given:


Focal length of the objective lens, = 140 cm


Focal length of the eyepiece, fe = 5 cm


Least distance of distinct vision, d = 25 cm


(a) When the telescope is in normal adjustment, its magnifying power is given as:


m = fo/fe …(1)


By putting the given values in equation (1), we get,


m = 140/5


m = 28


(b) When the final image is formed at d, the magnifying power m, of the telescope is given as:


…(2)


By putting the values in equation (2), we get,



m = 28 × (1 + 0.2)


m = 33.6


Hence, the magnification in case (a) is 28 and in case (b) is 33.6.


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