Question

A compound microscope has a magnification equal to 200 . If magnification due to objective is -4 , then the focal length of eye piece will be ?​

Answer 1

Answer:

focal length of eye piece = - 0.5

Explanation:

Given:

=> magnification of compound microscope = 200

=> magnification of objective = -4

=> focol length of eye piece = ?

Formula:

$m \: = \: (mo)(me) \: = ( \frac{l}{fo} )( \frac{d}{fe} )$

substitute "m" , "mo" values in above formula

where D = near point = 25cm

=> 200 = (-4)(25/fe)

=> 200 = (-100)/fe

=> fe = -100/200

=> fe = -0.5

The focal length of eye piece is = -0.5

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