In the following question, why do we use the formula l/fo(D/fe) and not L/fo(1+D/fe)?
The total magnification produced by a compound microscope is 20.The magnification produced by the eyepiece is 5.The microscope is focussed on a certain object.The distance between the objective and eyepiece is observed to be 14cm.If the least distance of distinct vision is 20cm calculate the focal length of the objective and the eyepiece.
Asked by Hppothen | 21st Feb, 2020, 02:41: PM
In Compound microscope, magnification me of eye piece when image is formed at near point is given by
me = 1 + ( D/fe ) .........................(1)
where D is the near point distance, fe is focal length of eye piece.
when image is formed at infinity, magnification me is given by , me = D/fe ...................(2)
When image is formed at infinity, it is comfortable to view and it gives less strain to the eye.
Hence for compound microscope , image is formed at infinity most of the times .
Hence we need to use the formula given in eqn.(2) for eye piece magnification.
Total magnification = mo me = 20
magnification of eyepiece = me= 5
Hence magnification of objective = mo = 4
magnification of eyepiece = me = D/fe = 20/fe = 5 , hence eyepiece focal length fe = 20/5 = 4 cm
magnification of objective = mo = L/fo = 4 , i.e. tube length L = 4fo
Distance between objective and eyepiece = fo + L + fe = 14 cm (given)
By substituting for L and fe in above expression, we get, fo + 4fo + 4 = 14 , i.e., 5 fo = 10
Hence focal length of objective, fo = 2 cm
Answered by Thiyagarajan K | 21st Feb, 2020, 09:42: PM
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