CBSE Class 12-science Answered
Optical instrument
Asked by tannudahiya80536 | 20 Dec, 2022, 01:13: PM
Expert Answer
It is required to get image at the least distance of clear vision D , i.e., D = 25 cm from eyepiece .
Let us use the lens equation to get the lens-to-object distance u for eyepiece
(1/v) - (1/u) = 1/f
Let us apply cartesian sign convention to all distances
where v = -25 cm is lens-to-image distance and f = 6.25 cm is focal length of eyepiece lens.
1/u = -(1/25) -(1/6.25) = -1/5
u = -5 cm , i.e., intermediate image is formed at a distance 5 cm to the left of eyepiece lens.
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Let us use the lens equation to get the lens-to-object distance uo for objective
(1/vo) - (1/uo) = 1/fo
Let us apply cartesian sign convention to all distances
where vo = (15-5 ) cm = 10 cm , is lens-to-image distance and fo = 2 cm is focal length of objective lens.
1/uo = (1/10) -(1/2) = -(4/10)
uo = -2.5 cm , i.e., object is placed 2.5 cm infront of the objective lens.
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Answers :-
(i) Distance of object to eye-piece = ( 2.5 + 15 ) cm = 17.5 cm
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(ii) Distance of object to objective lens = 2.5 cm
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(iii) Magnification of objective lens, mo = -10/2.5 = -4
Magnification of eyepiece lens, me = -25/-5 = 5
Overall magnification = mo × me = -4 × 5 = -20
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(iv) magnifying power of compound microscope can be increased by using
short focal length lenses for objective and eyepiece
.
Answered by Thiyagarajan K | 20 Dec, 2022, 08:44: PM
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