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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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(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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