A compound microscope consists of an objective of focal length 1 cm and eye piece of focal length 5 c separated by 12.2 cm. At what distance from the objective should an object be placed so that the final image is formed at least distance of distinct vision?

Asked by vishakhachandan026 | 23rd Feb, 2019, 06:15: PM

Expert Answer:

Focal length of objective, fo = 1 cmObject distance for objective, uo = ?Image distance for objective, vo = ?
Focal length of eye piece, fe = 5 cmObject distance for objective, ue = ?Image distance for objective, ve = -D = -25 cm (Least distance of distinct vision)
Distance between objective and eye piece, L = 12.2 cm
Since the image by objective is formed between the objective and the eye piece at a distance vo from objective and this image acts a virtual object for eye piece which is located at a distance ue from the eye piece,the length of microscope,L = |vo| + |ue| = 12.2 cm
Using the lens formula for eye piece.
1/fe = 1/ve - 1/ue
1/ue = 1/ve - 1/fe
1/ue = 1/-25 - 1/5 = -6/25
ue = -25/6 cm = -4.17 cm
We know that L = ​ |vo| + |ue| = 12.2 cmSo |vo| = 12.2 - ​ |ue| = 12.2 - 4.17 = 8.03 cm
Now using the lens formula for objective,1/fo = 1/vo - 1/uo
1/uo​ =  1/vo - 1/fo
1/uo = 1/8.03 - 1/1
1/uo = -0.875uo = -1.142 cm.
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Answered by Ankit K | 25th Feb, 2019, 10:33: AM