The total magnification produced by a compound microscope is 20 the magnification produced by the eyepiece is 5 the microscope is focused on a certain object the distance between objective and eyepiece is observed to be 14 cm at least distance of distinct vision is 20 centimetre calculate the focal length of the objective and the eyepiece
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The focal length of objective is 3.5 cm and the focal length of eyepiece is 5 cm.
Explanation:
- Focal length of objective:
From "m" = m0×me
m0 = m / me
= 20 / 5 = 4
Now, m0 = v0 / u0
= L / f0 =4
f0 = L / 4 = 14 / 4
= 3.5 cm
- Focal length of eyepiece.
Also, me = 1 + dfe = 5
dfe = 5−1 = 4
fe = d / 4 =20 / 4 = 5 cm
Hence the focal length of objective is 3.5 cm and the focal length of eyepiece is 5 cm.
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Find the focal length of a convex mirror whose radius of curvature is 32 cm ?
https://brainly.in/question/9214343
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Answer:
Given total magnification,m =20 m =mo X me
me =5 given so mo = 20/5 =4
me=ve/ue , 5=-20 /ue
So ue =-4 cm
1/fe= 1/ve-1/ue
1/fe =1/-20 -1/-4
So fe = 5 cm
Distance between objective and eyepiece is 14 cm
Vo + ue = 14
Vo =14-4= 10cm
Now mo= vo/uo
4= 10/uo
Uo =10/4cm
1/f=1/vo-1/uo
1/fo=1/10-1/10/4
So fo=2cm
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