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प्रश्न
A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at
- the least distance of distinct vision (25 cm), and
- infinity?
What is the magnifying power of the microscope in each case?
उत्तर
Focal length of the objective lens, f1 = 2.0 cm
Focal length of the eyepiece, f2 = 6.25 cm
Distance between the objective lens and the eyepiece, d = 15 cm
(a) Least distance of distinct vision, d' = 25
∴ Image distance for the eyepiece, v2 = −25 cm
Object distance for the eyepiece = u2
According to the lens formula, we have the relation:
`1/"v"_2 - 1/"u"_2 = 1/"f"_2`
`1/"u"_2 = 1/"v"_2 - 1/"f"_2`
`1/"u"_2 = 1/(-25) - 1/6.25`
`1/"u"_2 = (-1 - 4)/25`
`1/"u"_2 = (-5)/25`
`1/"u"_2 = (-1)/5`
∴ u2 = −5 cm
Image distance for the objective lens, v1 = d + u2 = 15 − 5 = 10 cm
Object distance for the objective lens = u1
According to the lens formula, we have the relation:
`1/"v"_1 - 1/"u"_1 = 1/"f"_1`
`1/"u"_1 = 1/"v"_1 - 1/"f"_1`
`1/"u"_1 = 1/10 - 1/2`
`1/"u"_1 = (1 - 5)/10`
`1/"u"_1 = (-4)/10`
u1 = `10/-4`
∴ u1 = −2.5 cm
Magnitude of the object distance, |u1| = 2.5 cm
The magnifying power of a compound microscope is given by the relation:
`"m" = "v"_1/|"u"_1| (1 + "d'"/"f"_2)`
= `10/2.5 (1+ 25/6.25)`
= 4 × (1 + 4)
= 20
Hence, the magnifying power of the microscope is 20.
(b) The final image is formed at infinity.
∴ Image distance for the eyepiece, v2 = ∞
Object distance for the eyepiece = u2
According to the lens formula, we have the relation:
`1/"v"_2 - 1/"u"_2 = 1/"f"_2`
`1/∞ - 1/"u"_2 = 1/6.25`
∴ u2 = −6.25 cm
Image distance for the objective lens, v1 = d + u2 = 15 − 6.25 = 8.75 cm
Object distance for the objective lens = u1
According to the lens formula, we have the relation:
`1/"v"_1 - 1/"u"_1 = 1/"f"_1`
`1/"u"_1 = 1/"v"_1 - 1/"f"_1`
`1/"u"_1 = 1/8.75 - 1/2.0`
`1/"u"_1 = (2 - 8.75)/17.5`
`1/"u"_1 = -6.75/17.5`
u1 = `-17.5/6.75`
∴ u1 = −2.59 cm
Magnitude of the object distance, |u1| = 2.59 cm
The magnifying power of a compound microscope is given by the relation:
`"m" = "v"_1/|"u"_1| (("d'")/|"u"_2|)`
= `8.75/2.59 xx 25/6.25`
= 13.51
Hence, the magnifying power of the microscope is 13.51.
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In a compound microscope an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. If the eye-piece has a focal length of 5 cm and the final image is formed at the near point, find the magnifying power of the microscope.
