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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?
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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.
संबंधित प्रश्न
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| Lenses | Power (D) | Aperture (cm) |
| L1 | 3 | 8 |
| L2 | 6 | 1 |
| L3 | 10 | 1 |
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| A compound microscope consists of two converging lenses. One of them, of smaller aperture and smaller focal length, is called objective and the other of slightly larger aperture and slightly larger focal length is called eye-piece. Both lenses are fitted in a tube with an arrangement to vary the distance between them. A tiny object is placed in front of the objective at a distance slightly greater than its focal length. The objective produces the image of the object which acts as an object for the eye-piece. The eye-piece, in turn, produces the final magnified image. |
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| A compound microscope consists of two converging lenses. One of them, of smaller aperture and smaller focal length, is called objective and the other of slightly larger aperture and slightly larger focal length is called eye-piece. Both lenses are fitted in a tube with an arrangement to vary the distance between them. A tiny object is placed in front of the objective at a distance slightly greater than its focal length. The objective produces the image of the object which acts as an object for the eye-piece. The eye-piece, in turn, produces the final magnified image. |
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