Advertisements
Advertisements
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?
Advertisements
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
Explain the basic differences between the construction and working of a telescope and a microscope
A giant refracting telescope has an objective lens of focal length 15 m. If an eye piece of focal length 1.0 cm is used, what is the angular magnification of the telescope ?
The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The microscope is focussed on a certain object. The distance between the objective and eyepiece is observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the focal length of the objective and the eye piece.
A person with a normal near point (25 cm) using a compound microscope with the objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
Does the magnifying power of a microscope depend on the colour of the light used? Justify your answer.
The focal length of the objective of a compound microscope if fo and its distance from the eyepiece is L. The object is placed at a distance u from the objective. For proper working of the instrument,
(a) L < u
(b) L > u
(c) fo < L < 2fo
(d) L > 2fo
A compound microscope forms an inverted image of an object. In which of the following cases it it likely to create difficulties?
The separation between the objective and the eyepiece of a compound microscope can be adjusted between 9.8 cm to 11.8 cm. If the focal lengths of the objective and the eyepiece are 1.0 cm and 6 cm respectively, find the range of the magnifying power if the image is always needed at 24 cm from the eye
An eye can distinguish between two points of an object if they are separated by more than 0.22 mm when the object is placed at 25 cm from the eye. The object is now seen by a compound microscope having a 20 D objective and 10 D eyepiece separated by a distance of 20 cm. The final image is formed at 25 cm from the eye. What is the minimum separation between two points of the object which can now be distinguished?
A compound microscope consists of an objective of focal length 1 cm and an eyepiece of focal length 5 cm. An object is placed at a distance of 0.5 cm from the objective. What should be the separation between the lenses so that the microscope projects an inverted real image of the object on a screen 30 cm behind the eyepiece?
compound microscope consists of two convex lenses of focal length 2 cm and 5 cm. When an object is kept at a distance of 2.1 cm from the objective, a virtual and magnified image is fonned 25 cm from the eye piece. Calculate the magnifying power of the microscope.
A convex lens of a focal length 5 cm is used as a simple microscope. Where should an object be placed so that the image formed by it lies at the least distance of distinct vision (D = 25 cm)?
In the case of a regular prism, in minimum deviation position, the angle made by the refracted ray (inside the prism) with the normal drawn to the refracting surface is ______.
A thin converging lens of focal length 5cm is used as a simple microscope. Calculate its magnifying power when image formed lies at:
- Infinity.
- Least distance of distinct vision (D = 25 cm).
Draw a ray diagram of compound microscope for the final image formed at least distance of distinct vision?
| 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. |
Which of the following is not correct in the context of a compound microscope?
The focal lengths of the objective and the eye-piece of a compound microscope are 1.0 cm and 2.5 cm respectively. Find the tube length of the microscope for obtaining a magnification of 300.
