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प्रश्न
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 ?
उत्तर
Let:
f0 = Focal length of the objective lens = 15 m = 1500 cm
fe = Focal length of the eye lens = 1.0 cm
Angular magnification of the giant refracting telescope is given by
`m_0=|f_0/(f_e)|`
`:.m_0=|1500/1|=1500`
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संबंधित 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?
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?
You are given the following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct a compound microscope?
| Lenses | Power (D) | Aperture (cm) |
| L1 | 3 | 8 |
| L2 | 6 | 1 |
| L3 | 10 | 1 |
An object is placed at a distance u from a simple microscope of focal length f. The angular magnification obtained depends
Can the image formed by a simple microscope be projected on a screen without using any additional lens or mirror?
The magnifying power of a converging lens used as a simple microscope is `(1+D/f).` A compound microscope is a combination of two such converging lenses. Why don't we have magnifying power `(1+D/f_0)(1+D/f_0)`?In other words, why can the objective not be treated as a simple microscope but the eyepiece can?
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?
How does the resolving power of a microscope change when
(i) the diameter of the objective lens is decreased?
(ii) the wavelength of the incident light is increased ?
Justify your answer in each case.
The near vision of an average person is 25 cm. To view an object with an angular magnification of 10, what should be the power of the microscope?
| 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. |
The focal lengths of the objective and eye-piece of a compound microscope are 1.2 cm and 3.0 cm respectively. The object is placed at a distance of 1.25 cm from the objective. If the final image is formed at infinity, the magnifying power of the microscope would be:
