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प्रश्न
Draw a labelled ray diagram of an astronomical telescope in the near point adjustment position. A giant refracting telescope at an observatory has an objective lens of focal length 15 m and an eyepiece of focal length 1.0 cm. If this telescope is used to view the Moon, find the diameter of the image of the Moon formed by the objective lens. The diameter of the Moon is `3.48 xx 10^6`m, and the radius of the lunar orbit is `3.48 xx 10^8`m.
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उत्तर
When the final image is formed at the least distance of distinct vision
Magnifying power,
`M = beta/alpha`
α and β are small
∴ `M = tan β/tan α` ...(i)
In ΔA'B'C2 ,tan β = `("A"'"B"')/("C"_2"B"')`
In ΔA'B'C1, tanα = `("A"'"B"')/("C"_1"B"')`
From equation (i), we have:
`M = ("A"'"B"')/("C"_2"B"') xx ("C"_1"B"')/("A"'"B"')`
`"M"= ("C"_1"B"')/("C"_2"B"')`
Here,
`C_1B' = +f_0`
`C_2B' = -u_e`
`M = f_0/-u_e` ...(ii)
Using the lens equation `(1/ν - 1/u = 1/f)` for the eyepieces, we get
`1/(-D) - 1/(-u_c) = 1/f_e`
`-1/D + 1/u_c = 1/f_e`
`1/u_e = 1/f_e + 1/D`
`f_0/u_e = f_0/f_e (1 + f_e/D)`
`(-f_0)/u_e = (-f_0)/f_e (1 + f_e/D)`
M = `-f_0/f_e (1 + f_e/D)`
Angular magnification is:
`m_0 = |(f_0)/(f_e)| = |1500/1| = 1500`
Where, `"f"_0` is the focal length of the objective lens and fe is the focal length of the eye piece
Given, the diameter of the moon = `3.48 xx 10^6`m
The radius of the lunar orbit = `3.8 xx 10^8`m
The diameter of the image of the moon formed by the objective lens is given by d = af0
d = `"Diameter of the moon"/"Radius of the lunar orbit" xx f_0`
d = `(3.48 xx 10^6)/(3.8 xx 10^8) xx 15 = 13.74` cm
संबंधित प्रश्न
Write two important advantages of reflecting telescope over a refracting telescope.
Draw a ray diagram showing the image formation of a distant object by a refracting telescope ?
"A telescope resolves whereas a microscope magnifies." Justify this statement ?
Define magnifying power of a telescope. Write its expression.
How is the working of a telescope different from that of a microscope?
The eyepiece of an astronomical telescope has a focal length of 10 cm. The telescope is focussed for normal vision of distant objects when the tube length is 1.0. m. Find the focal length of the objective and the magnifying power of the telescope.
(i) What is meant by resolving power of a telescope?
(ii) State any one method of increasing the resolving power of an astronomical telescope.
Draw a ray diagram of a refracting astronomical telescope when final image is formed at infinity. Also write the expression for its angular magnification (magnifying power).
The magnifying power of an astronomical telescope in normal adjustment is 2.9 and the objective and the eyepiece are separated by a distance of 150 cm. Find the focal lengths of the two lenses.
Read the passage given below and answer the question that follows.
| There are two types of optical instruments: Microscopes and Telescopes. Microscopes are used to magnify very tiny objects whereas telescopes are used to study distant objects. Both of them deploy convex lenses. In his telescope, Newton used a large parabolic mirror to collect light from the stars and reduce aberrations. |
- Rohit observed the launch of Chandrayan 3 with the help of an optical instrument. Name the instrument used by him.
- State any one advantage of a reflecting telescope over a refracting telescope.
- Which instrument is used to study the structure of a virus?
- What is the ability of an optical instrument to form enlarged images called?
-
What is the difference between a compound microscope and an astronomical telescope (refracting type), as far as their lenses are concerned?
Useful Constants & Relations:
| 1 | Charge of a proton | e | 1.6 × 10-19 C |
| 2 | Speed of light in vacuum | c | 3 × 108 ms-1 |
| 1 u = 931 MeV | |||
