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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
- For the telescope is in normal adjustment (i.e., when the final image is at infinity)? what is the separation between the objective lens and the eyepiece?
- If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
- What is the height of the final image of the tower if it is formed at 25 cm?
You are given the following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct an astronomical telescope ? Give reason
| Lenses | Power (D) | Aperture (cm) |
| L1 | 3 | 8 |
| L2 | 6 | 1 |
| L3 | 10 | 1 |
Write two important advantages of reflecting telescope over a refracting telescope.
Draw a labelled ray diagram of an astronomical telescope to show the image formation of a distant object. Write the main considerations required in selecting the objective and eyepiece lenses in order to have large magnifying power and high resolution of the telescope.
Write two important limitations of a refracting telescope over a reflecting-type telescope.
Define magnifying power of a telescope. Write its expression.
A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece lens of focal length 1.0 cm is used, find the angular magnification of the telescope. If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.42 × 106 m and the radius of the lunar orbit is 3.8 × 108 m.
Define the term ‘resolving power of a telescope’.
In an astronomical telescope in normal adjustment, a straight black line of length L is drawn on the objective lens. The eyepiece forms a real image of this line whose length is 𝑙. What is the angular magnification of the telescope?
In an astronomical refracting telescope, the final image formed by the eyepiece is:
