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प्रश्न
An astronomical telescope is to be designed to have a magnifying power of 50 in normal adjustment. If the length of the tube is 102 cm, find the powers of the objective and the eyepiece.
उत्तर
For the astronomical telescope,
Magnifying power, m = 50
Length of the tube, L = 102 cm
Let the focal length of objective and eye piece be f0 and fe respectively.
Now , using m = `f_0/f_e, we get :`
fo= 50fe ..(1)
And,
L = fo + fe =102 cm ...(2)
On substituting the value of fo from (1) in (2) . we get :
50 fe +fe =102
⇒ 51 fe = 102
⇒ fe = 2 cm = 0.02 m
And,
fo = 50 × 0.02 = 1 m
Power of the objective lens =`1/f_0` = 1 D And,
Power of the eye piece lens =`1/f_e = 1/0.02 = 50 D`
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| 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 | |||
Assertion: An astronomical telescope has an objective lens having large focal length.
Reason: Magnifying power of an astronomical telescope varies directly with focal length of the objective lens.
