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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.
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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 | |||
