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Question
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.
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Solution
Given: D = 150 cm, m = 2.9
fo = ?
Since, `m = f_o/f_e`
D = `f_o + f_e`
`2.9 = f_o/f_e`
`f_o = f_e xx 2.9` .....(i)
`f_o + f_e = 150` cm
From equation (i),
`2.9 f_e + f_e = 150`
`3.9f_e = 150`
fe = 38.46 cm
fo = 2.9 × 38.46
fo = 111.54 cm
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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 | |||
