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प्रश्न
A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
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उत्तर
Focal length of the objective lens, fo = 144 cm
Focal length of the eyepiece, fe = 6.0 cm
The magnifying power of the telescope is given as:
`"m" = ("f"_"o")/"f"_"e"`
= `144/6`
= 24
The separation between the objective lens and the eyepiece is calculated as:
fo + fe
= 144 + 6
= 150 cm
Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 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 | |||
