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प्रश्न
A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. Find the magnifying power of the telescope for viewing distant objects when
- the telescope is in normal adjustment,
- the final image is formed at the least distance of distinct vision.
उत्तर
- In normal adjustment:
Magnifying power (m) = `"f"_"o"/"f"_"e" = (140/5)` = 28 - When the final image is formed at the least distance of distinct vision (25 cm):
m = `"f"_"o"/"f"_"e" (1 +"f"_"e"/"D" ) = (28 xx 1.2) = 33.6`
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Useful Constants & Relations:
| 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.
