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Question
Draw a labeled ray diagram to obtain the real image formed by an astronomical telescope in normal adjustment position. Define its magnifying power.
Solution
In normal adjustment, the final image is formed at infinity.
Magnifying power or angular magnification of astronomical telescope: It is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye, by the object directly, when the final image and the object, both are at infinity.
Angular magnification,`M=beta/alpha`
β and α are very small.
`:.beta~~tan beta`
`alpha~~tanalpha`
`=>M=tanbeta/tanalpha`
I is the image formed by the objective. f0 and fe are the focal lengths of the objective and eyepiece, respectively.
Here,
`tanalpha=I/f_0`
`tan beta=I/-f_e`
(Distance of the image from the eyepiece is taken as negative.)
`:.M=((-I)/f_e)/(I/f_0)`
|
`M=(-f_0)/f_e` |
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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 | |||
