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Question
Draw a ray diagram for the formation of image of an object by an astronomical telescope, in normal adjustment. Obtain the expression for its magnifying power.
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Solution
The illustration displays an astronomy telescope used to provide angular magnification of distant objects. It sports a big diameter objective lens with a long focal length. The diameter and focal length of the eyepiece are both smaller than those of the objective lens. When light from a distant object enters the objective, an actual image is formed in the tube at its second focal point. The eyepiece enlarges this image, resulting in the final inverted image. The "tube length" or distance between the objective and the eyepiece, is the sum of the focal lengths of the two lenses when their innermost points meet.
The angle “β” that the picture subtends at the eye compared to the angle “α” that the object subends at the lens is the magnifying power or m.
`m = (beta/alpha)`
`beta = tanbeta = h/f_e`
Where h is image height and fe is focal length of eyepiece.
`alpha = tanalpha = h/f_o`
Where fo is focal length of objective lens.
Hence, we get magnifying power m as,
`m = (h"/"f_e)/(h"/"f_o) = f_o/f_e`
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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 | |||
