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Question
In an astronomical telescope in normal adjustment, a straight black line of length L is drawn on the objective lens. The eyepiece forms a real image of this line whose length is ๐. What is the angular magnification of the telescope?
Solution
Let fo and fe be the focal length of the objective and eyepiece respectively.
For normal adjustment, the distance from the objective to the eyepiece is fo + fe.
Taking the line on the objective as an object and the eyepiece as a lens.
u = -(fo + fe) and f = fe
`1/v = 1/([-{f_o + f_e}]) = 1/f_e` ⇒ v = `((f_o + f_e)/f_o)f_e`
Linear magnification (eyepiece) = `v/u = "Image size"/"Object size" = f_e/f_o = l/L`
∴ Angular magnification of the telescope
M = `f_o/f_e = L/l`
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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 | |||
