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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?
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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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