InterviewSolution
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InterviewSolution
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In an astronomical telescope in normal adjustment a straight black line of length `L` is drawn on inside part of objective lens. The eye piece forms a real image of this line. The length of this image is `I`. The magnification of the telescope isA. `L/l`B. `L/l + 1`C. `L/l-1`D. `(L + l)/(L -l)` |
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Answer» Correct Answer - A Let `f_(o)` and `f_(e)` be the focal lengths of the objective and eyepiece respectively. For normal adjustment, distance from the objective to the eyepiece (tube length ) `= f_(o) + f_(e)`. Treating the line of the objective as the object, and the eyepiece as the lens, `u = -(f_(o) + f_(e))` and `f = f_(e)` `:. (1)/(v) = (1)/(f_(e)) -(1)/(f_(o)+f_(e))= (f_(o))/((f_(e)+f_(e))f_(e))` or `v = ((f_(o) + f_(e))f_(e))/(f_(o))` Magnitifcation `= |v/u| = (f_(e))/(f_(o)) = ("image size")/("object size") = (I)/(O)` `:. (f_(o))/(f_(e)) = (L)/(l)` = magnification of telescope in normal adjustment. |
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