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 In normal adjustment `L = f_(0)+f_(e)` Treating line on obhective as object and eye-piece the lens `(1)/(v)-(1)/(u)=(1)/(f) rArr (1)/(v) - (1)/(-(f_(O)+f_(e)) = (1)/(f_(e))` `v = ((f_(O)+f_(e))f_(e))/(f_(O))` Magnification `= |(v)/(u)| = (f_(e))/(f_(O)) = ("image size")/("object size") = (l)/(L)` `(f_(O))/(f_(e)) = (L)/(l) =` magnification of telescope in normal adjustment |
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