The separation between the eye piece of focal length `0.3 m` and objective of focal length `0.4 m` of a microscope is `0.2 m`. The eye piece and objective are to be interchanged such that the angular magnification of the instrument remains the same. What is the new separation between the lenses ?
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For relaxed eye i.e., normal setting of microscope, final image is formed at infinity `m = -(v_(0))/(u_(0))((d)/(f_(e)))` …(i) If object lies close to focus of objective lenses, `u_(0) ~~ f_(0) and v_(0) = (L - f_(e))`, where `L` is distance between the two lenses. `:.` From (i), `m = (- (L - f_(e))d)/(f_(0)f_(e))` ...(ii) Let `L'` be the new distance between the two lenses when objective and eyes lens are interchanged. `:. m' = (- (L' - f_(0))d)/(f_(e)f_(0))` ...(iii) As `m' = m`, therefore, from (ii) and (iii), `L' - f_(0) = L - f_(e)` or `L' = L - f_(e) +f_(0) = 0.2 - 0.3 + 0.4 = 0.3 m`
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