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An infinitely long rod lies along the axis of a concave mirror of focal length f. The near end of the rod is distance `u gt f` from the mirror. Its image will have lengthA. `(f^(2))/(u-f)`B. `(uf)/(u-f)`C. `(f^(2))/(u+f)`D. `(uf)/(u+f)` |
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Answer» Correct Answer - A `u_(I) =- u,v_(1) =- ((uf)/(u-f))` `u_(2) = oo, v_(2) =- f` `|v_(2)|-|v_(1)| = (uf)/(u-f) -f = (uf-uf+f^(2))/(u-f) = (f^(2))/(u-f)` |
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