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When an small object is placed at a distance `x_(1)` and `x_(2)` from a lens on its principal axis, then real image and a virtual image are formed respectively having same magnitude of transverse magnification. Then the focal length of the lens is :A. `x_(1) -x_(2)`B. `(x_(1)-x_(2))/(2)`C. `(x_(1) +x_(2))/(2)`D. `x_(1) +x_(2)` |
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Answer» Correct Answer - C `m=(f)/(f+u)m_(1)=(f)/(f-x_(1))` `m_(2)=(f)/(f-x_(2))m_(1)=-m_(2)` `2f =(x_(1)+x_(2))f=(x_(1)+x_(2))/(2)`. |
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