A) 2.5 cm
B) 6 cm
C) 15 cm
D) 9 cm
Correct Answer: D
Solution :
If initially the objective (focal length Fo) forms the image at distance vo then \[{{v}_{o}}=\frac{{{u}_{o}}{{f}_{o}}}{{{u}_{o}}-{{f}_{o}}}=\frac{3\times 2}{3-2}=6\,cm\] Now as in case of lenses in contact \[\frac{1}{{{F}_{o}}}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}+\frac{1}{{{f}_{3}}}+.....=\frac{1}{{{f}_{1}}}+\frac{1}{{{{{F}'}}_{o}}}\] \[\frac{1}{v}-\frac{2}{(-15)}=\frac{(1-2)}{-10}\] So if one of the lens is removed, the focal length of the remaining lens system \[\frac{1}{{{{{F}'}}_{o}}}=\frac{1}{{{F}_{0}}}-\frac{1}{{{f}_{1}}}=\frac{1}{2}-\frac{1}{10}\] Þ \[{{{F}'}_{o}}=2.5\,cm\] This lens will form the image of same object at a distance \[{{{v}'}_{o}}\] such that \[{{{v}'}_{o}}=\frac{{{u}_{o}}{{{{F}'}}_{o}}}{{{u}_{o}}-{{{{F}'}}_{o}}}=\frac{3\times 2.5}{(3-2.5)}=15\,cm\] So to refocus the image, eye-piece must be moved by the same distance through which the image formed by the objective has shifted i.e. 15 ? 6 = 9 cm.
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