A) 2.4 and 12.0
B) 2.4 and 15.0
C) 2.3 and 12.0
D) 2.3 and 3.0
Correct Answer: A
Solution :
When final image is formed at infinity, length of the tube \[={{v}_{o}}+{{f}_{e}}\] \[\Rightarrow 15={{v}_{o}}+3\]\[\Rightarrow {{v}_{o}}=12\ cm\] For objective lens \[\frac{1}{{{f}_{o}}}=\frac{1}{{{v}_{o}}}-\frac{1}{{{u}_{o}}}\] \[\Rightarrow \frac{1}{(+2)}=\frac{1}{(+12)}-\frac{1}{{{u}_{o}}}\]\[\Rightarrow {{u}_{o}}=-\,2.4\,cm\]
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