A) 1.80 cm
B) 1.70 cm
C) 1.65 cm
D) 1.75 cm
Correct Answer: D
Solution :
\[{{f}_{o}}=1.6cm,\]\[{{f}_{e}}=2.5cm\] \[L={{v}_{o}}+{{v}_{e}}=21.7cm\] As final image is forming at infinity \[\therefore \] \[{{u}_{e}}={{f}_{e}}\] \[\therefore \] \[L={{v}_{o}}+{{f}_{e}}=21.7cm\] \[{{v}_{o}}=21.7-2.5=19.2cm\] For objective lens, \[\frac{1}{{{f}_{o}}}=\frac{1}{{{v}_{o}}}-\frac{1}{{{u}_{o}}}\] \[\frac{1}{1.6}=\frac{1}{19.2}-\frac{1}{{{u}_{o}}}\] \[\frac{1}{{{u}_{o}}}=\frac{1}{19.2}-\frac{1}{1.6}=\frac{1-12}{19.2}\] \[{{u}_{o}}=-\frac{19.2}{11}=-1.746cm\] \[=-1.75cm\]
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