A) \[\omega ={{\omega }_{0,}}\omega =2{{\omega }_{0,}}f=2f\]
B) \[\omega ={{\omega }_{0,}}\omega =2{{\omega }_{0,}}f=-2f\]
C) \[\omega ={{\omega }_{0,}}\omega =2{{\omega }_{0,}}f=f/2\]
D) \[\omega ={{\omega }_{0,}}\omega =2{{\omega }_{0,}}f=-f/2\]
Correct Answer: B
Solution :
The condition of achromatic combination is \[\frac{\omega }{\omega }=-\frac{f}{f}\] This is satisfied if \[\omega ={{\omega }_{0}}\,\omega =2{{\omega }_{0}}\] and \[f=2f\] So that, \[\frac{{{\omega }_{0}}}{2{{\omega }_{0}}}=\frac{-f}{-2f}\] \[\Rightarrow \] \[\frac{1}{2}=\frac{1}{2}\]
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