A) f
B) 2f
C) 3f
D) \[\frac{3}{2}f\]
Correct Answer: B
Solution :
This combinations will behave like a mirror of power. \[{{P}_{eq}}=2{{P}_{L}}+{{P}_{M}}\] \[{{P}_{eq}}=2\frac{1}{f}+0\] \[{{F}_{eq}}=-\frac{f}{2}\] so the behaviour will be like a mirror of focal length \[-\frac{f}{2}\] Using mirror equation\[\frac{1}{V}+\frac{1}{U}=\frac{1}{{{f}_{eq}}}\] \[\frac{1}{{{-}^{{}^{9}/{}_{3}}}}+\frac{1}{-a}=\frac{-1}{{}^{f}/{}_{2}}\] \[\frac{1}{a}=\frac{-2}{f}\] \[\]You need to login to perform this action.
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