A) \[3{{\mu }_{2}}-2{{\mu }_{1}}=1\]
B) \[{{\mu }_{1}}+{{\mu }_{2}}=3\]
C) \[2{{\mu }_{1}}-{{\mu }_{2}}=1\]
D) \[2{{\mu }_{2}}-{{\mu }_{1}}=1\]
Correct Answer: C
Solution :
\[{{f}_{1}}=2{{f}_{2}}\] \[\frac{1}{{{f}_{1}}}\,=\,\left( \frac{{{\mu }_{1}}-1}{1} \right)\frac{1}{R}\] \[\frac{1}{{{f}_{2}}}\,=\,\left( \frac{{{\mu }_{2}}-1}{1} \right)\left( -\frac{1}{R} \right)\] \[{{f}_{2}}=\,-\frac{R}{{{\mu }_{2}}-1};\,\,\,\,\,\,{{f}_{1}}\,=\,\frac{R}{{{\mu }_{2}}-1}\] \[{{f}_{1}}=\,2{{f}_{2}}\] \[\frac{R}{{{\mu }_{1}}-1}\,\,=\,\,\frac{2R}{{{\mu }_{2}}-1}\] \[{{\mu }_{2}}-1\,\,=\,\,\,2{{\mu }_{1}}-2\] \[1\,\,=\,\,2{{\mu }_{1}}\,-\,\,{{\mu }_{2}}\]You need to login to perform this action.
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