A) \[{{f}_{1}}+{{f}_{2}}\]
B) \[\frac{R}{{{\mu }_{2}}-{{\mu }_{1}}}\]
C) \[{{f}_{1}}-{{f}_{2}}\]
D) \[\frac{2{{f}_{1}}{{f}_{2}}}{{{f}_{1}}+{{f}_{2}}}\]
Correct Answer: B
Solution :
The focal length of the two lens is given \[\frac{1}{{{f}_{2}}}=({{\mu }_{2}}-1)\left( \frac{1}{{{R}_{2}}}-\frac{1}{\infty } \right)=\frac{{{\mu }_{2}}-1}{{{R}_{2}}}\] \[\frac{1}{{{f}_{1}}}=({{\mu }_{1}}-1)\left( \frac{1}{\infty }-\frac{1}{{{R}_{1}}} \right)=-\frac{{{\mu }_{1}}-1}{{{R}_{1}}}\] So, the focal length of the lens combination is \[\frac{{{f}_{2}}{{f}_{1}}}{{{f}_{2}}+{{f}_{1}}}=\frac{\left( \frac{{{R}_{2}}}{{{\mu }_{1}}-1} \right)\left( \frac{-{{R}_{1}}}{{{\mu }_{1}}-1} \right)}{\frac{{{R}_{2}}}{{{\mu }_{2}}-1}-\frac{{{R}_{1}}}{{{\mu }_{1}}-1}}=\frac{R}{{{\mu }_{2}}-{{\mu }_{1}}}\]
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