A) \[\frac{R}{2-({{\mu }_{1}}-{{\mu }_{2}})}\]
B) \[\frac{2R}{({{\mu }_{1}}-{{\mu }_{2}})}\]
C) \[\frac{R}{2({{\mu }_{1}}-{{\mu }_{2}})}\]
D) \[\frac{R}{{{\mu }_{1}}-{{\mu }_{2}}}\]
Correct Answer: D
Solution :
For 1st lens\[\frac{1}{{{f}_{1}}}=\left( \frac{{{\mu }_{1}}-1}{1} \right)\left( \frac{1}{\infty }-\frac{1}{-R} \right)=\frac{{{\mu }_{1}}-1}{R}\] for 2nd lens\[\frac{1}{{{f}_{2}}}=\left( \frac{{{\mu }_{2}}-1}{1} \right)\left( \frac{1}{-R}-0 \right)=-\frac{{{\mu }_{2}}-1}{R}\] \[\frac{1}{{{f}_{eq}}}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}\] \[\frac{1}{{{f}_{eq}}}=\frac{R}{{{\mu }_{1}}-1}+\frac{R}{-({{\mu }_{2}}-1)}\Rightarrow \frac{1}{{{f}_{eq}}}=\frac{R}{{{\mu }_{1}}-{{\mu }_{2}}}\] Hence \[{{f}_{eq}}=\frac{{{\mu }_{1}}-{{\mu }_{2}}}{R}\]You need to login to perform this action.
You will be redirected in
3 sec