A) \[\frac{R}{2({{\mu }_{1}}+{{\mu }_{2}})}\]
B) \[\frac{R}{2({{\mu }_{1}}-{{\mu }_{2}})}\]
C) \[\frac{R}{({{\mu }_{1}}-{{\mu }_{2}})}\]
D) \[\frac{2R}{({{\mu }_{2}}-{{\mu }_{1}})}\]
Correct Answer: C
Solution :
The combination of two lenses 1 and 2 is as shown in the figure. As \[\frac{1}{F}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}\] \[\therefore \] \[\frac{1}{F}=({{\mu }_{1}}-1)\left( \frac{1}{\infty }+\frac{1}{R} \right)+({{\mu }_{2}}-1)\left( \frac{1}{-R}-\frac{1}{\infty } \right)\] \[=\frac{{{\mu }_{1}}-1}{R}-\frac{{{\mu }_{2}}-1}{R}\] \[\frac{1}{F}=\frac{{{\mu }_{1}}-{{\mu }_{2}}}{R}\]or \[F=\frac{R}{{{\mu }_{1}}-{{\mu }_{2}}}\] Hence, the correction option is (c).You need to login to perform this action.
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