question_answer

A piano convex lens fits exactly into a piano concave lens. Their plane surfaces are parallel to each other. If the lenses are made of different materials of refractive indices\[{{\mu }_{1}}\] and \[{{\mu }_{2}}\] and R is the radius of curvature of the curved surface of the lenses, then focal length of the combination is

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).