question_answer

A plano-convex lens (focal length \[{{f}_{2}},\]refractive index \[{{\mu }_{2}},\] radius of curvature R) fits exactly into a Plano concave lens (focal length \[{{f}_{1}},\]refractive index \[{{\mu }_{1}},\]radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be:

A) \[{{f}_{1}}-{{f}_{2}}\]

B) \[\frac{R}{{{\mu }_{2}}-{{\mu }_{1}}}\]

C) \[\frac{2{{f}_{1}}{{f}_{2}}}{{{f}_{1}}+{{f}_{2}}}\]             

D) \[{{f}_{1}}+{{f}_{2}}\]

Correct Answer: B

Solution :

\[\frac{1}{F}-\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}-\frac{1-{{\mu }_{1}}}{R}+\frac{{{\mu }_{2}}-1}{R}\]

\[\frac{1}{F}=\frac{1-{{\mu }_{1}}+{{\mu }_{2}}-1}{R}\]

\[\frac{1}{F}=\frac{{{\mu }_{2}}-{{\mu }_{1}}}{R}\]

\[F=\frac{R}{{{\mu }_{2}}-{{\mu }_{1}}}\]