question_answer

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

A) \[\frac{R}{2\left( {{\mu }_{1}}+{{\mu }_{2}} \right)}\]         

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

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

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

Correct Answer: C

Solution :

It is combination of two lenses 1 and 2 \[\frac{1}{{{f}_{c}}}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}\] Using lens maker formula \[\frac{1}{{{f}_{1}}}=\left( {{\mu }_{2}}-1 \right)\left( \frac{1}{\infty }-\frac{1}{R} \right)=+\left[ \frac{{{\mu }_{1}}-1}{\operatorname{R}} \right]\] \[\frac{1}{{{f}_{2}}}=\left( {{\mu }_{2}}-1 \right)\left( -\frac{1}{R}-\frac{1}{\infty } \right)=-\left[ \frac{{{\mu }_{2}}-1}{\operatorname{R}} \right]\] \[\frac{1}{{{f}_{c}}}=\frac{{{\mu }_{1}}-1}{R}-\frac{\left( {{\mu }_{2}}-1 \right)}{R}\] \[\frac{1}{{{f}_{c}}}=\frac{{{\mu }_{1}}-1-{{\mu }_{2}}+1}{R}-\frac{{{\mu }_{2}}-{{\mu }_{1}}}{R}\] \[{{f}_{c}}=\frac{R}{{{\mu }_{1}}-{{\mu }_{2}}}\]