question_answer

Figure shows a concavo-convex lens \[{{\mu }_{2}}\] . What is the condition on the reflective indices so that the lens is diverging?     

A) \[2{{\mu }_{3}}<{{\mu }_{1}}+{{\mu }_{2}}\]

B) \[2{{\mu }_{3}}>{{\mu }_{1}}+{{\mu }_{2}}\]

C) \[{{\mu }_{3}}>2\left( {{\mu }_{1}}-{{\mu }_{2}} \right)\]

D) None of these          

Correct Answer: B

Solution :

[b] \[\frac{{{\mu }_{2}}}{V}-\frac{{{\mu }_{1}}}{-{{u}_{0}}}=\frac{{{\mu }_{2}}-{{\mu }_{1}}}{-2R}\] \[\frac{{{\mu }_{3}}}{{{V}_{f}}}-\frac{{{\mu }_{2}}}{V}=\frac{{{\mu }_{3}}-{{\mu }_{2}}}{-R}\] \[\frac{{{\mu }_{3}}}{{{V}_{f}}}+\frac{{{\mu }_{1}}}{{{u}_{0}}}=\frac{{{\mu }_{3}}-{{\mu }_{2}}}{-2R}+\frac{{{\mu }_{2}}-{{\mu }_{3}}}{R}\] \[\therefore \frac{{{\mu }_{1}}-{{\mu }_{2}}}{2}<{{\mu }_{3}}-{{\mu }_{2}}\Rightarrow {{\mu }_{1}}-{{\mu }_{2}}<2{{\mu }_{3}}-2{{\mu }_{2}}\] \[\Rightarrow {{\mu }_{1}}+{{\mu }_{2}}<2{{\mu }_{3}}\]