Two identical glass $$\left( {{\mu _g} = \frac{3}{2}} \right)$$  equi-convex lenses of focal length $$f$$ each are kept in contact. The space between the two lenses is filled with water $$\left( {{\mu _w} = \frac{4}{3}} \right).$$   The focal length of the combination is

Solution :

Consider the situation shown is figure.Let radius of curvature of lens surfaces is $$R.$$ The combination is equivalent to three lenses in contact.
$$\eqalign{ & \therefore \frac{1}{{{f_{eq}}}} = \frac{1}{{{f_1}}} + \frac{1}{{{f_2}}} + \frac{1}{{{f_3}}} \cr & = \frac{2}{{{f_1}}} + \frac{1}{{{f_2}}}\,\,\left( {\because {f_1} = {f_3}} \right) \cr & {\text{Now,}}\,\,\frac{1}{{{f_1}}} = \frac{1}{{{f_3}}} = \left( {\mu - 1} \right)\left( {\frac{2}{R}} \right) = \frac{1}{f} \cr & \frac{1}{{{f_2}}} = \left( {{\mu _w} - 1} \right)\left( { - \frac{2}{R}} \right) = \left( {\frac{4}{3} - 1} \right)\left( {\frac{{ - 2}}{R}} \right) \cr & = \left( {\frac{1}{3}} \right)\left( { - \frac{2}{R}} \right) = \left( {\frac{{ - 2}}{R}} \right)\left( {\frac{1}{{2\left( {\mu - 1} \right)}}} \right) \cr & \Rightarrow \frac{1}{{{f_2}}} = \left( { - \frac{1}{3}} \right)\left( {\frac{1}{{\frac{3}{2} - 1}}} \right)\left( {\frac{1}{f}} \right) = - \frac{2}{3} \times \frac{1}{f} \cr & \therefore \frac{1}{{{f_{eq}}}} = \frac{2}{f} - \frac{2}{3} \times \frac{1}{f} = \frac{{6 - 2}}{{3f}} = \frac{4}{{3f}} \Rightarrow {f_{eq}} = \frac{{3f}}{4} \cr} $$