Two identical glass $ \left( {{\mu _g} = 3/2} \right) $ equiconvex lenses of focal length f are kept in contact. The space between the two lenses is filled with water $ \left( {{\mu _w} = 4/3} \right) $ . The focal length of the combination is
A) $ 3f/4 $
B) $ f/3 $
C) $ f $
D) $ 4f/3 $

Hint
The focal length of any combination of lens and medium can be calculated using the
lens maker formula. The power of the lenses and water will add directly to give the focal length of the combination.

Formula used: $ \dfrac{1}{f}\, = \left( {\dfrac{{{\mu _2}}}{{{\mu _1}}}\, - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) $ where $ {\mu _1} $ and $ {\mu _2} $ are the refractive index of the surrounding medium and the lens respectively and $ {R_1} $ , $ {R_2} $ are the radius of curvature of the two surfaces of the lens

Complete step by step answer
First, let us start by determining the focal length of one lens considering that it is in the air
 $\Rightarrow \dfrac{1}{f}\, = \left( {\dfrac{{{\mu _2}}}{{{\mu _1}}}\, - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) $
Substituting the value of $ {\mu _2} = 3/2 $ and $ {\mu _1} = 1 $ , $ {R_1} = - {R_2} = R $ , we get
 $\Rightarrow \dfrac{1}{f}\, = \left( {\dfrac{3}{2}\, - 1} \right)\dfrac{2}{R} $
 $ \Rightarrow f = R $
Now since the spaces between the two lenses are filled with water, the water will also act as a lens. The radius of the curvature of this water lens will be due to the convex lenses which make the water lens surface concave. Then, we can calculate its focal length using the lens maker formula as
 $\Rightarrow \dfrac{1}{{{f_{water}}}} = \left( {\dfrac{4}{3} - 1} \right)\left( {\dfrac{{ - 2}}{R}} \right) $ which gives us
 $\Rightarrow \dfrac{1}{{{f_{water}}}} = \dfrac{1}{3} \times \dfrac{{ - 2}}{R} $
Substituting the value of $ f = R $ , we get
 $\Rightarrow {f_{water}} = - \dfrac{{3f}}{2} $
Since all the lenses are combined in series with respect to how the light passes through them, the power of the combination is equal to the sum of all the individual powers.
We know that the power of a lens in the inverse of its focal length so we can write the sum of the two lenses and the lens of water in between as:
 $\Rightarrow \dfrac{1}{{{f_{combination}}}} = \dfrac{1}{f} + \dfrac{1}{f} + \dfrac{2}{{ - 3f}} $
which gives us
$ {f_{combination}} = 3f/4 $ which corresponds to option (A).

Note
The trick in this question is to realize that the water between the lenses will act as a lens on its own and its radius of curvature will be concave but have the same radius as that of the lens. We must be careful to add the power of the 3 lenses and not the focal length directly for such combinations.