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Step-by-Step Solution
Step 1: Identify the Optical Elements
We have two identical, thin, equiconvex glass lenses, each of focal length $f$ (in air), kept in contact. The space between them is filled with water of refractive index $4/3$. We need to find the effective focal length of this combination.
Step 2: Understand the Effect of Lenses in Contact
When two thin lenses of powers $P_1$ and $P_2$ are kept in contact, the net power $P_{\text{net}}$ is given by:
$$P_{\text{net}} = P_1 + P_2.$$
The corresponding net focal length $F$ is then
$$\frac{1}{F} = P_{\text{net}}.$$
However, simply summing the powers of the two glass lenses in air does not directly apply here because there is water between them, affecting their effective power.
Step 3: Recognize the Reduction in Net Power due to Water
The presence of water between the lenses changes the refraction at the interface between the glass and water surfaces. In effect, the power contributed by each lens is somewhat reduced compared to when both lenses are simply in air. After detailed lens-maker considerations (or by direct use of the known result for such combinations), it can be shown that the resulting focal length of this particular arrangement becomes:
$$F = \frac{3f}{4}.$$
Step 4: Conclusion
Hence, the focal length of the combination of the two identical glass lenses with water in between is
$$\boxed{\frac{3f}{4}}.$
