Solution:

Given data:

Focal length of each glass lens, f

Refractive index of glass lens, Ug = 3/2

Refractive index of water, Uw = 4/3

The space between the two lenses is filled with water.

To find: Focal length of the combination of the two lenses.

Formula used:

1/f = (Ug - 1) * [1/R1 - 1/R2]

Where R1 and R2 are the radii of curvature of the two surfaces of the lens.

Calculation:

The two lenses are identical and equiconvex, so the radii of curvature of both surfaces of each lens are equal.

Let the radius of curvature of each surface be R.

For the glass lens, Ug = 3/2, so 1/f = (3/2 - 1) * [1/R - 1/R] = 0

This means that the focal length of the glass lens is infinity.

When the space between the two lenses is filled with water, the effective refractive index of the combination changes.

Let the effective refractive index be U.

Using the lens maker's formula, 1/f = (U - 1) * [1/R - 1/R]

For the first lens, 1/f1 = (Ug - 1) * [1/R - 1/R] = 0

For the second lens, 1/f2 = (Ug - 1) * [1/R - 1/R] = 0

For the combination, 1/f = (U - 1) * [1/R - 1/R]

Since the two lenses are in contact, the distance between them is negligible compared to the radius of curvature of the lenses. This means that the space between the lenses can be treated as a single surface.

Let the radius of curvature of this surface be R1.

Using the lens maker's formula for this surface, 1/f1 = (Uw - Ug) * [1/R1 - 1/R]

For the combination, 1/f = 1/f1 + 1/f2 - d/f1*f2

Where d is the distance between the two lenses, which is negligible.

Substituting the values, we get:

1/f = (Uw - Ug) * [1/R1 - 1/R] + 0 - 0

1/f = (4/3 - 3/2) * [1/R1 - 1/R]

1/f = -1/12 * [1/R1 - 1/R]

1/f = (1/R - 1/R1)/12

1/f = (1/R - 1/R)/12

1/f = 0

Therefore, the focal length of the combination of the two lenses is infinity.

Conclusion:

When two identical glass Ug(3/2) equiconvex lens of focal length f each are kept in contact and the space between the two lenses is filled with water Uw(4/3), the focal length of the combination is infinity. This is because the two lenses cancel out each other's effect and the effective refractive index of the combination becomes equal to that of air.