Introduction:
The refractive index of a material is a measure of how much it bends light as it passes through it. In this question, we are given the refractive indices of a material in both air and water, and we need to find the focal length of a convex lens when it is placed in water.

Given:
Refractive index of g in air and water in air is 3/2 and 4/3 respectively. Focal length of convex lens in air is 10cm.

Solution:

Step 1:
We know that the formula for the focal length of a lens is given by:

1/f = (n2 - n1) x (1/R1 - 1/R2)

Where n1 and n2 are the refractive indices of the two media on either side of the lens, R1 and R2 are the radii of curvature of the two surfaces of the lens, and f is the focal length of the lens.

Step 2:
Since we are given the focal length of the lens in air, we can calculate the radii of curvature of the two surfaces of the lens using the formula:

1/f = (n2 - n1) x (1/R1 - 1/R2)

R1 = (n1 x f)/(n2 - n1)

R2 = (n2 x f)/(n2 - n1)

Substituting the values of n1, n2, and f, we get:

R1 = (3/2 x 10)/(4/3 - 3/2) = 30 cm

R2 = (4/3 x 10)/(4/3 - 3/2) = -20 cm

Step 3:
Now that we have the radii of curvature of the two surfaces of the lens, we can use the same formula to calculate the focal length of the lens when it is placed in water:

1/f' = (n2' - n1') x (1/R1' - 1/R2')

Where n1' and n2' are the refractive indices of air and water respectively, and R1' and R2' are the radii of curvature of the two surfaces of the lens when it is placed in water.

Substituting the values, we get:

1/f' = (4/3 - 3/2) x (1/30 - (-1/20))

f' = -15 cm

Conclusion:
Therefore, the focal length of the convex lens when it is placed in water is -15 cm. This means that the lens will converge the light rays more strongly in water than in air, and the image formed by the lens will be closer to the lens in water than in air.