Given data:
Refractive index of the convex lens, n1 = 1.5
Focal length of the convex lens in air, f1 = 2 cm
Refractive index of the liquid, n2 = 1.25

To find: Focal length of the lens when immersed in the liquid, f2

Using the lens maker's formula, we can relate the focal length of a lens with its refractive index and the radii of curvature of its surfaces.

The lens maker's formula is given by:
1/f = (n - 1) * (1/R1 - 1/R2)

Where:
f = focal length of the lens
n = refractive index of the lens
R1 = radius of curvature of the first surface of the lens
R2 = radius of curvature of the second surface of the lens

For a convex lens, the radii of curvature are positive for both surfaces.

Since the lens in question is a convex lens, we can use the lens maker's formula to calculate the focal length of the lens in air.

Substituting the given values into the lens maker's formula, we have:
1/f1 = (n1 - 1) * (1/R1 - 1/R2)

Since the lens is a convex lens, the radius of curvature of both surfaces is positive. Therefore, 1/R1 and 1/R2 are positive.

Simplifying the equation, we get:
1/f1 = (1.5 - 1) * (1/R1 - 1/R2)
1/f1 = 0.5 * (1/R1 - 1/R2)
2/f1 = 1/R1 - 1/R2

Since the focal length in air, f1, is positive, the right-hand side of the equation must also be positive.

Now, when the lens is immersed in the liquid with a refractive index of n2 = 1.25, we can use the lens maker's formula again to calculate the new focal length, f2.

Substituting the given values into the lens maker's formula, we have:
1/f2 = (n2 - 1) * (1/R1 - 1/R2)

Simplifying the equation, we get:
1/f2 = (1.25 - 1) * (1/R1 - 1/R2)
1/f2 = 0.25 * (1/R1 - 1/R2)
4/f2 = 1/R1 - 1/R2

Since the refractive index of the liquid, n2, is less than the refractive index of the lens, n1, the focal length when immersed in the liquid, f2, will be smaller than the focal length in air, f1.

Comparing the equations for 2/f1 and 4/f2, we can see that the right-hand sides of the equations are the same.

Therefore, 2/f1 = 4/f2

Simplifying the equation, we get:
f2 = (2/f1) * 4
f2 = 8/f1

Substituting the value of f1, we have:
f2 = 8/2
f2 = 4 cm

Hence, the focal length of the lens when immersed in a liquid of refractive index 1.25