Answer:

Part 1: Finding the location of virtual image formed.

To find the location of the virtual image formed by the lens system, we can use the lens formula:

1/f = 1/u + 1/v

Where,
f = focal length of the lens system (since the lenses are in contact, their effective focal length is f1 + f2 = 30cm)
u = distance of the object from the first lens (given to be infinity)
v = distance of the image from the second lens

Substituting the values, we get:

1/30 = 0 + 1/v

v = 30 cm

Therefore, the virtual image is formed at a distance of 30 cm from the second lens.

Part 2: Finding the focal length of the equivalent lens.

To find the focal length of the equivalent lens, we can use the formula:

1/f = (n-1)(1/R1 - 1/R2)

Where,
n = refractive index of the lens material (assumed to be 1.5)
R1 and R2 = radii of curvature of the two lenses (assumed to be equal)

Since the two lenses are identical, their radii of curvature are equal and can be assumed to be R. Substituting the values, we get:

1/f = (1.5 - 1)(1/R - 1/R) = 0

Therefore, the equivalent lens has an infinite focal length.

Explanation:

When two convex lenses are placed in contact, their effective focal length is equal to the sum of their individual focal lengths. In this case, since the focal length of each lens is 15 cm, the effective focal length of the lens system is 30 cm.

The image formed by the first lens acts as the object for the second lens. Since the first lens forms a real image, the second lens sees this image as its object. The image formed by the second lens is therefore virtual and is located at a distance of 30 cm from the second lens.

To find the focal length of the equivalent lens, we can use the formula for the focal length of a single lens in terms of its radius of curvature. Since the two lenses are identical and have equal radii of curvature, we can assume them to be the same and simplify the formula accordingly. Substituting the values, we get the focal length of the equivalent lens to be infinite. This means that the equivalent lens has no power and does not converge or diverge light.