The given values are:
Object height, hₒ = 4 cm
Object distance, u = -27 cm (negative sign indicates that the object is placed in front of the lens)
Focal length, f = 18 cm

Using the lens formula, 1/f = 1/v - 1/u, where v is the image distance from the lens.

Substituting the given values, we get:
1/18 = 1/v - 1/-27

Simplifying the equation, we get:
v = 54 cm

Therefore, the image is formed at a distance of 54 cm from the lens.



To determine the nature of the image formed, we use the magnification formula, m = -v/u, where m is the magnification produced by the lens.

Substituting the given values, we get:
m = -54/-27

Simplifying the equation, we get:
m = 2

Since the value of magnification is positive, the image formed by the convex lens is erect.



To determine the size of the image formed, we use the magnification formula again, m = hᵢ/hₒ, where hᵢ is the height of the image formed.

Substituting the given values, we get:
2 = hᵢ/4

Simplifying the equation, we get:
hᵢ = 8 cm

Therefore, the size of the image formed is 8 cm.



In conclusion, when an object of height 4 cm is placed at a distance of 27 cm in front of a convex lens of focal length 18 cm, the image formed is at a distance of 54 cm from the lens, the image is erect, and the size of the image formed is 8 cm.