U = -10 cm

h1 = 4 cm

f = 20 cm

1/v – 1/u = 1/f

1/v – 1/-10 = 1/20

1/v = 1/20 – 1/20 = - (1/20)

V = -20 cm (Image is 20 cm in front of the convex lens)

m = v/u = 20/-10 = -2

m = h2/h1 = -2

h2/4 = -2

h2 = - 8 cm

Image is 8 cm in size and is real and inverted.

Given information:
- Object height (h) = 4 cm
- Distance of object from the lens (u) = 10 cm
- Focal length of the convex lens (f) = 20 cm

Find:
- Position of the image (v)
- Nature of the image (real or virtual)
- Size of the image (h')

Solution:

Step 1: Calculate the position of the image (v)
Using the lens formula:
1/f = 1/v - 1/u

Substituting the given values:
1/20 = 1/v - 1/10

Simplify the equation:
1/v = 1/20 + 1/10
1/v = (1 + 2)/20
1/v = 3/20

Taking the reciprocal:
v = 20/3 cm

So, the position of the image (v) is 20/3 cm.

Step 2: Determine the nature of the image
To determine the nature of the image, we can use the magnification formula:
m = -v/u

Substituting the given values:
m = -(20/3)/(10)
m = -2/3

Since the magnification (m) is negative, the image is inverted. Therefore, the nature of the image is real and inverted.

Step 3: Calculate the size of the image (h')
Using the magnification formula:
m = h'/h

Substituting the given values:
-2/3 = h'/4

Cross-multiplying:
-2 * 4 = 3 * h'
-8 = 3h'

Dividing both sides by 3:
h' = -8/3 cm

Since the height of the image (h') is negative, it indicates that the image is inverted.

Summary:
- The position of the image (v) is 20/3 cm.
- The nature of the image is real and inverted.
- The size of the image (h') is -8/3 cm, indicating an inverted image.

Therefore, the image formed by the convex lens is real, inverted, and smaller in size compared to the object.