Assume that

Height of the object (h_{1}) is 5 cm,

focal length (f) is 10 cm,

distance of the object (u) is – 20 cm

Then calculate Image distance (v) = ?, Height of the image (h_{2}) = ?,

and Magnification (M) = ?

Use the lens formula, 1/v-1/u=1/f

Then, 1/v=1/u + 1/f

1/v = 1/-20 + 1/10

1/v=(-1+2)/20 = 1/20

Therefore, v=20cm

The positive sign of the image distance shows that the image is formed at 20 cm on the other side of the lens.

Now, if M is magnification h_{2}/h_{1}= v/u

Then, h_{2}= (v/u) × h_{1}= 20/-20 × 5

h_{2}= -1 × 5 = -5cm

And M = v/u = 20/-20=-1

Since, there is a negative sign to the height of the image and the magnification, it means that the image is inverted and real. It is below the principal axis and is about the same size as the object.