Converging lens means a convex lens. As the distances given in the question are large, so we choose a scale of 1: 5, i.e., 1 cm represents 5 cm. Therefore, on this scale 5 cm high object, object distance of 25 cm and focal length of 10 cm can be represented by 1 cm high, 5 cm and 2 cm lines respectively. Now, we draw the ray diagram as follows:
(i) Draw a horizontal line to represent the principal axis of the convex lens.
(ii) Centre line is shown by DE.
(iii) Mark two foci F and F' on two sides of the lens, each at a distance of 2 cm from the lens.
(iv) Draw an arrow AB of height 1 cm on the left side of lens at a distance of 5 cm from the lens.
(v) Draw a line AD parallel to principal axis and then, allow it to pass straight through the focus (F') on the right side of the lens.
(vi) Draw a line from A to C (centre of the lens), which goes straight without deviation.
(vii) Let the two lines starting from A meet at A'.
(viii) Draw AB', perpendicular to the principal axis from A'.
(ix) Now AB', represents the real, but inverted image of the object AB.
(x) Then, measure CB' and A'B'. It is found that CB' = 3.3 cm and A'B' = 0.7 cm.

(xi) Thus the final position, nature and size of the image A'B' are:
         (a) Position of image A'B' = 3.3 cm × 5 = 16.5 cm from the lens on opposite side.
         (b) Nature of image A’B’: Real and inverted. 
         (c) Height of image A'B': 0.7 × 5 = 3.5 cm, i.e., image is smaller than the object.