An object is placed at a point X between the focal point F and the optical centre C of a diverging lens. If F…

Analyze the question: A real object is placed at point X, between the optical centre C and the focal point F of a diverging (concave) lens. We need to find where the image forms.

Concept: A diverging lens always forms a virtual, upright, diminished image on the same side as the object, for any real object position. The image always lies between the lens (optical centre) and the focal point on the object's side.

Using the lens formula with f negative (diverging lens):

1/v = 1/f − 1/u

Using the real-is-positive convention for the object distance u and f = −f:

1/v = −1/f + 1/u... more simply, with the Cartesian convention: u = −u, f = −f, so

1/v = 1/f + 1/u = −1/f − 1/u, so |v| = (f·u)/(f + u)

Solution:

• Since f + u > f, we get |v| = f·u/(f + u) < u (the image is closer to the lens than the object).
• Also |v| < f because u/(f + u) < 1.

So the image is nearer to the lens than the object X, and it's on the same side as the object. Example: f = 10 cm, u = 5 cm gives |v| = 50/15 = 3.3 cm. The image is at 3.3 cm, which lies between C (0) and X (5 cm).

So the image is formed between X and C.

Answer: B. X and C

Tip: For a diverging lens, the image is always virtual, upright, diminished, and located between the lens and the object (when the object is inside F) or between the lens and F (when the object is beyond F). Whatever the object position, the image is always closer to the lens than the object.