The image formed by a concave lens is

The correct answer is Option (2) → always virtual

A concave lens is a diverging lens. For a real object (light diverging from the object), the refracted rays diverge even more, so they never meet; they only appear to come from a point on the same side as the object. The image is therefore always virtual, erect and diminished, wherever the object is placed.

With the Cartesian sign convention, \( f = -F \) (F is the magnitude of the focal length) and a real object has \( u = -a \):

    \( \displaystyle \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = -\frac{1}{F} - \frac{1}{a} \)

    \( v \lt 0 \text{ for every } a \gt 0 \ \Rightarrow\ \text{the image is always virtual} \)

Why the other options are incorrect

• Option (1) always real — Incorrect. For a real object a concave lens never forms a real image.

• Option (3) certainly real if the object is real — Incorrect. For a real object the image is always virtual, never real.

• Option (4) real/virtual depending upon the position of the object from the lens — Incorrect. For a real object the position changes only the size and location of the image, not its nature (see the note below for a virtual object).

Note: The question does not state that the object is real. For a virtual object (a beam converging towards a point behind the lens, \( u = +a \)), \( 1/v = 1/a - 1/F \), so the image is real when the virtual object lies within the focal length (\( a \lt F \)) and virtual when it lies beyond it. Under that reading, Option (4) is also correct.