An object is placed at the focus of concave lens having focal length $$f$$. What is the magnification and distance of the image from the optical centre of the lens?

We wish to find the image distance $$v$$ and the linear magnification $$m$$ when an object is kept at the focus of a concave (diverging) lens whose focal length is $$f$$. Throughout we follow the Cartesian sign convention: distances measured to the left of the lens are taken negative, to the right positive; for a concave lens the focal length is negative.

The object is at the focus, so the magnitude of the object distance is $$f$$, but the object lies to the left of the lens. Therefore

$$u = -\,f.$$

For any thin lens we have the lens formula, which we first state:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}.$$

Now we substitute $$u = -\,f$$ and remember that for a concave lens $$f$$ itself is negative. Substituting, we get

$$\frac{1}{v} - \frac{1}{(-\,f)} = \frac{1}{(-\,f)}.$$

The second term simplifies because a minus sign in the denominator changes sign:

$$\frac{1}{v} + \frac{1}{f} = -\,\frac{1}{f}.$$

We now collect like terms:

$$\frac{1}{v} = -\,\frac{1}{f} - \frac{1}{f}.$$

Adding the right-hand side, we obtain

$$\frac{1}{v} = -\,\frac{2}{f}.$$

Taking the reciprocal of both sides gives

$$v = -\,\frac{f}{2}.$$

The negative sign again tells us that the image is formed on the same side as the object (to the left of the lens). Hence the distance of the image from the optical centre is

$$|v| = \frac{f}{2}.$$

Next, we calculate the magnification. The formula for linear magnification produced by a lens is

$$m = \frac{v}{u}.$$

Substituting $$v = -\,f/2$$ and $$u = -\,f$$, we have

$$m = \frac{-\,f/2}{-\,f} = \frac{1}{2}.$$

Thus the image is virtual, upright (positive magnification), half the size of the object, and situated at a distance $$f/2$$ from the lens on the same side as the object.

Hence, the correct answer is Option B.