The distance between an object and its virtual image of magnification…

The correct answer is Option (4) → \( -40\,\mathrm{cm} \)

A virtual image with magnification \( +1/2 \) is upright and diminished; a single lens forms such an image only if it is a diverging lens, so the focal length is negative. Use the new Cartesian sign convention with the object on the left, \( u \lt 0 \).

For a virtual, upright image \( m = v/u = 1/2 \), so \( v = u/2 \): the image lies on the same side as the object, between the object and the lens. The distance between object and image is

    \( \displaystyle |u| - |v| = |u| - \frac{|u|}{2} = 20\,\mathrm{cm} \;\Rightarrow\; |u| = 40\,\mathrm{cm} \)

    \( u = -40\,\mathrm{cm} \)
    \( v = -20\,\mathrm{cm} \)

Now the lens formula:

    \( \displaystyle \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = -\frac{1}{20} + \frac{1}{40} = -\frac{1}{40} \)

    \( \displaystyle f = -40\,\mathrm{cm} \)