A convex lens (of focal length 20 cm) and a concave mirror, having their principal axes along the same lines, are kept 80 cm apart from each other. The concave mirror is to the right of the convex lens. When an object is kept at a distance of 30 cm to the left of the convex lens, its image remains at the same position even if the concave mirror is removed. The maximum distance of the object for which this concave mirror, by itself would produce a virtual image would be:

We have a convex lens whose focal length is given as $$f_L = +20\ \text{cm}$$ (positive because it is converging). An object is placed at a distance of $$u_L = -30\ \text{cm}$$ to the left of the lens (negative according to the usual Cartesian sign convention in which all distances measured to the left of an optical element are negative, and those measured to the right are positive).

For a thin lens the lens formula is stated first:

$$\frac{1}{f_L} \;=\; \frac{1}{v_L} \;-\; \frac{1}{u_L}$$

Substituting the numerical values we get

$$\frac{1}{20} \;=\; \frac{1}{v_L} \;-\; \Bigl(\!-\,\frac{1}{30}\Bigr) \;=\; \frac{1}{v_L} \;+\; \frac{1}{30}$$

Re-arranging,

$$\frac{1}{v_L} \;=\; \frac{1}{20} - \frac{1}{30} \;=\; \frac{3 - 2}{60} \;=\; \frac{1}{60}$$

So,

$$v_L = +60\ \text{cm}$$

Thus the convex lens alone forms a real image at a point $$60\ \text{cm}$$ to the right of the lens.

The concave mirror is placed $$80\ \text{cm}$$ to the right of the lens, therefore the position of that real image with respect to the mirror is

$$u_M = -(80 - 60)\ \text{cm} = -20\ \text{cm}$$

(negative because the object for the mirror lies to its left, i.e. in front of the reflecting surface).

Now the problem statement says that the final image of the whole lens-mirror combination is still at the same place, $$20\ \text{cm}$$ in front of the mirror, even if the mirror is removed. This is only possible when, after reflection, the mirror sends the rays back to the very point from which they seemed to diverge; in other words, that point must be the centre of curvature of the concave mirror.

For a concave mirror, the centre of curvature lies at a distance $$2f_M$$ (twice the focal length) in front of the mirror; an object placed at that point produces its image at exactly the same point. Hence we write

$$2f_M = 20\ \text{cm}$$

so that

$$f_M = 10\ \text{cm}$$

Therefore the focal length of the concave mirror is $$10\ \text{cm}.$$

When the concave mirror is used alone, it produces a virtual (erect) image only when the object lies between the pole and the principal focus, i.e. when the object distance satisfies

$$0 < u_M < f_M$$

Consequently, the largest object distance that still gives a virtual image is exactly equal to the focal length, $$f_M = 10\ \text{cm}.$$ Any object placed farther away than this will yield a real image.

Hence, the correct answer is Option D.