A person has been using spectacles of power $$-1.0$$ diopter for distant vision and a separate reading glass of power $$2.0$$ diopters. What is the least distance of distinct vision for this person:

The least distance of distinct vision (near point) is obtained from the data of the reading (converging) glasses, because these spectacles are chosen so that an object kept at the ordinary reading distance appears at the patient’s near point.

Step 1: Write the basic formulas
Power: $$P = \frac{1}{f}$$  (with $$f$$ in metres)
Lens formula: $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ $$-(1)$$
Here $$u$$ = object distance from the lens, $$v$$ = image distance from the lens (measured from the same side as the light is incident; a virtual image has negative $$v$$).

Step 2: Focal length of the reading glass
Given power of reading glass, $$P_r = +2.0\ \text{D}$$.
Therefore $$f_r = \frac{1}{P_r} = \frac{1}{2.0}\ \text{m} = 0.50\ \text{m}$$.

Step 3: Choose the usual reading distance
While reading, a book is normally held at about $$25\ \text{cm}$$ from the spectacles.
Hence for the lens, $$u = -25\ \text{cm} = -0.25\ \text{m}$$ (negative by the Cartesian sign convention).

Step 4: Locate the virtual image (near point) using Eq. $$(1)$$
Substituting $$f = +0.50\ \text{m}$$ and $$u = -0.25\ \text{m}$$:
$$\frac{1}{0.50} = \frac{1}{v} - \left(-\frac{1}{0.25}\right)$$
$$2 = \frac{1}{v} + 4$$
$$\frac{1}{v} = 2 - 4 = -2$$
$$v = -0.50\ \text{m}$$.

The negative sign confirms that the image is virtual and lies on the same side as the object (towards the eye). The magnitude gives the distance of the near point:

$$|v| = 0.50\ \text{m} = 50\ \text{cm}$$.

Step 5: State the result
The person’s least distance of distinct vision is $$50\ \text{cm}$$.

Therefore, the correct option is Option D (50 cm).