Two thin convex lenses of focal lengths 30 cm and 10 cm are placed coaxially, 10 cm apart. The power of this combination is :

The equivalent focal length of two thin lenses separated by a distance is found with the lens-combination formula
$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1\,f_2}$$
where
$$f_1, f_2$$ are the focal lengths of the individual lenses (signs taken from the Cartesian convention) and $$d$$ is the separation between their optical centres.

All three quantities are first expressed in the same unit. Working in centimetres:
$$f_1 = 30\; \text{cm},\; f_2 = 10\; \text{cm},\; d = 10\; \text{cm}$$

Substituting into the formula gives
$$\frac{1}{F} = \frac{1}{30} + \frac{1}{10} - \frac{10}{30 \times 10}$$

Simplify each term:
$$\frac{1}{30} = 0.0333\ldots,\quad \frac{1}{10} = 0.1,\quad \frac{10}{30 \times 10} = \frac{10}{300} = 0.0333\ldots$$

Add and subtract term by term:
$$\frac{1}{F} = 0.0333\ldots + 0.1 - 0.0333\ldots = 0.1$$

Thus
$$F = \frac{1}{0.1} = 10\; \text{cm}$$

Convert the focal length to metres for power calculation:
$$F = 10\; \text{cm} = 0.10\; \text{m}$$

Lens power is defined as
$$P = \frac{1}{F\;(\text{in metres})}$$
so
$$P = \frac{1}{0.10} = 10\; \text{dioptre}$$

Therefore, the power of the lens combination is $$10\; \text{D}$$, which corresponds to Option D.