In Young's double-slit experiment, the distance between the two identical slits is 6.1 times larger than the slit width. Then the number of intensity maxima observed within the central maximum of the single-slit diffraction pattern is:

In Young's double-slit experiment, we have two identical slits, and the distance between the centers of these slits is denoted by $$d$$. The width of each slit is denoted by $$a$$. According to the problem, the distance between the slits is 6.1 times larger than the slit width, so we can write:

$$ d = 6.1a $$

We need to find the number of intensity maxima observed within the central maximum of the single-slit diffraction pattern. The central maximum of the single-slit diffraction pattern is the region between the first minima on either side of the central peak. For a single slit of width $$a$$, the first minimum occurs at an angle $$\theta$$ where:

$$ a \sin \theta = \lambda $$

Here, $$\lambda$$ is the wavelength of light. Therefore, the angular positions of the first minima are at $$\sin \theta = \pm \frac{\lambda}{a}$$. The central maximum extends from $$\sin \theta = -\frac{\lambda}{a}$$ to $$\sin \theta = \frac{\lambda}{a}$$.

In the double-slit experiment, the interference maxima occur at angles where the path difference is an integer multiple of the wavelength. The condition for the $$m$$-th interference maximum is:

$$ d \sin \theta = m \lambda $$

where $$m$$ is an integer (the order of the maximum). To find which interference maxima lie within the central maximum of the diffraction pattern, we require that $$\sin \theta$$ satisfies:

$$ -\frac{\lambda}{a} \leq \sin \theta \leq \frac{\lambda}{a} $$

Substituting $$\sin \theta = \frac{m \lambda}{d}$$ from the interference condition, we get:

$$ -\frac{\lambda}{a} \leq \frac{m \lambda}{d} \leq \frac{\lambda}{a} $$

Since $$\lambda > 0$$, we can divide all parts of the inequality by $$\lambda$$:

$$ -\frac{1}{a} \leq \frac{m}{d} \leq \frac{1}{a} $$

Multiplying through by $$d$$:

$$ -\frac{d}{a} \leq m \leq \frac{d}{a} $$

Given that $$d = 6.1a$$, we substitute $$\frac{d}{a} = 6.1$$:

$$ -6.1 \leq m \leq 6.1 $$

Since $$m$$ must be an integer, the possible values of $$m$$ are:

$$ m = -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 $$

This gives 13 values: from $$-6$$ to $$6$$ inclusive.

However, the central interference maximum at $$m = 0$$ is the brightest and is considered separately in some contexts. The problem asks for the number of intensity maxima within the central maximum of the single-slit diffraction pattern, and in many interpretations, this excludes the central interference maximum ($$m = 0$$) but includes all other maxima inside the central diffraction envelope. Therefore, we exclude $$m = 0$$ and count the remaining maxima.

The values excluding $$m = 0$$ are:

$$ m = -6, -5, -4, -3, -2, -1, 1, 2, 3, 4, 5, 6 $$

This gives 12 maxima: six on the left side ($$m = -6$$ to $$m = -1$$) and six on the right side ($$m = 1$$ to $$m = 6$$).

It is important to verify that these maxima are visible and not suppressed by diffraction minima. The diffraction minima occur at $$a \sin \theta = n \lambda$$ for $$n = \pm 1, \pm 2, \pm 3, \ldots$$. The interference maxima at $$d \sin \theta = m \lambda$$ coincide with diffraction minima when $$\frac{d}{a} = \frac{m}{n}$$, which is $$\frac{6.1}{1} = \frac{61}{10}$$. For $$m$$ and $$n$$ integers, this would require $$m$$ to be a multiple of 61 and $$n$$ a multiple of 10. Within the range $$|m| \leq 6$$, the only possible $$m$$ is 0 (when $$n = 0$$), but $$n = 0$$ corresponds to the central maximum, not a minimum. Thus, no interference maximum in this range coincides with a diffraction minimum, and all 12 maxima are visible within the central diffraction envelope.

Therefore, the number of intensity maxima observed within the central maximum of the single-slit diffraction pattern, excluding the central interference maximum, is 12.

Hence, the correct answer is Option D.