In an experiment of single slit diffraction pattern, first minimum for red light coincides with first maximum of some other wavelength. If wavelength of red light is 6600 $$\text{Å}$$, then wavelength of first maximum will be:

In a single slit diffraction experiment, the position of minima and maxima are determined by specific conditions. For the first minimum of red light, the condition is given by:

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

where $$ a $$ is the slit width, $$ \theta $$ is the angle from the central maximum, $$ \lambda $$ is the wavelength, and $$ n $$ is the order of the minimum. For the first minimum, $$ n = 1 $$. Given the wavelength of red light $$ \lambda_r = 6600 \text{Å} $$, we have:

$$ a \sin \theta = 1 \cdot 6600 = 6600 \text{Å} \quad \text{(Equation 1)} $$

For the first maximum of another light with wavelength $$ \lambda $$, the condition is approximately:

$$ a \sin \theta = \left( m + \frac{1}{2} \right) \lambda $$

where $$ m $$ is the order of the maximum. For the first maximum, $$ m = 1 $$, so:

$$ a \sin \theta = \left( 1 + \frac{1}{2} \right) \lambda = \frac{3}{2} \lambda \quad \text{(Equation 2)} $$

The problem states that the first minimum for red light coincides with the first maximum of the other light. This means the angular position $$ \theta $$ is the same for both, so $$ \sin \theta $$ is identical. Therefore, the expressions for $$ a \sin \theta $$ from Equation 1 and Equation 2 must be equal:

$$ 6600 = \frac{3}{2} \lambda $$

Solving for $$ \lambda $$:

$$ \lambda = 6600 \times \frac{2}{3} $$

Performing the multiplication:

$$ \lambda = \frac{6600 \times 2}{3} = \frac{13200}{3} = 4400 \text{Å} $$

Thus, the wavelength of the other light for which the first maximum coincides with the first minimum of red light is 4400 Å.

Comparing with the options:

A. 3300 Å

B. 4400 Å

C. 5500 Å

D. 6600 Å

Hence, the correct answer is Option B.