Two light beams of intensities in the ratio of $$9 : 4$$ are allowed to interfere. The ratio of the intensity of maxima and minima will be :

Two light beams with intensities in the ratio $$I_1 : I_2 = 9 : 4$$. The resultant intensity in interference is given by $$I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$$, where $$\phi$$ is the phase difference.

Maximum intensity occurs when $$\cos\phi = 1$$: $$I_{\max} = I_1 + I_2 + 2\sqrt{I_1 I_2} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2$$.

Minimum intensity occurs when $$\cos\phi = -1$$: $$I_{\min} = I_1 + I_2 - 2\sqrt{I_1 I_2} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^2$$.

Since $$I_1 : I_2 = 9 : 4$$, we have $$\sqrt{I_1} : \sqrt{I_2} = 3 : 2$$, and hence $$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2} = \frac{(3 + 2)^2}{(3 - 2)^2} = \frac{25}{1}$$. The ratio of intensity of maxima to minima is $$25 : 1$$. The correct answer is Option D.