Two coherent monochromatic light beams of intensities $$4I$$ and $$9I$$are superimposed. The difference between the maximum and minimum intensities in the resulting interference pattern is $$xI$$. The value of x is __________.

Let the intensities of the two coherent beams be $$I_1 = 4I$$ and $$I_2 = 9I$$.

For two coherent sources, the resultant intensity at any point is given by the interference formula
$$I = I_1 + I_2 + 2\sqrt{I_1 I_2}\,\cos\phi$$
where $$\phi$$ is the phase difference between the beams.

Maximum intensity ($$\phi = 0$$, constructive interference):
$$I_{\max} = I_1 + I_2 + 2\sqrt{I_1 I_2}$$

Minimum intensity ($$\phi = \pi$$, destructive interference):
$$I_{\min} = I_1 + I_2 - 2\sqrt{I_1 I_2}$$

The required quantity is the difference between these two:
$$I_{\max} - I_{\min} = \left(I_1 + I_2 + 2\sqrt{I_1 I_2}\right) \;-\; \left(I_1 + I_2 - 2\sqrt{I_1 I_2}\right)$$

Simplifying gives
$$I_{\max} - I_{\min} = 4\sqrt{I_1 I_2}$$

Substitute $$I_1 = 4I$$ and $$I_2 = 9I$$:
$$\sqrt{I_1 I_2} = \sqrt{(4I)(9I)} = \sqrt{36I^2} = 6I$$

Hence
$$I_{\max} - I_{\min} = 4 \times 6I = 24I$$

The problem states that this difference equals $$xI$$, so
$$x = 24$$

Therefore, the value of $$x$$ is $$24$$.