Two coherent monochromatic light beams of intensities I and 4I are superimposed. The difference between maximum and minimum possible intensities in the resulting beam is $$xI$$. The value of $$x$$ is ___________.

We need to find the difference between maximum and minimum intensities when two coherent beams of intensities $$I$$ and $$4I$$ are superimposed.

For two coherent light beams with intensities $$I_1$$ and $$I_2$$, the maximum intensity for constructive interference is given by $$I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$$ and the minimum intensity for destructive interference is given by $$I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$$.

Substituting $$I_1 = I$$ and $$I_2 = 4I$$ into the expression for the maximum intensity gives $$I_{\max} = (\sqrt{I} + \sqrt{4I})^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$$, while substituting into the minimum intensity yields $$I_{\min} = (\sqrt{4I} - \sqrt{I})^2 = (2\sqrt{I} - \sqrt{I})^2 = (\sqrt{I})^2 = I$$.

The difference between these extremes is $$I_{\max} - I_{\min} = 9I - I = 8I$$, which shows that $$x = 8$$.

The correct answer is 8.