1. For two coherent light beams with intensities $I_1$ and $I_2$, the resultant intensity is

$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$

where $\phi$ is the phase difference.

2. The maximum intensity occurs when $\cos\phi = 1$:

$I_{\max} = I_1 + I_2 + 2\sqrt{I_1 I_2}$

3. The minimum intensity occurs when $\cos\phi = -1$:

$I_{\min} = I_1 + I_2 - 2\sqrt{I_1 I_2}$

4. Here,

$I_1 = I, \qquad I_2 = 4I$

So,

$\sqrt{I_1 I_2} = \sqrt{I\cdot 4I} = 2I$

5. Now calculate maximum intensity:

$I_{\max} = I + 4I + 2(2I) = 5I + 4I = 9I$

6. Calculate minimum intensity:

$I_{\min} = I + 4I - 2(2I) = 5I - 4I = I$

7. Therefore, the difference is

$I_{\max} - I_{\min} = 9I - I = 8I$

Given this difference is $xI$, we get

$xI = 8I \Rightarrow x = 8$

Final Answer

$\boxed{8}$

Comparison with stored correct answer

Stored correct answer = $8$

My derived answer also equals $8$, so they agree.