The two coherent monochromatic light beams of intensities I and 41 are superposed. The maximum and minimum possible intensities in the resulting beams are:

The correct option is 9 I and I (represented as "9 1 and I" in the options).

Step-by-Step Explanation:

When two coherent light waves of intensities $I_1$ and $I_2$ superimpose, the resultant intensity depends on the phase difference between the waves. The maximum and minimum resultant intensities are determined by constructive and destructive interference, respectively.

The formula for the maximum intensity ($I_{\text{max}}$) is given by:
$I_{\text{max}}={\left(\sqrt{I_1}+\sqrt{I_2}\right)}^2$
Similarly, the formula for the minimum intensity ($I_{\text{min}}$) is given by:
$I_{\text{min}}={\left(\sqrt{I_1}-\sqrt{I_2}\right)}^2$

Given in the problem:
Intensity of the first beam, $I_1=I$
Intensity of the second beam, $I_2=4I$

Let us calculate the square roots of the intensities:
$\sqrt{I_1}=\sqrt{I}$
$\sqrt{I_2}=\sqrt{4I}=2\sqrt{I}$

Now, let's find the maximum possible intensity ($I_{\text{max}}$):
$I_{\text{max}}={\left(\sqrt{I}+2\sqrt{I}\right)}^2$
$I_{\text{max}}={\left(3\sqrt{I}\right)}^2$
$I_{\text{max}}=9I$

Next, let's find the minimum possible intensity ($I_{\text{min}}$):
$I_{\text{min}}={\left(2\sqrt{I}-\sqrt{I}\right)}^2$
$I_{\text{min}}={\left(\sqrt{I}\right)}^2$
$I_{\text{min}}=I$

Thus, the maximum and minimum possible intensities in the resulting beams are 9 I and I.