Two light beams of intensities 4I and 9I interfere on a screen. The...

The resultant intensity is given by ${I}_{net} = {I}_{1} + {I}_{2} + 2\sqrt{{I}_{1}}\sqrt{{I}_{2}}cosϕ$

Now, for {I}_\max, phase angle is $ϕ = 0$ and for {I}_\min, phase angle is $ϕ = \pi$.

Thus, {I}_\max = {\left(\sqrt{{I}_{1}} + \sqrt{{I}_{2}}\right)}^{2} = {\left(\sqrt{9I} + \sqrt{4I}\right)}^{2} = 25I

And {I}_\min = {\left(\sqrt{{I}_{1}} - \sqrt{{I}_{2}}\right)}^{2} = {\left(\sqrt{9I} - \sqrt{4I}\right)}^{2} = I

The difference of resultant intensities is {I}_\max - {I}_\min = 25I - I = 24I