Intensity In Interference

Question

Two coherent light waves of equal amplitude superpose at a point on a screen where the path difference corresponds to a phase difference of 90 degrees between them. What is the resultant intensity expressed in terms of the maximum intensity?

RankGuru · Accepted Answer

When two coherent waves of equal intensity $I_0$ overlap, the resultant intensity varies with their phase difference $\phi$ as $I=4I_0\cos^2\!\left(\frac{\phi}{2}ight)$, and the maximum $I_{max}=4I_0$ occurs when $\phi=0$. Here the phase difference is $\phi=90°=\frac{\pi}{2}$, so $\frac{\phi}{2}=45°$ and $\cos^2(45°)=\left(\frac{1}{\sqrt2}ight)^2=\frac{1}{2}$. Therefore $I=4I_0	imes\frac{1}{2}=2I_0=\frac{I_{max}}{2}$. It is worth noting that the resultant is not the simple sum $2I_0$ of the two individual intensities; interference redistributes energy, so the combined brightness depends on phase, ranging from $4I_0$ at full reinforcement down to zero at complete cancellation. The cosine-squared dependence guarantees energy conservation overall, since the excess brightness piled into the maxima is exactly balanced by the darkness of the minima across the pattern. The value $I_{max}/4$ is wrong because it uses $\cos^2(\phi)$ rather than $\cos^2(\phi/2)$. The value $I_{max}$ is wrong as it corresponds to zero phase difference, full constructive interference. The value $3I_{max}/4$ is wrong since it misapplies the half-angle and would need $\phi=60°$. This NCERT intensity relation underlies the smooth cosine-squared brightness profile of fringes. A sanity check confirms a quarter-wave path difference gives exactly half the peak brightness, the natural midway point of the fringe.

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