Question:medium

Two coherent light waves, each of intensity \( I_0 \), superpose and produce an interference pattern on a screen. Obtain the expression for the resultant intensity at a point where the phase difference between the waves is \( \phi \). Write its maximum and minimum possible values.

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Interference intensity varies with phase difference \( \phi \) as \( I = 4I_0 \cos^2\left(\frac{\phi}{2}\right) \). Always remember: maxima occur when waves are in phase, and minima when out of phase by \( \pi \).
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Solution and Explanation

Consider two interfering waves with identical amplitude \( A \), and intensity \( I_0 \propto A^2 \). The combined amplitude at a location with a phase difference \( \phi \) is calculated as: \[A_R = 2A \cos\left(\frac{\phi}{2}\right)\Rightarrow I = A_R^2 = 4A^2 \cos^2\left(\frac{\phi}{2}\right)\Rightarrow I = 4I_0 \cos^2\left(\frac{\phi}{2}\right)\] Combined Intensity: \[\boxed{I = 4I_0 \cos^2\left(\frac{\phi}{2}\right)}\] Peak Intensity: For phase differences \( \phi = 0, 2\pi, 4\pi, \ldots \), where \( \cos\left(\frac{\phi}{2}\right) = 1 \): \[I_{\max} = 4I_0\] Lowest Intensity: For phase differences \( \phi = \pi, 3\pi, \ldots \), where \( \cos\left(\frac{\phi}{2}\right) = 0 \): \[I_{\min} = 0\]
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