Two identical light waves having phase difference $\phi$ propagate in same direction. When they superpose,…

Two identical light waves having phase difference $\phi$ propagate in same direction. When they superpose, the intensity of resultant wave is proportional to
  1. $\cos ^2\left(\frac{\phi}{4}\right)$
  2. $\cos ^2\left(\frac{\phi}{3}\right)$
  3. $\cos ^2\left(\frac{\phi}{2}\right)$
  4. $\cos ^2 \phi$.

Solution

$A^2=a_1^2+a_i^2+2 a_1 a_2 \cos \phi$, where $A$ is amplitude of resultant wave and given that, $a_1=a_2=a$, where, a is amplitude of individual wave. $\begin{array}{ll} \therefore \quad & A^2=2 a^2(1+\cos \phi)=2 a^2\left(1+2 \cos ^2 \frac{\phi}{2}-1\right) \\ & \Rightarrow A^2 \propto \cos ^2 \cdot \frac{\phi}{2} \end{array}$
Now, $I \propto \mathrm{~A}^2$ $\therefore \quad I \propto A^2 \propto \cos ^2 \frac{\phi}{2} \Rightarrow I \propto \cos ^2 \frac{\phi}{2}$ /

Asked in: MHT CET 2024 (11 May Shift 1)

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