Q118 / 150 · MHT CET 2024 (11 May Shift 1) · Physics
Two identical light waves having phase difference $\phi$ propagate in same direction. When they superpose, the intensity of resultant wave is proportional to
- A$\cos ^2\left(\frac{\phi}{4}\right)$
- B$\cos ^2\left(\frac{\phi}{3}\right)$
- Correct answerC$\cos ^2\left(\frac{\phi}{2}\right)$
- D$\cos ^2 \phi$.
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, $I \propto \mathrm{~A}^2$ $\therefore \quad I \propto A^2 \propto \cos ^2 \frac{\phi}{2} \Rightarrow I \propto \cos ^2 \frac{\phi}{2}$
Now, $I \propto \mathrm{~A}^2$ $\therefore \quad I \propto A^2 \propto \cos ^2 \frac{\phi}{2} \Rightarrow I \propto \cos ^2 \frac{\phi}{2}$
Hence, the correct answer is option: 'C'.