MHT CET · Physics · Wave Optics
\(\mathrm{A}\) and \(\mathrm{B}\) are two interfering sources where \(\mathrm{A}\) is ahead in phase by \(54^{\circ}\) relative to \(B\). The observation is taken from point \(\mathrm{P}\) such that \(\mathrm{PB}-\mathrm{PA}=2.5 \lambda\). Then the phase difference between the waves from \(A\) and \(B\) reaching point \(\mathrm{P}\) is (in rad)
- A \(3.5 \pi\)
- B \(4.3 \pi\)
- C \(5.3 \pi\)
- D \(5.8 \pi\)
Answer & Solution
Correct Answer
(C) \(5.3 \pi\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Total phase difference }=\phi_1+\phi_2 \\ & \begin{aligned} \phi_1 & =54 \times \frac{\pi}{180}=0.3 \pi \\ \phi_2 & =\frac{2 \pi}{\lambda} \times(\mathrm{PB}-\mathrm{PA}) \\ & =\frac{2 \pi}{\lambda} \times 2.5 \lambda=5 \pi \\ \therefore \quad & \phi_1+\phi_2=5 \pi+0.3 \pi=5.3 \pi\end{aligned}\end{aligned}\)
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