$\mathrm{A}$ and $\mathrm{B}$ are two interfering sources where $\mathrm{A}$ is ahead in phase by…

$\mathrm{A}$ 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)
  1. $3.5 \pi$
  2. $4.3 \pi$
  3. $5.3 \pi$
  4. $5.8 \pi$

Solution

$\begin{aligned} & \text { Total phase difference }=\phi_1+\phi_2 \\ & \begin{aligned} \phi_1 & =54 \times \frac{\pi}{180}=0.3 \pi \\ \phi_2 & =\frac{2 \pi}{\lambda} \times(\mathrm{PB}-\mathrm{PA}) \\ & =\frac{2 \pi}{\lambda} \times 2.5 \lambda=5 \pi \\ \therefore \quad & \phi_1+\phi_2=5 \pi+0.3 \pi=5.3 \pi\end{aligned}\end{aligned}$

Asked in: MHT CET 2023 (13 May Shift 1)

Practice more Wave Optics questions on Aicharya