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(C) The total phase difference $\Delta \phi$ at point $P$ is given by the sum of the initial phase difference between the sources and the phase differ

Vedclass · Accepted Answer

$5.3$

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(C) The total phase difference $\Delta \phi$ at point $P$ is given by the sum of the initial phase difference between the sources and the phase difference due to the path difference.Let $\phi_1$ be the initial phase difference between the sources $A$ and $B$. Given $\phi_1 = 54^{\circ}$.Converting to radians: $\phi_1 = 54 	imes \frac{\pi}{180} = 0.3 \pi 	ext{ rad}$.Let $\phi_2$ be the phase difference due to the path difference $\Delta x = PB - PA = 2.5 \lambda$.The formula for phase difference due to path difference is $\phi_2 = \frac{2 \pi}{\lambda} 	imes \Delta x$.Substituting the values: $\phi_2 = \frac{2 \pi}{\lambda} 	imes 2.5 \lambda = 5 \pi 	ext{ rad}$.Since source $A$ is ahead in phase, the total phase difference is $\Delta \phi = \phi_2 + \phi_1 = 5 \pi + 0.3 \pi = 5.3 \pi 	ext{ rad}$.

$A$ and $B$ are two interfering sources where $A$ is ahead in phase by $54^{\circ}$ relative to $B$. The observation is taken from point $P$ such that $PB-PA=2.5 \lambda$. Then the phase difference between the waves from $A$ and $B$ reaching point $P$ is (in rad) (in $\pi$)

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