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(A) The total phase difference at point $P$ is the sum of the initial phase difference and the phase difference caused by the path difference.Initial

Vedclass · Accepted Answer

$156$

Vedclass · Answer

(A) The total phase difference at point $P$ is the sum of the initial phase difference and the phase difference caused by the path difference.Initial phase difference $\phi_i = 66^\circ$ (where $A$ is ahead of $B$).Path difference $\Delta x = PB - PA = \lambda / 4$.The phase difference due to path difference is given by $\phi_p = \frac{2\pi}{\lambda} 	imes \Delta x = \frac{360^\circ}{\lambda} 	imes \frac{\lambda}{4} = 90^\circ$.Since $A$ is ahead of $B$ by $66^\circ$,and the path $PB$ is longer than $PA$ by $\lambda/4$,the wave from $B$ lags behind the wave from $A$ due to the path difference.The total phase difference $\Delta \phi = 66^\circ + 90^\circ = 156^\circ$.

Among the two interfering monochromatic sources $A$ and $B$; $A$ is ahead of $B$ in phase by $66^\circ$. If the observation is taken from point $P$,such that $PB - PA = \lambda / 4$. Then the phase difference between the waves from $A$ and $B$ reaching $P$ is.....$^\circ$.

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