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(D) Given: Amplitude of each wave = $a$. Wavelength $\lambda = 1 \ m$. Path difference $\Delta x = AP - BP = 50 \ cm = 0.5 \ m$.The phase difference d

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$a$

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(D) Given: Amplitude of each wave = $a$. Wavelength $\lambda = 1 \ m$. Path difference $\Delta x = AP - BP = 50 \ cm = 0.5 \ m$.The phase difference due to path difference is $\phi_{path} = \frac{2\pi}{\lambda} 	imes \Delta x = \frac{2\pi}{1} 	imes 0.5 = \pi$.Since source $A$ is ahead of $B$ by $\frac{\pi}{3}$,the total phase difference $\Delta \phi$ between the two waves at point $P$ is $\Delta \phi = \phi_{path} - \frac{\pi}{3} = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.The resultant amplitude $R$ is given by $R = \sqrt{a^2 + a^2 + 2a^2 \cos(\Delta \phi)}$.Substituting $\Delta \phi = \frac{2\pi}{3}$:$R = \sqrt{2a^2 + 2a^2 \cos(120^\circ)} = \sqrt{2a^2 + 2a^2(-0.5)} = \sqrt{2a^2 - a^2} = \sqrt{a^2} = a$.

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