Two identical waves 1 and 2,each of intensity | vedclass

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(D) The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula:$\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$From

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$2 I_0$

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(D) The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula:$\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$From the graph,the path difference between the two waves is $\Delta x = \frac{\lambda}{4}$.Substituting this value:$\Delta \phi = \frac{2 \pi}{\lambda} 	imes \frac{\lambda}{4} = \frac{\pi}{2}$The resultant intensity $I$ of two waves with equal intensity $I_0$ and phase difference $\phi$ is given by:$I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos \phi = 2I_0(1 + \cos \phi)$Substituting $\phi = \frac{\pi}{2}$:$I = 2I_0(1 + \cos \frac{\pi}{2})$Since $\cos \frac{\pi}{2} = 0$,we get:$I = 2I_0(1 + 0) = 2I_0$

Two identical waves $1$ and $2$,each of intensity $I_0$,are superimposed. The resulting intensity is:

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