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(A) The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula $\phi = \frac{2\pi}{\lambda} \Delta x$.For the first point

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$K/2$

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(A) The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula $\phi = \frac{2\pi}{\lambda} \Delta x$.For the first point,the path difference is $\Delta x_1 = \lambda$. Thus,the phase difference is $\phi_1 = \frac{2\pi}{\lambda} 	imes \lambda = 2\pi$.The intensity is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$. Since the sources are identical,$I_1 = I_2 = I_0$,so $I = 2I_0 + 2I_0 \cos \phi = 4I_0 \cos^2(\phi/2)$.For the first point: $K = 4I_0 \cos^2(2\pi/2) = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0$. Therefore,$I_0 = K/4$.For the second point,the path difference is $\Delta x_2 = \lambda/4$. The phase difference is $\phi_2 = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{4} = \frac{\pi}{2}$.The intensity at this point is $I' = 4I_0 \cos^2(\phi_2/2) = 4I_0 \cos^2(\pi/4) = 4I_0 (1/\sqrt{2})^2 = 4I_0 (1/2) = 2I_0$.Substituting $I_0 = K/4$,we get $I' = 2(K/4) = K/2$ units.

Waves emitted from two identical sources produce an intensity of $K$ units at a point on the screen. If the path difference between these two waves is $\lambda$,calculate the intensity at a point on the screen where the path difference is $\lambda/4$.

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