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(B) The initial phase difference is $\phi_0 = \pi / 4$.The path difference is $\Delta = d \sin 	heta$.The phase difference due to path difference is

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$2I_0$

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(B) The initial phase difference is $\phi_0 = \pi / 4$.The path difference is $\Delta = d \sin 	heta$.The phase difference due to path difference is $\phi' = \frac{2\pi}{\lambda} \Delta = \frac{2\pi}{\lambda} (d \sin 	heta)$.Substituting $d = \lambda / 4$ and $	heta = 30^o$:$\phi' = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{4} 	imes \sin(30^o) = \frac{\pi}{2} 	imes \frac{1}{2} = \frac{\pi}{4}$.The total phase difference is $\phi = \phi_0 + \phi' = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$.The intensity is given by $I = 4I_0 \cos^2(\phi / 2)$.Substituting $\phi = \pi / 2$:$I = 4I_0 \cos^2(\pi / 4) = 4I_0 	imes (1 / \sqrt{2})^2 = 4I_0 	imes (1 / 2) = 2I_0$.

In a Young's double-slit experiment,the distance between the two slits is $d = \lambda / 4$,where $\lambda$ is the wavelength of the light used. The initial phase difference is $\pi / 4$. What is the intensity at $\theta = 30^o$?

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