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(C) The path difference is given by $\Delta p = \frac{xd}{D}$.Given $x = 4 	imes 10^{-5} \, m$,$d = 0.25 \, cm = 2.5 	imes 10^{-3} \, m$,and $D = 10

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$3I_0 / 4$

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(C) The path difference is given by $\Delta p = \frac{xd}{D}$.Given $x = 4 	imes 10^{-5} \, m$,$d = 0.25 \, cm = 2.5 	imes 10^{-3} \, m$,and $D = 100 \, cm = 1 \, m$.$\Delta p = \frac{(4 	imes 10^{-5} \, m) 	imes (2.5 	imes 10^{-3} \, m)}{1 \, m} = 10^{-7} \, m$.The phase difference $\phi$ is given by $\phi = \frac{2\pi}{\lambda} \Delta p$.Given $\lambda = 6000 \mathring{A} = 6 	imes 10^{-7} \, m$.$\phi = \frac{2\pi}{6 	imes 10^{-7}} 	imes 10^{-7} = \frac{\pi}{3} = 60^{\circ}$.The resultant intensity $I$ is given by $I = I_0 \cos^2(\phi / 2)$.$I = I_0 \cos^2(60^{\circ} / 2) = I_0 \cos^2(30^{\circ})$.Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we have $I = I_0 	imes (\frac{\sqrt{3}}{2})^2 = I_0 	imes \frac{3}{4} = \frac{3I_0}{4}$.

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