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(B) Given: Slit separation $d = 2 \ mm = 2 	imes 10^{-3} \ m$, Screen distance $D = 10 \ m$, Wavelength $\lambda = 6000 \ \mathring{A} = 6 	imes 10^

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$I_0$

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(B) Given: Slit separation $d = 2 \ mm = 2 	imes 10^{-3} \ m$, Screen distance $D = 10 \ m$, Wavelength $\lambda = 6000 \ \mathring{A} = 6 	imes 10^{-7} \ m$.In $Young's$ double slit experiment, the intensity at any point on the screen is given by $I = 4 I_0 \cos^2(\frac{\phi}{2})$, where $\phi$ is the phase difference.The path difference $\Delta x$ at a position $y$ on the screen is $\Delta x = \frac{yd}{D}$.For a point in front of one of the slits, $y = \frac{d}{2}$.Thus, $\Delta x = \frac{(d/2)d}{D} = \frac{d^2}{2D}$.Substituting the values: $\Delta x = \frac{(2 	imes 10^{-3})^2}{2 	imes 10} = \frac{4 	imes 10^{-6}}{20} = 2 	imes 10^{-7} \ m$.The phase difference $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{6 	imes 10^{-7}} 	imes 2 	imes 10^{-7} = \frac{2\pi}{3}$.The intensity $I = I_{max} \cos^2(\frac{\phi}{2}) = (4 I_0) \cos^2(\frac{2\pi/3}{2}) = 4 I_0 \cos^2(\frac{\pi}{3})$.Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$, we have $I = 4 I_0 (\frac{1}{2})^2 = 4 I_0 	imes \frac{1}{4} = I_0$.

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