Diffraction by two slits of width a
Two slits separated by a distance $d$ that each have a width $a$ display a diffraction pattern that is a product of the double slit interference pattern and the single slit diffraction pattern. We define $\delta = 2\pi d\sin(\theta)/\lambda$ and $\beta = 2\pi a\sin(\theta )/\lambda$. To determine the total amplitude, we integrate over both slits. $$A=\frac{A_0}{a}\int\limits_{-\frac{d}{2}-\frac{a}{2}}^{-\frac{d}{2}+\frac{a}{2}}\exp \left(i\omega t + i\frac{\beta y}{a}\right)dy+\frac{A_0}{a}\int\limits_{\frac{d}{2}-\frac{a}{2}}^{\frac{d}{2}+\frac{a}{2}}\exp \left(i\omega t + i\frac{\beta y}{a}\right)dy$$Performing the integration gives, $$ A=\frac{A_0}{i\beta}e^{i\omega t} \left( e^{-i\frac{\delta}{2}}e^{i\frac{\beta}{2}} - e^{-i\frac{\delta}{2}}e^{-i\frac{\beta}{2}} +e^{i\frac{\delta}{2}}e^{i\frac{\beta}{2}} - e^{i\frac{\delta}{2}}e^{-i\frac{\beta}{2}} \right)$$This can be written as, $$A=\frac{A_0}{i\beta}e^{i\omega t} \left( e^{i\frac{\beta}{2}} - e^{-i\frac{\beta}{2}} \right)\left( e^{i\frac{\delta}{2}} + e^{i\frac{\delta}{2}} \right)=\frac{2A_0}{\beta/2}e^{i\omega t}\cos(\delta/2)\sin(\beta/2)$$The intensity is proportional to the squared amplitude, $$I=I_0\frac{\sin^2(\beta/2)}{(\beta/2)^2}\cos^2(\delta/2)$$
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