= Solution
Let each slit alone give local <light intensity> $I_s$, and assume equal coherent illumination. In <Fraunhofer diffraction>, the two <wave amplitudes> differ in <wave phase> by $\delta=2\pi s\sin\theta/\lambda$. Adding the <wave amplitudes> before squaring gives
$$
I(\theta)=I_s|1+e^{i\delta}|^2=4I_s\cos^2\left(\frac{\pi s\sin\theta}{\lambda}\right).
$$
For a flat screen, $\sin\theta=x/\sqrt{d^2+x^2}$. In the paraxial region $|x|\ll d$, the <double-slit interference> profile is therefore
$$
\boxed{I(x)=4I_s\cos^2\left(\frac{\pi sx}{\lambda d}\right),\qquad \Delta x=\frac{\lambda d}{s}.}
$$
The given $d\gg s^2/\lambda$ justifies the far-field expansion. Slit widths have been neglected; a finite width multiplies this profile by the single-slit <diffraction envelope>. At large angles the one-slit <light intensity> and geometric spreading must also be retained, rather than treating $I_s$ as constant.
Back to article page
