= Solution
For two identical narrow slits under coherent normal illumination, with $\lambda$ the <wavelength> in the propagation medium, let $I_s$ be the intensity contributed by either slit alone in the relevant angular range. At far-field angle $\theta$, the path difference is $\Delta=s\sin\theta$ and the <phase> difference is $2\pi\Delta/\lambda$. Adding the two complex amplitudes gives
$$
\boxed{I(\theta)=2I_s\left[1+\cos\left(\frac{2\pi s\sin\theta}{\lambda}\right)\right]=4I_s\cos^2\left(\frac{\pi s\sin\theta}{\lambda}\right)}.
$$
For a screen perpendicular to the optical axis, $\tan\theta=z/d$. Thus the angular fringe factor uses $\sin\theta=z/\sqrt{d^2+z^2}$; in the paraxial region $|z|\ll d$ it reduces to
$$
\boxed{I(z)\simeq4I_s\cos^2\left(\frac{\pi sz}{\lambda d}\right),\qquad\Delta z=\frac{\lambda d}{s}}.
$$
The supplied large-distance condition gives the <Fraunhofer diffraction> regime for the slit separation; the further replacement $\sin\theta\simeq z/d$ is a separate small-angle approximation. An unspecified finite slit width would add a single-slit envelope, and $I_s$ can then vary with angle. The formula above gives the interference factor in the ideal narrow-slit, equal-amplitude model; no absolute source intensity is supplied.
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