For Fraunhofer diffraction, the intensity of a single-slit setup obeys
$I(\theta) = I_{o}\left [\frac{\sin(\frac{\pi a}{\lambda}\sin\theta)}{\frac{\pi a}{\lambda}\sin\theta} \right ]^2$.
where $a$ is the slit-width, $\lambda$ is the wavelength of light from the source, and $I_{o}$ is the intensity of the source. Whereas for a double-slit,
$I(\theta) = 4I_{o}\cos^2(\frac{\pi d}{\lambda}\sin\theta)\left [\frac{\sin(\frac{\pi a}{\lambda}\sin\theta)}{\frac{\pi a}{\lambda}\sin\theta} \right ]^2$
with $d$ being the slit-separation distance. My question is where does the factor of 4 come from? This seems to imply that the central maximum from the same source of light in the case of a double-slit would have it's central maxima 4 times higher. But from constructive interference I would only expect a factor of 2 from the superposition principle. Is there an intuitive reason behind this?