What is the intuitive reasoning behind a factor of 4 in the Fraunhofer double-slit intensity equation?

For Fraunhofer diffraction, the intensity of a single-slit scenario is described by the following equation:

$I(\theta) = I_{o}\left [\frac{\sin(\frac{\pi a}{\lambda}\sin(\theta))}{\frac{\pi a}{\lambda}\sin(\theta)} \right ]^2$.

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?

• intensity is amplitude squared. Interference is amplitudes (unless it's the HBT). – JEB Sep 18 '18 at 2:14
• Then why isn't I_o squared? – Hale Bays Sep 18 '18 at 3:11

I think anther way to think about it is through conservation of energy. Units of intensity are $\text{W}/\text{m}^{2} = \text{J}/\text{s}\cdot\text{m}^{2}$, so intensity is the energy per area per second of the light.