# 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 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?

• intensity is amplitude squared. Interference is amplitudes (unless it's the HBT).
– JEB
Sep 18, 2018 at 2:14

I think another 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.