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?

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

1 Answer 1


As already pointed out, the interference takes place between the amplitudes, and intensity is proportional to the square of the amplitude.

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.

When we think in terms of energy in the double-slit, you expect 2x times the intensity at the bright fringes (because there are two openings/"sources"), but there is also another 2x factor because energy that is canceled at the dark fringes has to be displaced to the bright fringes by conservation of energy.

In total, that makes 4x times the intensity at the bright spots.


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