# What is the difference between the single and double slit interference pattern of white light?

I believe for a single slit, the central maximum appears white while the other orders of maxima create a spectra in the order of the wavelengths of the components of white light.

Is this the same for a double slit? Why/ why not?

I disagree with the preceding answers

First consider the single slit of width $d$. Each wave length comprised in the white light gives an angular intensity distribution proportional to $sinc^2 (d sin(\theta) / \lambda)$, where $sinc x =sin \pi x / \pi x$. All wavelengths contribute to $\theta=0$ so the center of the pattern is white. As you see this distribution is wider for larger lambda. Going outward the distribution will be zero for blue light first. At this point red light will dominate. Further out, the blue minimum occurs and red light dominates, etc.

For two slit separated by $D$ you have to multiply this distribution with $cos^2 (\pi D sin \theta / \lambda$ ). Similar conclusions hold.

If you're comfortable with Fourier transforms, a concise explanation is that the diffraction pattern is the Fourier transform of the aperture. If your aperture is a vertical slit (i.e. a rect function in one dimension) then you get the FT of that, a sinc function. If your aperture is two slits, then at the image plane you have two sinc functions with different phases, causing interference in the direction of their separation. You end up with a sinc envelope but have interference all along it.

Bear in mind this is for fields, so take the absolute value squared for intensity.

Similar behaviour happens in the case of the double slit experiment. Since the path difference between the light passing through each slit is defined as:

$$\text{Path Difference} = \frac{xd}{D}$$

Where $x$ is distance from center of screen, $D$ is distance to screen and $d$ is the separation of the slits, we can see that for the central maxima, the path difference is 0 so none of the wavelengths are cancelled by partial interference.

For the other bands, there will be a a gradient of colors in the order of the wavelengths of the white light as at each point in the band there will be certain wavelengths with destructive interference that are not visible.

So similar to the single slit, the central maxima will be white because the light passes from each slit with no path difference but as we go to the outer bands, the partial interference from the path difference cancels the effect of some wavelengths, creating a gradient of colors in each band.

Both of these scenarios deal with the diffraction of light waves. When a light wave passes through a slit that is smaller than the wavelength (don't ask why it is wavelength), then the wave will spread out similar to the ripples in a pond caused by a disturbance.

In the single slit this creates a single bright band since the magnitude of the wave decreases as it moves farther from the slit

However, in a double slit experiment, light passes through both slits and the resulting waves interfere with one another causing light and dark bands to appear.

For white light fringes of color will appear around the band(s) due to the varying indices of refraction of different frequencies (see thin films).

I hope this was hlpful