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?
[Physics] the difference between the single and double slit interference pattern of white light
diffractiondouble-slit-experimentinterferenceoptics
Related Solutions
I boil everything down to Fourier transforms and I'll refer to the useful Wikipedia article several times. I won't care about pre-factors in the Fourier transforms for the sake of simplicity. You can look them up in the referred transforms.
First, consider the single slit with width a, it has the function rect(ax) and the Fourier transform sinc(x/a) (No. 201). Square it (Intensity) and you get the diffraction pattern.
Second, the double slit. It is basically a convolution (I denote it by ** and multiplication by *) of two delta functions δ(x-d/2)+δ(x+d/2) for the positions of the slits spaced by d and the box function rect(ax) giving the single slit(s): f(x)=(δ(x-d/2)+δ(x+d/2))**rect(ax).
Now the FT of the two delta functions is either directly the reverse of No. 304 or you can compute it as the sum of two shifted delta functions (shift theorem No. 102 and delta function No. 302). So FT[(δ(x-d/2)+δ(x+d/2))]=cos(dx).
By the convolution theorem (No 108) we get that FT[g(x)**h(x)]=FT[g(x)]*FT[h(x)] and thus the resulting FT[f(x)]=cos(dx)*sinc(x/a). Since the pattern displays the intensity which is the square of the field we have to square everything and get the resulting interference pattern.
Now, computing the minima is just maths and property of the sinc² function resp. the cos². It turns out that the sinc²function has minima wheresin² has minima (i.e. where sin is zero, with π periodicity) except for zero because of the definition of sinc that is 1 there. This explains the double width of the central maximum.
The `cos², however, has equally spaced minima (π periodicity again)
If you have infinitely small slits, you just get a uniform fourier transform and thus the double slit would be just the cos².
It is not refraction which you need to consider it is the change in the length of the optical path.
If you have a ray of light incident on a parallel sided piece of glass all that happens is that the ray suffers a lateral shift but the ray would still be travelling in the same direction as shown in the diagram below.
Since the wavelength of light is less in the glass for a given thickness of glass would have more wavelength in it than the same thickness of air.
Best Answer
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.