Why do the distances between the fringes depend on the wavelength? In the double-slit experiement, with the monochromatic light, the longer the wavelength, the more the distances between the fringes. Can somebody explain why?
 A: Prerequisites :- You need to know that the higher(or lower, as both the directions are symmetric) you go up the screen, you start accumulating more and more path difference(or equivalently, phase difference) between the two sources. You can prove this by using,
$$\Delta x = d \sin(\theta) \qquad \text{or} \qquad k\Delta x = \Delta \phi =kd \sin(\theta)= \frac{2\pi d \sin(\theta)}{\lambda}$$
,where $\Delta x$ is the path difference, $\Delta \phi$ is the phase difference, $k$($=\frac{2\pi}{\lambda}$) is the angular wave number, $\lambda$ is the wavelength. As you can see, that if we increase the height the path difference increases.

Answer :- Let's say that for a certain wavelength($\lambda_1$), you need to go at height $h_1$ up the screen to achieve a phase difference of $\Delta \phi$(corresponding path difference $\Delta x_1=\frac{\lambda_1\Delta\phi}{2\pi}$), and for another wavelength($\lambda_2$), you need to go up till height $h_2$ to achieve the same phase difference of $\Delta \phi$(corresponding path difference $\Delta x_2 =\frac{\lambda_2\Delta\phi}{2\pi}$). Now if $\lambda_2> \lambda_1$, that means $\Delta x_2>\Delta x_1$. So that means you need to go higher up on the screen in the 2nd case, thus $h_2>h_1$. So for longer wavelengths, you need to achieve larger path differences, for which you need to span a larger height on the screen. Thus for larger wavelengths, everything gets stretched out and so do the fringe widths.
Image credits :- YDSE - coherence of light from slits and width of slits
A: Intuitively speaking, the reason behind this is that for creating the same phase difference for light with larger wavelength, the path difference needs to be more than that for light with a shorter wavelength, so the fringe width would be more for a larger wavelength.
Speaking in terms of formulas, the path difference can be approximated to be $\frac{dy}{D}$, where $d$ is the spacing between the slits and $D$ is the distance between slits and the screen. 
To create a phase difference of $2\pi$ radians, the path difference needs to be $\lambda$. From this, the fringe width can be calculated to be $\frac{\lambda D}{d}$. 
So fringe width is directly proportional to wavelenght.
