First, let’s take a look at how Young’s double slit experiment is to be carry out.

We’ve a monochromatic light source on the left, through the first slit and diffraction, the light waves propagates to the second slit, where the double slit is. Further diffraction on the double slit produces 2 coherent light wave, where the light waves will undergo interference with each other, where constructive interference produces light fringes on the screen (marked with n=0, 1, 2…) while destructive interference produces dark fringes.

The necessary condition for constructive interference to occur, is having the 2 coherent light source coincides in phase. When 2 light waves are in phase, their phase difference = nλ, where λ is the wavelength, n is a constant (0,1,2…)

From the figure above, the difference of light paths traveled by 2 light waves can be determined from the red triangle drawn, where it’s equal to asinθ.

So mathematically, asinθn = nλ.

Also from the triangle, when θ is sufficiently small and x is much lesser than D, we can approximate that sinθ = x/D.
Bringing the equations together, we have,
sinθn = x/D = nλ/a

Rearranging the equation, we have,
Distance between n-th bright fringe with 0th bright fringe, = nλD/a OR,
Successive distance between bright fringes, x = λD/a
where λ = wavelength of light source
D = distance between double slit and screen
a = distance between double slit

Similarly for dark fringe, destructive interference occurs when the 2 coherent light wave are in anti-phase, i.e. phase difference = (n+1/2)λ. Using similar technique, we get
Distance between n-th dark fringe with center = (n+1/2)λD/a

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