Wave Optics: Young's Double Slit Experiment

Why fringes form

Light from two narrow, closely-spaced slits (separation $d$) spreads out and overlaps on a screen a distance $D$ away. At a point on the screen a height $y$ from the centre, the two waves have traveled slightly different distances, a path difference that, for $D \gg d$, works out to:

Bright and dark fringes

Where the path difference is a whole number of wavelengths, the two waves arrive in phase and reinforce: a bright fringe. Where it's a half-integer number of wavelengths, they arrive out of phase and cancel: a dark fringe:

Fringe width

Setting $\Delta = n\lambda$ and solving for $y$ gives the position of the $n$th bright fringe, $y_n = \tfrac{n\lambda D}{d}$. Consecutive bright fringes are therefore evenly spaced, a distance apart called the fringe width:

Worked example

Two slits are 0.5 mm apart, the screen is 1 m away, and the light has a wavelength of 500 nm. Find the fringe width and the position of the 3rd bright fringe.

Solution: Convert to metres first: $d = 0.5\times10^{-3}\text{ m}$, $\lambda = 500\times10^{-9}\text{ m}$, $D=1\text{ m}$:

The $n$th bright fringe sits at $n$ fringe-widths from the centre, so the 3rd bright fringe is at: