Double-Slit Interference

The angles of the maxima follow from $d\cdot{}\sin{}(\alpha{}_{k}) = k\cdot{}\lambda{}$: $\sin{}(\alpha{}_{k}) =\frac{ k\cdot{}\lambda{}}{d}$ with order $k = 0$, 1, 2, ... The zeroth maximum always lies straight ahead in the centre. Example: $d = 0.1 mm$ and $\lambda{} = 600 nm$ give for $k = 1$ the value $\sin{} \alpha{} =\frac{ 6\times{}10^{-7}}{10^{-4}} = 0.006$, so $\alpha{} \approx{} 0.34$°. The position on a screen at distance L follows from geometry: $x = L\cdot{}\tan{} \alpha{}$, for small angles $x \approx{}\frac{ k\cdot{}\lambda{}\cdot{}L}{d}$, here 12 mm. The maxima are then equally spaced. Convert all lengths consistently to metres; mixing nm, µm and mm is the most common source of error.

The path difference Δs is the difference in distance the waves travel from the two slits to a point on the screen; for distant points $\Delta{}s = d\cdot{}\sin{} \alpha{}$. It determines the relative timing of the two waves there. If Δs is a whole multiple of the wavelength (0, λ, 2λ, ...), crest meets crest: constructive interference, a bright maximum. If Δs is a half-integer multiple (λ/2, $\frac{3\lambda{}}{2}$, ...), crest meets trough and the waves cancel: a dark minimum. Everything in between produces intermediate brightness. That light plus light yields darkness at the dark places is the decisive wave argument; particle trajectories alone could never explain it.

You measure lengths and convert back to nanometres; that is the core of the Young experiment. Illuminate the double slit with the light to be measured, ideally a laser, and measure three quantities: the slit separation d, the screen distance L and the distance x of the first maximum from the centre. For small angles $\lambda{} =\frac{ d\cdot{}x}{(k\cdot{}L)}$. Example: $d = 0.2 mm$, $L = 2 m$, first maximum at $x = 6.4 mm$: $\lambda{} =\frac{ 2\times{}10^{-4} \times{} 6.4\times{}10^{-3}}{2} = 6.4\times{}10^{-7} m = 640 nm$, red light. It becomes more accurate if you average over several orders or measure the distance across many maxima and divide.

From $\sin{} \alpha{} =\frac{ k\cdot{}\lambda{}}{d}$ both trends can be read off directly. A larger wavelength spreads the pattern: red light (about 650 nm) produces maxima farther apart than blue (about 450 nm); with white light this creates coloured fringes, blue on the inside, red on the outside. A smaller slit separation also spreads the pattern, so a finer double slit separates better. Conversely, for large d the maxima crowd together until they merge, which is why you never observe interference from two windows. In addition $\sin{} \alpha{} \le{} 1$ limits the number of orders to $k \le{}\frac{ d}{\lambda{}}$: for $d = 4 \mu{}m$ and $\lambda{} = 500 nm$ only orders up to $k = 8$ exist.

Yes, and it is one of the most important experiments of quantum physics. Sending electrons through a double slit produces the same fringe pattern on the detector as light, described by the same formula $d\cdot{}\sin{} \alpha{} = k\cdot{}\lambda{}$, only with the de Broglie wavelength $\lambda{} =\frac{ h}{p}$ of the electron. The astonishing part: the pattern builds up even when the electrons fly one at a time; each electron in a sense interferes with itself. If, however, you measure which slit the electron passes, the pattern disappears. The experiment thus demonstrates wave-particle duality and the role of measurement, and in final exams it is the standard bridge from optics to quantum physics.