Two-slit interference

Key ideas

• When light (or sound) passes through two narrow slits, each slit acts as a source, and the two sets of waves superpose on the far side.
• What each point on a screen receives depends on the path difference — how much further the wave from one slit travelled than from the other:
  • path difference $= n\lambda$ → waves arrive in phase → constructive → a bright fringe (or loud point)
  • path difference $= (n + \tfrac{1}{2})\lambda$ → arrive out of phase → destructive → a dark fringe (quiet point)
• The bright fringes ("orders") sit at angles given by:

$d\sin\theta = n\lambda$

• Each variable, with sub-bullets:
  • $d$ — the slit separation (m)
  • $\theta$ — the angle from the straight-through direction to the $n$th bright fringe
  • $n$ — the order: $0$ (centre), $1, 2, \dots$
  • $\lambda$ — the wavelength (m)
• On a screen a distance $L$ away, with fringe spacing $x$, the small-angle version applies:

$n\lambda = \frac{dx}{L}$

• valid because $L$ is metres while $d$ is fractions of a millimetre, so $\sin\theta \approx \tan\theta = \dfrac{x}{L}$

Tips

• Say what the path difference is, not just the formula. Merit/Excellence explanations need "the wave from the lower slit travels one whole wavelength further, so crests still meet crests" — not just $d\sin\theta = n\lambda$.
• Moving the screen further away ($L\uparrow$) or using longer wavelength ($\lambda\uparrow$) spreads the fringes; wider slit spacing ($d\uparrow$) squeezes them. Check your answer moves the right way.