Interference and Young's double-slit experimentAQA A-Level Physics: Revision notes

Interference occurs when waves from two sources overlap and superpose. The result at a point depends on the path difference, the difference in the distances from the point to the two sources.

For two sources emitting in phase:

• constructive interference (bright fringe, loud sound) when the path difference is a whole number of wavelengths: $n\lambda$
• destructive interference (dark fringe, quiet) when the path difference is an odd number of half wavelengths: $\left(n+\tfrac{1}{2}\right)\lambda$

Worked example. At a point 4.00 m and 4.75 m from two in-phase speakers, with $\lambda = 0.50$ m, the path difference is 0.75 m = 1.5 wavelengths, so there is destructive interference: a quiet spot.

Two sources are coherent if they emit waves with the same frequency (wavelength) and a constant phase difference. Only coherent sources give a stable interference pattern: two separate lamps, or two separate signal generators, do not, because the phase difference changes randomly.

In Young's experiment, coherence is obtained by using one source to illuminate both slits. A laser provides intense, monochromatic (single wavelength), coherent light, so clear fringes form.

Laser safety (how a laser works is not needed): never look along the beam, because the eye lens focuses it onto the retina and can cause permanent damage. Use a low-power laser, do not point it at people, remove reflective objects, end the beam at a screen and display a warning sign.

Light passes through two narrow slits a distance $s$ apart. Each slit diffracts the light and the overlapping waves interfere, giving equally spaced bright and dark fringes on a screen a distance $D$ away. Each bright fringe is where the path difference is $n\lambda$.

For $D \gg s$ the fringe spacing (distance between adjacent bright fringes) is

$w = \frac{\lambda D}{s}$

So $w$ increases with larger wavelength, larger $D$ or smaller $s$.

Worked example. $\lambda = 650$ nm, $s = 0.40$ mm, $D = 2.5$ m: $w = \dfrac{650\times10^{-9}\times2.5}{0.40\times10^{-3}} = 4.1\times10^{-3}$ m = 4.1 mm. Halving $D$ halves $w$ to 2.0 mm.

With white light (a source plus a narrow slit in front of the double slit), $w \propto \lambda$ so each colour has its own fringe spacing. The central fringe is white (zero path difference for all wavelengths) and the other fringes are short spectra, violet nearest the centre and red furthest, which overlap at larger distances.

Sound: two loudspeakers fed by the same signal generator give loud and quiet positions where the path difference is $n\lambda$ or $(n+\tfrac12)\lambda$.

Microwaves: two coherent transmitters, or one transmitter and a pair of slits in a metal sheet, give maxima and minima found with a detector. The pattern is much larger than for light because microwaves have much longer wavelengths.

Newton favoured a particle model of light. Young's experiment (1801) gave strong evidence for a wave model. In the 1860s Maxwell showed that light is an electromagnetic wave, and Hertz's radio waves (1887) supported this. In the early 1900s the photoelectric effect needed light to arrive as photons, so light shows both wave and particle behaviour. Models change when new evidence appears.

Required practical 2: investigate interference using Young's slits and a diffraction grating. For double slits, measure $D$ with a metre rule or tape, measure across several fringes (for example ten) and divide to get $w$ with a smaller percentage uncertainty, and use $\lambda = ws/D$. Use a laser safely. (The grating is covered in the next subtopic.)