Diffraction and diffraction gratingsAQA A-Level Physics: Revision notes

Diffraction is the spreading of a wave as it passes through a gap or around an obstacle. It is most noticeable when the gap width is similar to the wavelength; a gap much wider than the wavelength gives little spreading.

Monochromatic light through a single narrow slit forms a pattern on a distant screen:

• a wide, bright central maximum
• much dimmer, narrower maxima on either side, separated by dark minima.

The central maximum is about twice as wide as the others. (You do not need to draw intensity–angle graphs.)

The width of the central maximum depends on the wavelength $\lambda$ and the slit width $a$ qualitatively as $\lambda/a$:

• a narrower slit gives a wider central maximum (more diffraction)
• a longer wavelength gives a wider central maximum
• a wider slit gives a narrower, brighter central maximum, until little diffraction is seen.

With white light the central maximum is white, because every wavelength has a maximum at the centre. The side maxima are coloured, with blue/violet nearer the centre and red furthest out, because red light has the longest wavelength and diffracts the most.

A diffraction grating is a plate with a very large number of equally spaced parallel slits. If it has $N$ lines per metre, the slit spacing is $d = 1/N$. For example 600 lines per mm is $6.0\times10^{5}$ lines m⁻¹, so $d = 1.67\times10^{-6}$ m.

Monochromatic light at normal incidence gives sharp bright maxima at angles $\theta$ on both sides of the straight-through direction ($n = 0$). Compared with a double slit the maxima are much narrower, brighter and further apart, so measurements are more precise.

Plane waves arrive in phase at every slit. Consider the waves from two adjacent slits, spacing $d$, travelling at angle $\theta$ to the straight-through direction. The wave from one slit travels an extra distance to the screen. In the right-angled triangle with hypotenuse $d$, the path difference is $d\sin\theta$.

A bright maximum needs the waves to be in phase, so the path difference is a whole number of wavelengths:

$d\sin\theta = n\lambda$

where $n = 0, 1, 2, \ldots$ is the order. Since $\sin\theta \le 1$, the highest order is $n \le d/\lambda$.

Worked example 1. 600 lines per mm, $\lambda = 589$ nm. For $n=1$: $\sin\theta = 589\times10^{-9}/1.67\times10^{-6} = 0.353$, so $\theta = 20.7°$. The highest order is $d/\lambda = 2.8$, so $n = 2$ at most.

Worked example 2. 300 lines per mm, second order at 22.3°: $\lambda = \dfrac{3.33\times10^{-6}\sin 22.3°}{2} = 6.3\times10^{-7}$ m.

White light: the $n=0$ maximum is white; each higher order is a spectrum with violet nearest the centre and red furthest away. Adjacent orders may overlap at large angles.

Applications: analysing line spectra to identify elements (including in stars), and X-ray diffraction by the regular layers in crystals. (Use of a spectrometer is not tested.)