Diffraction and diffraction gratingsEdexcel A-Level Physics: Revision notes
Section 1
Diffraction
Diffraction is the spreading of waves as they pass through a gap or round an obstacle. It is most pronounced when the width of the gap is about the same size as the wavelength.
- Gap much larger than λ: waves pass through with very little spreading, leaving a clear shadow
- Gap about λ: strong spreading into the geometric shadow, wavefronts become almost circular
For a fixed gap, a longer wavelength diffracts more. In a ripple tank, raising the frequency at constant speed shortens λ and so reduces diffraction.
Diffraction does not change the wavelength, frequency or speed of the waves. Only the direction of travel spreads out.
Section 2
Huygens' construction
Huygens' construction explains wave propagation. Every point on a wavefront acts as a source of secondary wavelets, which spread out in all forward directions at the wave speed. The new wavefront is the envelope of the wavelets.
At a gap, only the wavelets from points in the gap contribute. Near the edges nothing is added from outside, so the wavefront curves round into the geometric shadow. This is diffraction.
Section 3
The diffraction grating
A diffraction grating is a plate with a very large number of equally spaced parallel slits, typically hundreds of lines per millimetre. The slit spacing is d = 1 ÷ (lines per metre). For 300 lines per mm, d = 1 ÷ 300 000 = 3.33 × 10⁻⁶ m.
Light diffracted by each slit overlaps and interferes. In certain directions the waves from adjacent slits arrive in phase and give sharp, bright maxima. Between them there is destructive interference. Many slits make the maxima narrow and bright.
Section 4
The grating equation nλ = d sinθ
A maximum is formed at angle θ to the normal where the path difference between adjacent slits is a whole number of wavelengths:
nλ = d sinθ
where n = 0, 1, 2, ... is the order. The n = 0 maximum is straight ahead.
The highest order occurs as sinθ approaches 1, so n is the largest whole number not exceeding d/λ.
Worked example: d = 1.67 × 10⁻⁶ m and λ = 633 nm gives sinθ = 0.380 for n = 1, so θ = 22.3°. Since d/λ = 2.6, the highest order is n = 2.
Check that sinθ is less than 1. If a calculation gives a value above 1, that order does not exist.
Section 5
White light and the spectrum
With white light the n = 0 maximum is white, because the path difference is zero for every wavelength. For n ≥ 1, sinθ = nλ/d depends on λ, so each order is a spectrum with violet nearest the centre and red furthest out.
Higher-order spectra are wider and can overlap: the third-order violet (about 1200 nm path difference) lies beyond the start of the second-order red at about 1400 nm.
If the grating has more lines per mm (smaller d), the maxima are spread further apart.
Section 6
Core practical: wavelength of light with a grating
- Fix the grating perpendicular to a laser beam, a known distance D from the screen.
- Measure the distance between maxima of the same order on both sides of the centre, then halve it to get x.
- Calculate θ = tan⁻¹(x/D).
- Use λ = d sinθ ÷ n.
- Repeat for several orders and plot sinθ against n. The gradient is λ/d.
Measuring across both sides avoids having to find the centre exactly and gives a smaller percentage uncertainty.
Safety: never look into a laser beam or its reflections, and keep the beam below head height.
Do not use the screen distance as the grating-to-maximum distance in sinθ. Use θ = tan⁻¹(x/D) first, then take the sine.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Diffraction and diffraction gratings
- In a ripple tank, a vibrating dipper produces plane water waves of wavelength 1.5 cm that travel towards two barriers with an adjustable gap between them. The speed of the waves in the tank is constant, and the frequency of the dipper can be changed.Use Huygens' construction to explain why the waves spread into the region behind the edges of the barriers.2 marks
- A student shines a red laser of wavelength 633 nm at normal incidence onto a diffraction grating with 600 lines per millimetre. A series of bright spots is seen on a distant screen, either side of a central bright spot.Calculate the angle between the central maximum and the first-order maximum.2 marks
- In a core practical, a student determines the wavelength of light from a laser using a diffraction grating with 300 lines per millimetre. The grating is held perpendicular to the beam and the screen is 1.00 m from it. The student measures the distance between the two second-order maxima, one on each side of the central maximum, as 0.850 m.Calculate the wavelength of the laser light.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).