This is an alternative or additional method. The pattern obtained is easier to see, since the bright fringes (maxima) are well-defined ‘spots’. however, the mathematics involved is a little more involved, and students must use trigonometry to find the diffraction angle \(\theta\).

In the film, Christina sets up the apparatus vertically, which both requires less space in the lab and reduces concerns over laser safety. However, you will still need to be cautious for stray reflections.

The relationship here is

\(n\lambda = d \sin\theta\)

where:

- \(n\) = the order of the maximum, with \(n=0\) as the central maximum, \(n=1\) for the ‘first order’ to either side, and so on.
- \(\lambda\) = wavelength of light
- \(d\) = slit spacing. Usually a diffraction grating slide states the number of lines per mm. For example: 300 lines/mm implies 300,000 slits/m, so \(d = 1/300,000~\mathrm{m}\).
- \(\theta\) = angle between the straight-through direction (helpfully marked by the zero-order maximum) and the maximum being investigated. This must be found using trigonometry:

As stated in the film (at 3’15”), we cannot use the small angle approximation here, since the angles are too large. Hence:

\(\theta = \tan^{-1}(\frac{x}{D})\)

Using the measurements in the film, Alom’s miraculous ‘off-the-cuff’ calculation went like this:

- Total distance = 191 mm, so \(x\) = 191/2.
- \(D\) = 225 mm
- So \(\theta = \tan^{-1}((191/2)/225) = 22.9985\) ≈ 23.0°

Now, to calculate the wavelength:

\(n\lambda = d\sin\theta\) or \(\lambda = \frac{d\sin\theta}{n}\)

Since we measured two fringes either side of the central maximum (for accuracy):

- \(n = 1\)
- \(d = \frac{1}{300,000}~\mathrm{m}\)
- \(\lambda = (\frac{1}{300,000}\sin 22.9985)/2 = 6.5118 \times 10^{-7} = 651~\mathrm{nm}\) (to 3 sf).

Again, students can compare this with the manufacturer’s stated value; the students should be able to assess their uncertainty to check that the manufacturer’s value falls within their uncertainty range. The same discussions as above can be had about how to minimise uncertainties, and what would happen if a different colour laser were to be used.