Diffraction Grating Central Maximum Width

This is from AQA A-level Physics 2019 Paper 1, Question 3.4, I don't understand the result from the mark-scheme and examiner report.

The context is a white light source incident on a single slit, then passes through a filter to make it a single wavelength of light (650 nm, red) before being diffracted by a diffraction grating.

The filter is then revealed to allow a range of light (600nm to 700 nm, still considered red light) to pass through, and it's asked how this impacts the appearance of the maxima.

There are 4 markscheme points:
- Central maximum unchanged in width
- Broader maxima/range of angles for each maximum/order
- Gradually getting broader/more spread out for greater order maxima
- Part of third order maximum suppressed at long wavelengths (for  >
90°)

Specifically confused about the first bit, where it says the central maximum's width is unchanged. From what I can tell, a maxima's width is considered to be the distance between the 2 adjacent minima. The position of maxima and minima are predictable via the equation dsin(θ) = nλ for principle maxima or dsin(θ) = (n+0.5)λ for principle minima (I think). Surely the position of the minima are changed such that the central maximum's width changes?

The Examiner report says "Many students realised that the range of wavelengths would cause the point maxima
associated with a diffraction grating to increase. What was less well understood was that this would not occur with the central maximum, as all wavelengths have a maximum at an angle of zero to the normal of the grating. There were several marking points for students to access, but the award of both marks was rarely seen"

Reply 1

Ferret!

This question had me very confused when I did it. My explanation may well be wrong but here goes!

When the light leaves the diffraction grating, regardless of if the light has a range of wavelengths, the path difference will be the same for when this light reaches the centre of the screen. I think then that the waves will superpose like before at the centre. However, phase difference is now not constant at all with the range of wavelengths/frequencies, so the diffraction pattern will be more dispersed at the subsidiary maxima. This dispersion will decrease the darker fringes size.

Reply 2

Ethanity007

Central max: all wavelength have zero path diff. Thus regardless of which wavelengths, they will meet constructively at the centre.

However, at higher orders maximas, diff wavelengths will meet constructively at slightly diff positions which explains the spread at higher order maxima