Trap: counting orders, fringes and gaps
Between the 1st and 5th bright fringe there are 4 gaps, not 5, so x = (separation)/4. A grating showing orders up to n_max gives 2n_max + 1 maxima in total. The nth bright fringe corresponds to path difference nλ, but the nth dark fringe corresponds to path difference (n − ½)λ.
x = (distance between mth and pth bright fringe)/(p − m) · total grating maxima = 2n_max + 1What this actually means
Count gaps, not fringes. This single habit fixes most double-slit arithmetic errors. From the first to the fifth fringe is four gaps, so a 2.5 mm span gives x = 0.625 mm.
For the number of fringes fitting into a region of width W, use W/x and then round down, because a partial fringe does not count.
For gratings the central maximum is shared, so do not double it. Orders 0, ±1, ±2, ±3 give seven maxima, not eight.
The dark-fringe indexing catches people out. The first dark fringe sits at 0.5λ, the second at 1.5λ, the third at 2.5λ. So the nth minimum is at (n − ½)λ, which is the same set as (n + ½)λ starting from n = 0.
When a question asks how many maxima are detected as a probe moves from one point to another, work out the path difference at each end in wavelengths and count the whole numbers strictly between them.
Dividing by the number of fringes instead of the number of gaps, or double-counting the central maximum on a grating.
Prove it — watch it be true
- Mark the first and fifth bright fringes on the screen intensity profile.
- Confirm the measured span divided by four, not five, matches the λD/a readout.
- Drag the path-difference probe from the centre outwards and count how many whole-number wavelength crossings occur before a chosen point.