Path difference, measured in wavelengths

$$\frac{\Delta L}{\lambda} = \begin{cases} \text{integer} & \text{constructive} \\ \text{half-integer} & \text{destructive}\end{cases}$$

Noticing that one path is longer and calling the result destructive because the paths differ at all is the error, and so is calling a small difference constructive because it is small. Equal paths are just the case where the ratio is zero, which is one constructive case among infinitely many.

For two slits separated by $d$, the geometry converts an angle into a path difference:

$$d\sin\theta = m\lambda \quad \text{(maxima)}, \qquad m = 0, \pm 1, \pm 2, \ldots$$

Note which way round this is: for a double slit with separation $d$ the equation gives maxima, while the identical-looking single-slit equation with width $a$ gives minima. Name what you are locating before you use the number.

The familiar linear expression

$$y = \frac{m\lambda L}{d}$$

comes from replacing $\sin\theta$ and $\tan\theta$ with $\theta$, which holds only for angles under roughly ten degrees. Applied to high orders, or to a grating with closely spaced lines, it drifts badly or lands the fringe somewhere the setup could never produce.

So the safe order of operations is:

1. Start from $d\sin\theta = m\lambda$, which always holds.
2. Solve for $\theta$.
3. If $\theta$ is under about ten degrees, $y = m\lambda L/d$ is fine.
4. Otherwise get the screen position from $y = L\tan\theta$.

A grating with a few thousand lines per centimetre puts even the first order well outside the small-angle range, so for gratings expect large angles and go through $\theta$ every time.

Two bookkeeping errors produce plausible-looking wrong numbers rather than obvious ones.

The order starts at zero. The central maximum is $m = 0$, the first bright fringe on either side is $m = 1$, and the third bright fringe out from the centre is $m = 3$. Counting the central fringe as number one shifts every order, and since $y \propto m$, every position afterwards is off by a fixed ratio. Label the pattern $0, 1, 2, 3$ outward on each side before computing anything, and check against the fact that adjacent maxima are spaced by $\lambda L/d$.

$d$ and $L$ are different lengths by orders of magnitude.

1. $d$ is the centre-to-centre separation of the two slits, typically micrometres.
2. $L$ is the slit-to-screen distance, typically a metre or two.
3. The width of each individual slit is a third quantity, and it governs the diffraction envelope rather than the fringe spacing.

Swapping them gives a fringe spacing in kilometres or in nanometres, so size-check the answer: on a classroom screen the fringes land in millimetres or centimetres.

A grating obeys the same condition, $d\sin\theta = m\lambda$. So for equal $d$, the maxima appear at exactly the same angles as a double slit's. Adding slits does not move them.

What changes is sharpness: with thousands of slits, the light interferes destructively at every angle except the ones that satisfy the condition, so the maxima become very narrow and bright. That is why gratings measure wavelength so well. Expecting more slits to smear the pattern out has it backwards.

What genuinely spreads a real grating's orders far apart is its very small $d$: a few thousand lines per centimetre makes $\lambda/d$ large, which pushes the angles out.

And the colour order is the opposite of a prism's. Solving $\sin\theta = m\lambda/d$ shows $\theta$ rising with $\lambda$, so red sits at the outer edge of each order and violet nearest the centre. A prism bends violet most because $n$ is larger for violet, which is a different mechanism with a different result. One more grating fact worth having: the central maximum in white light stays white, since every wavelength has zero path difference there.