Diffraction grating — condition for maxima: Worked examples

Intuition What this page is for

The parent note Diffraction grating — condition for maxima gave you the one rule that runs everything: $d sin θ_{m} = mλ .$ Here $θ_{m}$ just means "the angle $θ$ of the maximum whose order is $m$ " — we tack the little subscript $m$ on to remind us which bright spot we're pointing at (the $m = 1$ one, the $m = 2$ one, and so on). When there's no ambiguity we drop the subscript and write plain $θ$ .

Here we don't learn new physics — we hunt down every kind of question that rule can be bent into, and solve one of each. If you can do all of these, no grating problem can surprise you.

Before we start, let us re-earn every symbol so a reader arriving cold can follow:

$d$ = the centre-to-centre distance between two neighbouring slits, measured in metres. Picture two tiny gaps in a fence; $d$ is the gap-to-gap step.
$θ$ = the angle the outgoing ray makes with the straight-ahead direction (the "normal", a line perpendicular to the grating). Straight ahead is $θ = 0$ ; hard to the side is $θ = 9 0^{\circ}$ . With a subscript, $θ_{m}$ is the angle of the $m$ -th maximum.
$λ$ (lambda) = the wavelength, the length of one full crest-to-crest ripple of the light. See Wavelength and the visible spectrum.
$m$ = the order, a whole counting number $0, \pm 1, \pm 2, \dots$ . It counts how many whole wavelengths of extra path separate neighbouring rays.
$sin θ$ = "opposite over hypotenuse" of a right triangle. We use it (and not $tan$ or $cos$ ) because the extra path a ray travels is the side of a right triangle opposite the angle $θ$ , with $d$ as the hypotenuse — exactly what the parent note drew.

Intuition Signs, negative orders and the two answers of arcsine

Every worked example below solves for a positive $θ$ , but the physics is symmetric left–right, so we must be honest about the signs:

Negative orders. Because $m$ ranges over $0, \pm 1, \pm 2, \dots$ , each positive maximum has a mirror twin. If $m = + 1$ sits at $+ θ$ , then $m = - 1$ sits at $- θ$ — the same angle on the other side of the centre. That is why counting problems (Cell D) always give an odd total $2 m_{m a x} + 1$ .
The two answers of arcsine. When you write $θ = sin^{- 1} (x)$ , remember $sin$ takes the same value at $θ$ and at $18 0^{\circ} - θ$ . For a grating we only keep the branch $- 9 0^{\circ} \leq θ \leq + 9 0^{\circ}$ , because a ray leaving the grating must point into the forward half-space (it cannot travel backwards through the grating). So we always take the "principal" arcsine and never the obtuse partner.
The ceiling. $sin θ$ can never exceed $1$ nor drop below $- 1$ , so $- 9 0^{\circ} \leq θ \leq + 9 0^{\circ}$ is the entire physically allowed range. Any $m$ forcing $∣ mλ / d ∣ > 1$ has no real angle — that order simply does not exist.

Keep this in mind: whenever we report " $θ_{1} = 19. 3^{\circ}$ ", there is silently a matching $- 19. 3^{\circ}$ for $m = - 1$ .

The scenario matrix

Every grating question is one (or a blend) of these cells. Each worked example below is tagged with the cell it kills.

Cell	What makes it different	Example
A. Forward — find angle	Given $d, λ, m$ ; solve for $θ$	Ex 1
B. Backward — find $d$	Given $θ, λ, m$ ; solve for spacing / lines-per-mm	Ex 2
C. Backward — find $λ$	Given $θ, d, m$ ; solve for wavelength	Ex 3
D. Counting orders	How many maxima fit; uses $sin θ \leq 1$ and $\pm m$ symmetry	Ex 4
E. Degenerate: $m = 0$	Zero order — all colours overlap, no dispersion	Ex 5
F. Limiting: order at exactly $9 0^{\circ}$	$sin θ = 1$ boundary; the "just barely fits / just fails" case	Ex 6
G. Two wavelengths / dispersion	Angular separation of two colours in same order	Ex 7
H. Real-world word problem	Starlight spectrometer, must extract the numbers	Ex 8
I. Exam twist — overlapping orders	A colour's order $m$ lands at the same angle as another colour's order $m^{'}$	Ex 9

Cells covered: A B C D E F G H I — all of them.

Cell A — forward: find the angle

Worked example Example 1 (Cell A)

Green light, $λ = 550 nm$ , strikes a grating of 600 lines/mm. Where is the first-order ( $m = 1$ ) maximum?

Forecast: With $d$ only a few wavelengths wide, do you expect a tiny angle (a few degrees) or a big one (tens of degrees)? Guess, then read on.

Step 1 — get $d$ in metres. $d = \frac{1}{600 mm ^{- 1}} = \frac{1}{600} mm = 1.667 \times 1 0^{- 6} m .$ Why this step? The equation needs the physical spacing, not "lines per mm". We invert, and convert mm→m so $d$ and $λ$ share SI units (see the "lines vs spacing" mistake in the parent note).

Step 2 — rearrange the grating rule for $sin θ$ . $sin θ_{1} = \frac{mλ}{d} = \frac{1 \cdot 550 \times 1 0 ^{- 9}}{1.667 \times 1 0 ^{- 6}} = 0.330.$ Why this step? $θ$ is the unknown; isolating $sin θ$ is the only algebra needed. Look at the step figure below — $mλ / d$ is literally the "opposite ÷ hypotenuse" ratio.

Step 3 — undo the sine. $θ_{1} = sin^{- 1} (0.330) = 19. 3^{\circ} .$ Why this step? $sin^{- 1}$ (arcsine) answers "which angle has this sine?" — it undoes the sine to reveal the angle. We keep the principal value in $[- 9 0^{\circ}, + 9 0^{\circ}]$ ; the twin maximum $m = - 1$ sits at $- 19. 3^{\circ}$ .

Verify: Plug back: $d sin θ = 1.667 \times 1 0^{- 6} \times sin (19. 3^{\circ}) = 1.667 \times 1 0^{- 6} \times 0.330 = 5.5 \times 1 0^{- 7} m = 550 nm = 1 \cdot λ$ . ✓ Angle is large (tens of degrees) — the small-angle shortcut would have failed.

Cell B — backward: find the spacing

Worked example Example 2 (Cell B)

A red laser, $λ = 633 nm$ , produces its third-order maximum at $θ = 35. 0^{\circ}$ . Find the grating spacing $d$ and the number of lines per mm.

Forecast: Higher order at a modest angle means the slits are relatively... wide or narrow? Guess.

Step 1 — solve the rule for $d$ . $d = \frac{mλ}{s i n θ} = \frac{3 \cdot 633 \times 1 0 ^{- 9}}{s i n 3 5 ^{\circ}} .$ Why this step? Now $d$ is the unknown, so we divide both sides by $sin θ$ .

Step 2 — evaluate. $sin 3 5^{\circ} = 0.5736, d = \frac{1.899 \times 1 0 ^{- 6}}{0.5736} = 3.31 \times 1 0^{- 6} m .$ Why this step? We turn the symbolic answer into a number: compute $sin 3 5^{\circ}$ from the calculator, form the numerator $mλ = 3 \times 633 nm$ , then divide, so we can read off a physical spacing.

Step 3 — convert to lines per mm. $lines/mm = \frac{1}{d} = \frac{1}{3.31 \times 1 0 ^{- 3} mm} \approx 302 lines/mm .$ Why this step? Lines-per-mm is the reciprocal of the spacing expressed in mm.

Verify: $d sin θ = 3.31 \times 1 0^{- 6} \times 0.5736 = 1.90 \times 1 0^{- 6} m$ , and $mλ = 3 \times 633 \times 1 0^{- 9} = 1.90 \times 1 0^{- 6} m$ . ✓ Slits are fairly wide (≈300/mm), which is why order 3 sits at a modest 35°.

Cell C — backward: find the wavelength

Worked example Example 3 (Cell C)

An unknown spectral line appears at second order ( $m = 2$ ), $θ = 28. 0^{\circ}$ , on a grating of 400 lines/mm. What wavelength is it, and is it visible?

Forecast: Visible light runs ~400–700 nm. Will your answer land inside that band?

Step 1 — get $d$ . $d = \frac{1}{400 mm ^{- 1}} = 2.5 \times 1 0^{- 6} m .$ Why this step? The equation is written in terms of the physical spacing $d$ , but the data comes as "lines per mm"; we invert to convert that ruling density into a metre-scale gap so $d$ and $λ$ share SI units.

Step 2 — solve the rule for $λ$ . $λ = \frac{d s i n θ}{m} = \frac{2.5 \times 1 0 ^{- 6} \times s i n 2 8 ^{\circ}}{2} .$ Why this step? $λ$ is now the unknown; divide the path difference $d sin θ$ by the order $m$ .

Step 3 — evaluate. $sin 2 8^{\circ} = 0.4695, λ = \frac{2.5 \times 1 0 ^{- 6} \times 0.4695}{2} = 5.87 \times 1 0^{- 7} m = 587 nm .$

Verify: $587 nm$ sits in the yellow of the visible spectrum — it's the famous sodium line. Plug back: $d sin θ = 2.5 \times 1 0^{- 6} \times 0.4695 = 1.174 \times 1 0^{- 6}$ ; $mλ = 2 \times 5.87 \times 1 0^{- 7} = 1.174 \times 1 0^{- 6}$ . ✓

Cell D — counting orders

Worked example Example 4 (Cell D)

Blue light $λ = 450 nm$ on a grating of 800 lines/mm. How many bright maxima appear in total?

Forecast: Guess the total before computing. (Remember maxima appear on both sides!)

Step 1 — get $d$ . $d = \frac{1}{800 mm ^{- 1}} = 1.25 \times 1 0^{- 6} m .$ Why this step? The counting formula uses the physical spacing $d$ ; the data is a ruling density in lines/mm, so we invert (and read it in metres) to get the gap that actually enters $d / λ$ .

Step 2 — find the largest possible order. $\frac{d}{λ} = \frac{1.25 \times 1 0 ^{- 6}}{450 \times 1 0 ^{- 9}} = 2.78.$ Why this step? Setting $sin θ$ to its physical ceiling of $1$ gives $m_{m a x} = ⌊ d / λ ⌋$ ; any bigger $m$ demands $sin θ > 1$ , which no real angle can supply.

Step 3 — floor and count both sides. $m_{m a x} = ⌊ 2.78 ⌋ = 2.$ Orders present: $m = - 2, - 1, 0, + 1, + 2$ → 5 maxima. Why this step? The pattern is symmetric about the centre: every positive order $+ m$ has a negative twin $- m$ at the mirror angle, and $m = 0$ is the single central spot — giving $2 m_{m a x} + 1$ .

Verify: Check order 2 is real: $sin θ_{2} = 2 \times 450 \times 1 0^{- 9} /1.25 \times 1 0^{- 6} = 0.72 \leq 1$ . ✓ Check order 3 is impossible: $3 \times 450/1250 = 1.08 > 1$ → no solution. ✓ Count $= 2 \times 2 + 1 = 5$ . ✓

Cell E — the degenerate zero order

Worked example Example 5 (Cell E)

White light (all wavelengths 400–700 nm) hits any grating. Where is the zero-order ( $m = 0$ ) maximum, and what colour is it?

Forecast: Does the zero-order angle depend on colour? Guess yes or no.

Step 1 — put $m = 0$ into the rule. $d sin θ_{0} = 0 \cdot λ = 0 \Rightarrow sin θ_{0} = 0 \Rightarrow θ_{0} = 0^{\circ} .$ Why this step? With zero extra path difference, every wavelength interferes constructively straight ahead.

Step 2 — reason about colour. Because $θ_{0} = 0$ is independent of $λ$ , all colours land on the same spot and recombine. Why this step? No wavelength gets bent, so no dispersion happens at the centre.

What to look at in the figure: the red arrow straight ahead is the $m = 0$ ray — notice it is a single arrow carrying every colour at once, because all wavelengths pile onto $θ = 0$ . Now follow the black arrows fanning up and down: those are the $m = \pm 1$ rays, and here the colours split apart (blue nearer the centre, red further out) because for $m \geq 1$ the angle finally depends on $λ$ . The picture is the whole point of this cell: the centre is white; the fans are coloured.

Verify: The equation gives $θ_{0} = 0$ for $λ = 400$ nm and for $λ = 700$ nm alike, so the spread between them is $0^{\circ}$ — a single white central maximum. ✓ (Only for $m \geq 1$ do colours fan out into spectra.)

Cell F — the limiting order at exactly 90°

Worked example Example 6 (Cell F)

A grating has spacing $d = 1200 nm$ . For $λ = 600 nm$ , does the second-order maximum exist? Where does it appear, and what happens if you nudge $λ$ slightly larger?

Forecast: $d / λ = 2$ exactly — a razor-edge case. Will order 2 be visible, invisible, or grazing the edge?

Step 1 — compute $sin θ_{2}$ . $sin θ_{2} = \frac{mλ}{d} = \frac{2 \times 600}{1200} = 1.000.$ Why this step? We test the order against the ceiling $sin θ \leq 1$ .

Step 2 — interpret $sin θ = 1$ . $θ_{2} = sin^{- 1} (1) = 9 0^{\circ} .$ Why this step? $θ = 9 0^{\circ}$ means the ray skims along the grating surface — mathematically it exists but is infinitely spread and never actually reaches a screen in front. It is the exact boundary of visibility.

Step 3 — nudge the wavelength. For $λ = 610$ nm: $sin θ_{2} = 2 \times 610/1200 = 1.017 > 1$ . No real angle — order 2 vanishes. Why this step? This shows the boundary is a genuine cliff: just past $d / λ = m$ , that order disappears.

Verify: At $λ = 600$ nm, $m_{m a x} = ⌊ 1200/600 ⌋ = 2$ (edge counts). At $λ = 610$ nm, $⌊ 1200/610 ⌋ = ⌊ 1.97 ⌋ = 1$ — order 2 is gone. ✓

Cell G — dispersion: separating two colours

Worked example Example 7 (Cell G)

A grating of 500 lines/mm is lit by two lines: $λ_{1} = 486 nm$ (blue) and $λ_{2} = 656 nm$ (red), both from hydrogen. Find the angular separation of these two lines in the first order.

Forecast: Which colour bends more — the shorter blue or the longer red? Guess before computing.

Step 1 — get $d$ . $d = \frac{1}{500 mm ^{- 1}} = 2.0 \times 1 0^{- 6} m .$

Step 2 — angle of each colour ( $m = 1$ ). $sin θ_{1, blue} = \frac{486 \times 1 0 ^{- 9}}{2.0 \times 1 0 ^{- 6}} = 0.243 \Rightarrow θ = 14.0 6^{\circ} .$ $sin θ_{1, red} = \frac{656 \times 1 0 ^{- 9}}{2.0 \times 1 0 ^{- 6}} = 0.328 \Rightarrow θ = 19.1 4^{\circ} .$ Why this step? Same equation applied twice; the only change is $λ$ , so the longer wavelength must bend to a larger angle.

Step 3 — subtract. $Δ θ = 19.1 4^{\circ} - 14.0 6^{\circ} = 5.0 8^{\circ} .$ Why this step? "Angular separation" is literally the gap between the two directions, so we subtract the smaller angle from the larger to get how far apart the two coloured spots sit.

Verify: Red ( $656$ ) bends more than blue ( $486$ ) because $sin θ \propto λ$ — consistent with a grating fanning red furthest, opposite to a raindrop but correct here. Separation $\approx 5. 1^{\circ}$ is comfortably resolvable; this is the basis of Resolving power and spectrometers. ✓

Cell H — real-world word problem

Worked example Example 8 (Cell H)

An astronomer feeds starlight into a spectrometer whose grating has 1200 lines/mm. A calcium absorption line appears in the first order at a measured angle of $θ = 28. 4^{\circ}$ . What is the wavelength, and does it match the known Ca II K line at $\approx 393.4 nm$ ?

Forecast: 1200 lines/mm is a fine grating (small $d$ ). Will first order sit at a small or large angle for near-UV light? Guess.

Step 1 — extract the numbers. Let $N_{ℓ} = 1200 lines/mm$ be the ruling density (the "lines per mm" quoted for the grating), $m = 1$ , $θ = 28. 4^{\circ}$ . Why this step? Word problems hide the variables in prose — name them first. Note $N_{ℓ}$ is lines per length; the spacing $d$ is its reciprocal, exactly as in every example above.

Step 2 — get $d$ . $d = \frac{1}{N _{ℓ}} = \frac{1}{1200 mm ^{- 1}} = 8.333 \times 1 0^{- 7} m .$

Step 3 — solve for $λ$ . $λ = \frac{d s i n θ}{m} = \frac{8.333 \times 1 0 ^{- 7} \times s i n 28. 4 ^{\circ}}{1} .$ $sin 28. 4^{\circ} = 0.4756, λ = 8.333 \times 1 0^{- 7} \times 0.4756 = 3.96 \times 1 0^{- 7} m = 396 nm .$

Verify: $396 nm$ is in the violet/near-UV. It is closest to the Ca II H line at $\approx 396.8 nm$ , not the K line at $393.4 nm$ — a genuinely useful distinction, since the H and K lines are the two famous calcium absorption features and lie only $\approx 3 nm$ apart. Plug back: $d sin θ = 8.333 \times 1 0^{- 7} \times 0.4756 = 3.96 \times 1 0^{- 7} m = 1 \cdot λ$ . ✓

Cell I — exam twist: overlapping orders

Worked example Example 9 (Cell I)

On a grating, red light $λ_{R} = 700 nm$ in its second order happens to overlap (same angle) with a shorter wavelength $λ_{x}$ in its third order. What is $λ_{x}$ , and what colour is it?

Forecast: For the same angle, does the higher order need a longer or shorter wavelength? Guess.

Step 1 — write the overlap condition. Same $d$ , same $θ$ , so the value of $d sin θ$ is identical for both rays. Writing the grating rule once per ray: $d sin θ = m_{R} λ_{R} and d sin θ = m_{x} λ_{x},$ so the left-hand sides match and $m_{R} λ_{R} = m_{x} λ_{x} \Rightarrow 2 \times 700 = 3 \times λ_{x} .$ Why this step? "Same angle" means $d sin θ$ is common to both — set the two $mλ$ products equal. The $d$ and $sin θ$ cancel, so the grating spacing never even enters!

Step 2 — solve. $λ_{x} = \frac{2 \times 700}{3} = 466.7 nm .$ Why this step? Straight algebra: divide the known product $m_{R} λ_{R}$ by the other order $m_{x} = 3$ .

Step 3 — identify. $466.7 nm$ is blue in the visible band. Why this step? We convert the bare number into a physical colour so the answer is meaningful, not just arithmetic.

Verify: Check the equal path difference — pick any grating, say $d = 2000$ nm. Red order 2: $sin θ = 2 \times 700/2000 = 0.700$ . Blue order 3: $sin θ = 3 \times 466.7/2000 = 0.700$ . Identical angle ✓. Higher order needs the shorter wavelength — as forecast-checked.

Recall Self-test the whole matrix

Which cell would each of these be? "Find lines/mm from a measured angle" ::: Cell B (backward for $d$ ). "How many spots on the screen?" ::: Cell D (counting orders, remember $\pm m$ ). "Does order 4 exist?" ::: Cell F (limiting / boundary at $sin θ \leq 1$ ). "Two spectral lines' angular gap" ::: Cell G (dispersion). "Colour x order 3 overlaps red order 2" ::: Cell I (overlapping orders).

Connections

Diffraction grating — condition for maxima — the parent rule every example above bends.
Interference and path difference — why $d sin θ = mλ$ means "crests line up".
Young's double-slit experiment — the $N = 2$ ancestor of the same condition.
Resolving power and spectrometers — Example 7's dispersion is how spectrometers separate lines.
Wavelength and the visible spectrum — used to judge whether answers land in visible light.