ARCHETYPE: Diffraction, Resolution & Grating Numericals

The last gym drilled interference from two slits. This one opens the aperture to every diffraction numerical a wave-optics paper can throw at you — and there are three distinct families, forever being mistaken for one another. There is the single slit, whose finite width spreads light into a broad central band with dark minima at $a\sin\theta = m\lambda$ . There is the diffraction grating, thousands of slits together, whose sharp bright orders sit at $d\sin\theta = n\lambda$ . And there is resolution, the diffraction limit $\theta = 1.22\lambda/D$ that decides the finest detail an instrument can show. Same wave physics, three formulas, and a paper-setter's favourite trick is to hand you numbers that fit two of them and see which you reach for. So this is a sorting problem as much as an arithmetic one. The decisive question on every diffraction numerical is: single slit, grating, or aperture? Get that right and the formula is forced; get it wrong and a correct calculation answers the wrong question. The single slit gives minima at $a\sin\theta = m\lambda$ ; the grating gives maxima at $d\sin\theta = n\lambda$ — one letter apart, opposite meanings. The grating's spacing is the reciprocal of its line density ( $d = 1/N_{\text{lines per metre}}$ ), its highest visible order is capped by $\sin\theta \le 1$ (so $n_{\max} = \lfloor d/\lambda\rfloor$ ), and its power to split close wavelengths is $R = nN$ . This article is the drill: a master recipe that routes each question to its family, two derivations, and four fully worked examples — a $500$ -lines-per-millimetre grating, its order limit, a single-slit central width, and the number of grating lines needed to split the sodium doublet — every number carried to the last digit and independently recomputed. Master the sort, and diffraction stops being a guessing game.

Key takeaways

Three families, three formulas: single slit minima $a\sin\theta = m\lambda$ ; grating maxima $d\sin\theta = n\lambda$ ; resolution limit $\theta = 1.22\lambda/D$ (circular) or $\lambda/a$ (slit).
The grating spacing is the reciprocal of the line density: $500$ lines per mm gives $d = 2\times10^{-6}\ \mathrm{m}$ .
Orders are capped by $\sin\theta \le 1$ , so the highest visible order is $n_{\max} = \lfloor d/\lambda\rfloor$ — here $\lfloor 3.33\rfloor = 3$ .
The single-slit central maximum has linear width $2\lambda D/a$ (twice the side fringes) and angular width $2\lambda/a$ .
A grating's wavelength resolving power is $R = \lambda/\Delta\lambda = nN$ — splitting the sodium doublet needs about $983$ lines in first order.
Key idea: decide single slit vs grating vs aperture first; then $a\sin\theta=m\lambda$ (minima), $d\sin\theta=n\lambda$ (maxima, capped at $n_{\max}=\lfloor d/\lambda\rfloor$ ), and $\theta=1.22\lambda/D$ each solve their own family.

The core insight

Every diffraction numerical belongs to one of three families — the single slit ( $a\sin\theta = m\lambda$ minima, central width $2\lambda D/a$ ), the grating ( $d\sin\theta = n\lambda$ maxima, with $d = 1/N_\text{lines}$ , $n_{\max} = \lfloor d/\lambda\rfloor$ , resolving power $R = nN$ ), and resolution ( $\theta = 1.22\lambda/D$ ) — so the whole gym is won by first sorting the question into its family, then applying that family's one formula.

Where this lives today

The diffraction grating is the workhorse of modern spectroscopy: the instrument that reads a star's composition from its spectrum, that identifies pollutants by their absorption lines, and that splits laser light in every optical spectrum analyser is a grating obeying $d\sin\theta = n\lambda$ . Your DVD and Blu-ray discs are reflective gratings — tilt one under a lamp and you see the orders as rainbow bands, their angles set by the track spacing. Fibre-optic networks route dozens of wavelength channels with grating-based multiplexers, and the same $d\sin\theta = n\lambda$ tuned to X-ray wavelengths and atomic-scale $d$ is how crystallographers read the structure of proteins and drugs. Single-slit diffraction sets the spreading of every laser pointer and the beam width of every radio dish, and the resolution limit $1.22\lambda/D$ governs how sharp a camera, a telescope or a microscope can ever be. The three families drilled here as exam arithmetic are the daily tools of physics and engineering.

The setup

The physics is the sorting. All three families come from the same fact — a wave spreads when it meets an aperture — but the aperture's shape decides the formula. Read the question and place it: A single slit of width $a$ produces a broad central bright band with dark minima where the wavelets across the slit cancel in pairs, at $a\sin\theta = m\lambda$ ( $m = \pm1, \pm2,\dots$ ; $m=0$ is the bright centre, not a minimum). A grating of many slits spaced $d$ apart produces razor-sharp bright maxima where every slit adds in phase, at $d\sin\theta = n\lambda$ . A circular aperture of diameter $D$ limits an instrument's resolution to $\theta = 1.22\lambda/D$ . The trap is that the single-slit and grating equations look identical — $a\sin\theta = m\lambda$ versus $d\sin\theta = n\lambda$ — but one locates dark fringes and the other bright orders. Decide which aperture you have before you write a single number.

Figure

A diffraction grating sends light into sharp bright orders at angles set by d sinθ = nλ, with the highest visible order capped where the sine would exceed one

The master relations

The grating spacing is the reciprocal of its line density. A grating ruled with $N_L$ lines per metre has

d = \frac{1}{N_L}, \qquad d\sin\theta = n\lambda, \qquad n_{\max} = \left\lfloor \frac{d}{\lambda}\right\rfloor. \qquad (1)

where: $d$ — the line spacing; $n$ — the order; $\theta$ — the angle of the $n$ -th bright order. Because $\sin\theta$ cannot exceed $1$ , the grating equation demands $n \le d/\lambda$ , so the highest order actually visible is the whole number $n_{\max} = \lfloor d/\lambda\rfloor$ . A grating of $500$ lines per mm has $d = 1/(500\times10^3) = 2\times10^{-6}\ \mathrm{m}$ ; at $\lambda = 600\ \mathrm{nm}$ that gives $d/\lambda = 3.33$ , so $n_{\max} = 3$ . The single slit has a broad central maximum flanked by dimmer side fringes; its minima and the width of the bright centre are

a\sin\theta = m\lambda, \qquad \text{angular width} = \frac{2\lambda}{a}, \qquad \text{linear width} = \frac{2\lambda D}{a}. \qquad (2)

where: $a$ — slit width; $m = \pm1, \pm2,\dots$ orders the minima; the central maximum spans from the $m=-1$ minimum to the $m=+1$ minimum, so it is twice as wide as any side fringe. And resolution, the third family, caps how close two point sources or spectral lines may be:

\theta_{\min} = 1.22\frac{\lambda}{D} \ \text{(circular)}, \qquad R = \frac{\lambda}{\Delta\lambda} = nN \ \text{(grating)}. \qquad (3)

where: $\theta_{\min}$ — the angular resolution of a circular aperture of diameter $D$ ; $R$ — the grating's power to separate wavelengths, growing with order $n$ and the number of illuminated lines $N$ . (A $0.1\ \mathrm{m}$ telescope at $550\ \mathrm{nm}$ resolves $1.22\lambda/D = 6.71\times10^{-6}\ \mathrm{rad}$ .)

Derivation AGrating angles and the maximum order

The goal is to turn $d\sin\theta = n\lambda$ into a working routine: get $d$ from the line density, solve for each order's angle, and find how many orders exist before $\sin\theta$ would exceed $1$ . The strategy is direct algebra on the grating equation; the only subtlety is the physical cap $\sin\theta \le 1$ that limits the orders. No calculus.

Step 1. Convert line density to spacing. A grating with $500$ lines per millimetre has $500\times10^3 = 5\times10^5$ lines per metre, so $d = 1/(5\times10^5) = 2\times10^{-6}\ \mathrm{m}$ .

Step 2. Solve for each order at $\lambda = 600\ \mathrm{nm}$ : $\sin\theta = n\lambda/d = n(600\times10^{-9})/(2\times10^{-6}) = 0.3\,n$ . So $n=1$ : $\sin\theta = 0.3$ , $\theta = 17.5^\circ$ ; $n=2$ : $\sin\theta = 0.6$ , $\theta = 36.9^\circ$ ; $n=3$ : $\sin\theta = 0.9$ , $\theta = 64.2^\circ$ .

Step 3. Find the cap. For $n=4$ , $\sin\theta = 1.2 > 1$ — impossible, no fourth order. Formally $n \le d/\lambda = 2\times10^{-6}/600\times10^{-9} = 3.33$ , so the highest order is $n_{\max} = \lfloor 3.33\rfloor = 3$ .

Step 4. Read the pattern: the orders spread apart as $n$ climbs (the angles jump $17.5^\circ \to 36.9^\circ \to 64.2^\circ$ , not evenly), because $\theta = \sin^{-1}(0.3n)$ is nonlinear near the $\sin\theta = 1$ ceiling. Verified: $d = 2\times10^{-6}\ \mathrm{m}$ , angles $17.5^\circ/36.9^\circ/64.2^\circ$ , and $n_{\max} = 3$ are recomputed in verify_hs2-grating-diffraction-problems.py.

Go deeper · Derivation B · 1 minThe single-slit central-maximum width

Derivation BThe single-slit central-maximum width

The goal is the width of the single slit's bright central band, both angular ( $2\lambda/a$ ) and linear on a screen ( $2\lambda D/a$ ) — and to see why it is twice as wide as the side fringes. The strategy is to locate the first minima on either side of centre and measure between them. This is the single-slit companion to the grating routine, and the numbers feed the worked example.

Step 1. Locate the first minima. The single-slit minima are at $a\sin\theta = m\lambda$ . The first on each side of the centre are $m = +1$ and $m = -1$ , at $\sin\theta = \pm\lambda/a$ , i.e. angles $\pm\lambda/a$ for small angles.

Step 2. The central maximum fills the gap between them, so its angular width is $\theta_{+1} - \theta_{-1} = 2\lambda/a$ .

Step 3. Convert to a length on a screen a distance $D$ away: linear width $= D \times (\text{angular width}) = 2\lambda D/a$ . Each side fringe spans only from one minimum to the next, a width $\lambda D/a$ — half the centre. The central band is twice as wide (and far brighter).

Step 4. Plug numbers: $a = 0.2\ \mathrm{mm}$ , $\lambda = 500\ \mathrm{nm}$ , $D = 2\ \mathrm{m}$ . Linear width $= 2(500\times10^{-9})(2)/(2\times10^{-4}) = 10^{-2}\ \mathrm{m} = 10\ \mathrm{mm}$ ; each first minimum sits $5\ \mathrm{mm}$ from the centre; angular width $= 2\lambda/a = 5\times10^{-3}\ \mathrm{rad}$ . Verified: central linear width $10\ \mathrm{mm}$ , half-width $5\ \mathrm{mm}$ , and angular width $5\times10^{-3}\ \mathrm{rad}$ are recomputed in verify_hs2-grating-diffraction-problems.py.

Figure

The single-slit intensity envelope, a tall central maximum of linear width 2 lambda D over a flanked by much smaller side lobes

The variants examiners actually pose

Grating angles. Get $d$ from the line density, then $\sin\theta = n\lambda/d$ for each order, stopping at $n_{\max} = \lfloor d/\lambda\rfloor$ . Watch that later orders spread nonlinearly.

Maximum order and total number of orders. $n_{\max} = \lfloor d/\lambda\rfloor$ gives the highest order on one side; the total count of bright fringes is $2n_{\max} + 1$ (both sides plus the centre).

Single-slit widths. Central linear width $2\lambda D/a$ , side fringes half that; narrower slit gives a wider pattern (inverse relation).

Grating resolving power. To split two wavelengths $\Delta\lambda$ apart, supply $R = \lambda/\Delta\lambda$ from $R = nN$ . In first order you need $N \ge \lambda/\Delta\lambda$ illuminated lines.

Angular resolution. For a circular aperture (telescope, eye) use $\theta_{\min} = 1.22\lambda/D$ ; a slit uses $\lambda/a$ . Convert an angle to a length with the distance.

Trap table and limiting cases

pitfall or case	what actually happens	the fix
single slit vs grating	$a\sin\theta = m\lambda$ gives minima; $d\sin\theta = n\lambda$ gives maxima	one letter, opposite meaning — decide the aperture first
grating spacing	$d = 1/(\text{lines per metre})$ , so $500/\mathrm{mm} \Rightarrow 2\times10^{-6}\ \mathrm{m}$	convert line density to metres
a fourth order at $500/\mathrm{mm}$ , $600\ \mathrm{nm}$	none: $\sin\theta = 1.2 > 1$	cap orders at $n_{\max} = \lfloor d/\lambda\rfloor$
central maximum width	$2\lambda D/a$ — twice the side fringes	not $\lambda D/a$
narrower slit	wider pattern	width $\propto 1/a$ , the inverse of intuition
resolving two wavelengths	$R = nN$ ; need $N \ge \lambda/\Delta\lambda$	more lines or higher order
resolving two points	$\theta = 1.22\lambda/D$ (circular only)	the $1.22$ is for a round aperture, not a slit

Common trapThe grating equation gives maxima; the single-slit equation gives minima

The two most-confused formulas in wave optics are $d\sin\theta = n\lambda$ and $a\sin\theta = m\lambda$ . They differ by one letter and by everything that matters: the grating equation locates bright orders (every slit in phase), while the single-slit equation locates dark minima (wavelets cancelling across the slit). If a question gives "lines per mm," it is a grating and you want maxima; if it gives a single slit width and asks for the dark fringe or the central width, you want the minima formula. A second habit worth building: always convert a line density to a spacing in metres ( $d = 1/N_\text{lines per metre}$ ) before substituting, and always check that your solved $\sin\theta \le 1$ — an order with $\sin\theta > 1$ does not exist. Getting the family wrong turns a correct calculation into a confident wrong answer.

Worked examples

1 · Board level — grating orders (CBSE and AP Physics 2 pattern)

Light of wavelength $600\ \mathrm{nm}$ falls normally on a diffraction grating of $500$ lines per millimetre. Find the angles of the first three bright orders.

Step 1. Spacing: $d = 1/(500\ \text{per mm}) = 1/(5\times10^5\ \text{per m}) = 2\times10^{-6}\ \mathrm{m}$ .

Step 2. Grating equation: $\sin\theta = n\lambda/d = n(600\times10^{-9})/(2\times10^{-6}) = 0.3\,n$ .

Step 3. Solve each order: $n=1$ : $\sin\theta = 0.3 \Rightarrow \theta = \boxed{17.5^\circ}$ ; $n=2$ : $\sin\theta = 0.6 \Rightarrow \theta = \boxed{36.9^\circ}$ ; $n=3$ : $\sin\theta = 0.9 \Rightarrow \theta = \boxed{64.2^\circ}$ .

Step 4. Note the widening gaps — $17.5^\circ$ , then $19.4^\circ$ more, then $27.3^\circ$ more — a signature of the nonlinear $\sin^{-1}$ near its ceiling. A fourth order would need $\sin\theta = 1.2$ and does not exist.

2 · Board level — the maximum order (ICSE and JEE pattern)

For the same grating ( $500$ lines per mm) and light ( $600\ \mathrm{nm}$ ), how many bright orders can be seen in total?

Step 1. Highest order on one side: $n_{\max} = \lfloor d/\lambda\rfloor = \lfloor (2\times10^{-6})/(600\times10^{-9})\rfloor = \lfloor 3.33\rfloor = \boxed{3}$ .

Step 2. So orders $n = 1, 2, 3$ appear on each side of the centre.

Step 3. Total bright fringes: both sides plus the central maximum, $2n_{\max} + 1 = 2(3) + 1 = \boxed{7}$ .

Step 4. Sanity: $n=4$ would need $\sin\theta = 4(0.3) = 1.2 > 1$ — confirming the cap at $3$ .

3 · Competitive level — single-slit central width (JEE-Advanced and AP pattern)

A single slit of width $0.2\ \mathrm{mm}$ is lit by $500\ \mathrm{nm}$ light. On a screen $2.0\ \mathrm{m}$ away, how wide is the central bright band?

Step 1. Identify the family: single slit, width $a = 0.2\ \mathrm{mm} = 2\times10^{-4}\ \mathrm{m}$ — use the minima and central-width formula, not the grating equation.

Step 2. Central linear width, equation (2): $2\lambda D/a = 2(500\times10^{-9})(2.0)/(2\times10^{-4})$ .

Step 3. Compute: numerator $= 2\times10^{-6}$ ; divide by $2\times10^{-4}$ : $= 1.0\times10^{-2}\ \mathrm{m} = \boxed{10\ \mathrm{mm}}$ . Each first minimum sits $5\ \mathrm{mm}$ from centre; the angular width is $2\lambda/a = 5\times10^{-3}\ \mathrm{rad}$ .

Step 4. Check the direction of the trend: halving the slit to $0.1\ \mathrm{mm}$ would double the central band to $20\ \mathrm{mm}$ — a narrower slit spreads light more. Timing note: about 60 seconds. The whole task is spotting "single slit, central width" and writing $2\lambda D/a$ . Do not use $d\sin\theta = n\lambda$ here — that is the grating, a different family.

4 · Competitive level — resolving the sodium doublet (JEE and India-depth pattern)

A grating is to resolve the two sodium lines at $589.0$ and $589.6\ \mathrm{nm}$ in the first order. How many lines must be illuminated?

Step 1. Family: this is grating resolving power, $R = nN$ . Note the mean wavelength $589.3\ \mathrm{nm}$ and the split $\Delta\lambda = 0.6\ \mathrm{nm}$ .

Step 2. Required resolving power: $R = \lambda/\Delta\lambda = 589.3/0.6 \approx \boxed{982}$ .

Step 3. In first order $n = 1$ , so $R = N$ . You need $N \ge 982$ , and since $N$ is a whole number of lines, $\boxed{N \ge 983}$ illuminated lines.

Step 4. Board-scope caveat: this resolving-power calculation is examined in India (CBSE/JEE/NEET) but is beyond AP Physics 2 and A-Level 9702, which carry the grating angles ( $d\sin\theta = n\lambda$ ) and single-slit widths but not the Rayleigh-style resolving power. Know which board you are sitting. Timing note: about 60 seconds. "Resolve two close wavelengths" always means $R = \lambda/\Delta\lambda = nN$ — do not confuse it with the angular resolution $1.22\lambda/D$ , which separates two directions, not two colours.

Exam traps

Mixing the single-slit and grating equations. $a\sin\theta = m\lambda$ gives single-slit minima; $d\sin\theta = n\lambda$ gives grating maxima. They look alike and mean opposite things. Decide the aperture — single slit, or many-slit grating — before choosing.

Forgetting to invert the line density. A grating quoted as " $500$ lines per mm" has spacing $d = 1/(500\ \text{per mm}) = 2\times10^{-6}\ \mathrm{m}$ , not $500\ \mathrm{mm}$ or $500\ \mu\mathrm{m}$ . Convert to lines per metre, then take the reciprocal.

Allowing an impossible order. The grating equation only has solutions with $\sin\theta \le 1$ , so orders above $n_{\max} = \lfloor d/\lambda\rfloor$ do not exist. Always check; a $500/\mathrm{mm}$ grating at $600\ \mathrm{nm}$ stops at the third order.

Halving the central maximum. The single-slit central band is $2\lambda D/a$ wide — twice the side fringes. Using $\lambda D/a$ (the half-width, or a side-fringe width) is a common factor-of-two slip.

Confusing the two "resolutions." Grating resolving power $R = nN$ separates wavelengths; the Rayleigh limit $1.22\lambda/D$ separates directions (and its $1.22$ is for circular apertures only). Different questions, different formulas.

Board scope. Single-slit and grating numericals are examined on CBSE, ICSE, JEE-Main, JEE-Advanced, NEET, AP Physics 2 and A-Level 9702. The resolving-power numericals (both $R = nN$ and $1.22\lambda/D$ ) are India-depth — AP-2 and 9702 do not assess Rayleigh-style resolution, so an AP or A-Level student should master the grating angles and single-slit widths but can skip example 4's resolving-power step.

Origins · 1 minErnst Abbe and how this was discovered

Ernst Abbe

1840–1905 · Jena

Ernst Abbe was the physicist who put the microscope on a rigorous diffraction footing and, with it, tied together the three families of this article. Working for the optical workshop of Carl Zeiss in Jena, he showed in 1873 that an image forms by the recombination of diffracted light, and that the finest detail a microscope can resolve is limited to about $d \approx \lambda/(2\,NA)$ — the diffraction grating, the single slit and the resolution limit all appearing as facets of one theory of image formation. His insight transformed lens-making from craft to science, and Zeiss microscopes built to Abbe's designs dominated the field. A committed social reformer, Abbe later took over the Zeiss enterprise and, through the Carl Zeiss Foundation, introduced paid holidays and profit-sharing in the 1890s and, in 1900, the eight-hour day — remarkable for the era. His name survives in the Abbe number of glasses, the Abbe condition for aberration-free imaging, and the resolution limit that still bounds every light microscope.

History

1785

— David Rittenhouse builds one of the first diffraction gratings, ruling fine wires to split light into orders.

1821

— Joseph von Fraunhofer rules precision gratings and uses them to measure the wavelengths of the solar spectral lines.

1873

— Ernst Abbe publishes his diffraction theory of image formation and the microscope resolution limit.

1882

— Henry Rowland's ruling engines produce gratings of unprecedented quality, revolutionizing spectroscopy.

1912

— Max von Laue shows that crystals diffract X-rays like three-dimensional gratings, opening crystallography.

Going further

Texts. NCERT Physics Class 12, Ch. 10 ("Wave Optics"), single-slit diffraction and grating sections · H. C. Verma, Concepts of Physics Vol. 1, "Light Waves," diffraction problems · Irodov, Problems in General Physics, diffraction and grating set · Cambridge A-Level 9702 (single slit and diffraction grating).

Provenance. Standard exam-archive problems. Recompute: $d = 1/(5\times10^5) = 2\times10^{-6}\ \mathrm{m}$ ; grating angles $\sin\theta = 0.3n$ → $17.5^\circ, 36.9^\circ, 64.2^\circ$ ( $n=1,2,3$ ); $n_{\max} = \lfloor 3.33\rfloor = 3$ , total orders $2n_{\max}+1 = 7$ ; single-slit central width $2\lambda D/a = 10\ \mathrm{mm}$ , half-width $5\ \mathrm{mm}$ , angular $5\times10^{-3}\ \mathrm{rad}$ ( $a=0.2\ \mathrm{mm}$ , $500\ \mathrm{nm}$ , $D=2\ \mathrm{m}$ ); telescope $1.22\lambda/D = 6.71\times10^{-6}\ \mathrm{rad}$ ; sodium doublet $R = 589.3/0.6 \approx 982 \Rightarrow N \ge 983$ .

Prerequisites. Single-Slit Diffraction and Resolving Power, the Rayleigh Criterion & Gratings-Lite — the concept articles this gym drills. Leads to. the university seams Diffraction Grating and the Grating Equation, Single-Slit Diffraction, and Rayleigh Resolution Criterion.

Go deeper (UG). The full grating line-shape (why more lines sharpen the orders and give $R = nN$ ) and the exact single-slit intensity pattern are developed in Diffraction Grating and the Grating Equation and Single-Slit Diffraction; the resolution limit is treated in Rayleigh Resolution Criterion. Enrichment only.

How do I instantly tell a single-slit problem from a grating problem?

Look at what the aperture is. If the problem describes one slit of a given width $a$ and asks about the central band, the dark fringes, or the spreading, it is single-slit: use $a\sin\theta = m\lambda$ for the minima and $2\lambda D/a$ for the central width. If it mentions "lines per mm," "a grating," or "many slits" and asks for bright orders or their angles, it is a grating: use $d\sin\theta = n\lambda$ for the maxima. The tell is the word "lines" (grating) versus a single "width" (slit), and whether the question wants bright orders or dark fringes. Circle the aperture description before you compute — that one decision fixes the whole solution.

Why is there a maximum order for a grating but not for a double slit?

There is a maximum order for both — it is just rarely binding for a wide double slit. The grating equation $d\sin\theta = n\lambda$ can only be satisfied while $\sin\theta \le 1$ , so $n \le d/\lambda$ ; the highest order is $n_{\max} = \lfloor d/\lambda\rfloor$ . For a fine grating (small $d$ , say $2\ \mu\mathrm{m}$ ) only a handful of orders fit before $\sin\theta$ would exceed $1$ — here just three. A double slit usually has a much larger separation $d$ , so $d/\lambda$ is huge and dozens of orders appear across a small angle, which is why the cap is not emphasised there. The physics is identical; the grating simply makes the ceiling visible because its $d$ is small.

Why does a narrower slit make a wider diffraction pattern?

Because the spreading angle is inversely proportional to the slit width. The central maximum has angular width $2\lambda/a$ : shrink $a$ and the angle grows. Physically, forcing a wave through a tighter opening gives it less room to define a direction, so it fans out more — a manifestation of the same trade-off that appears everywhere in wave physics (confine a wave in space and its range of directions widens). At the extreme, a slit far narrower than the wavelength radiates almost hemispherically. So a $0.1\ \mathrm{mm}$ slit spreads light twice as widely as a $0.2\ \mathrm{mm}$ one. This is the opposite of the everyday intuition that a smaller hole makes a smaller spot, and examiners test it constantly.

What is the difference between a grating's resolving power and a telescope's resolution?

They resolve different things. A grating's resolving power $R = nN$ measures how close in wavelength two spectral lines can be and still be separated — it is a pure number, "one part in $R$ ." A telescope's resolution $1.22\lambda/D$ measures how close in angle two point sources can be and still be seen as two — it is an angle in radians. One is about colour, the other about direction. A grating with $R = 1000$ splits wavelengths $\lambda/1000$ apart; a telescope with a $0.1\ \mathrm{m}$ aperture splits stars a few microradians apart. Both come from diffraction, but they answer different questions with different formulas, and swapping them is a classic trap.

How many significant figures, and where do people lose marks in this gym?

Carry four, present two or three. The recurring losses: (1) forgetting to invert the line density — writing $d = 500$ instead of $d = 1/(5\times10^5) = 2\times10^{-6}\ \mathrm{m}$ ; (2) letting an impossible order through by not checking $\sin\theta \le 1$ ; (3) the factor-of-two slip on the central maximum ( $2\lambda D/a$ , not $\lambda D/a$ ); and (4) unit mixing — keep nm, mm and m consistent by converting all to metres first. On angle questions, remember the grating angles are not evenly spaced because you invert a sine, so compute each $\theta = \sin^{-1}(n\lambda/d)$ separately rather than multiplying the first angle.

Do I need the resolving-power formulas if I sit AP or A-Level?

Not for those boards. AP Physics 2 and A-Level 9702 examine the grating angles ( $d\sin\theta = n\lambda$ ), the maximum order, and single-slit diffraction (central width, minima), so master those thoroughly. But the resolving-power formulas — the grating's $R = nN$ and the Rayleigh limit $1.22\lambda/D$ — are beyond both syllabuses; they are India-facing (CBSE, JEE, NEET). If your board is AP or A-Level, you can safely skip example 4's resolving-power step and the $1.22\lambda/D$ family, and concentrate your practice on the grating angles and single-slit widths that your paper actually tests. If you sit an Indian board, learn all three families.