Diffraction grating — condition for maxima: Question bank

Intuition How to use this page

These are the conceptual traps the grating equation $d sin θ = mλ$ sets for you — sign traps, "which condition am I using" traps, and boundary cases where the tidy formula quietly stops making sense. Cover the answer, commit to a reason (not just a verdict), then reveal. If you can only guess "true/false" without a because, you don't own the idea yet.

Symbols used throughout, so no line surprises you:

$d$ = centre-to-centre spacing between neighbouring slits (metres).
$θ$ = angle of an outgoing beam measured from the normal (the line straight out, perpendicular to the grating).
$m$ = the order, a whole number $0, \pm 1, \pm 2, \dots$ counting how many wavelengths of extra path separate neighbouring slits.
$λ$ = wavelength; $N$ = total number of slits.

See the parent for the full derivation: Diffraction grating — condition for maxima.

True or false — justify

True or false: "A grating with more slits produces maxima at new angles that a two-slit setup would not show."

False — the angles of the principal maxima depend only on

d

λ

m

, not on how many slits there are; adding slits makes the same maxima sharper and brighter, not relocated. See Young's double-slit experiment.

True or false: "The central (

m = 0

) maximum is white for white-light input."

True — at

m = 0

the condition

d sin θ = 0

is satisfied at

θ = 0

for every wavelength, so all colours overlap and recombine into white.

True or false: "Doubling the wavelength always doubles the angle of the first-order maximum."

False — it doubles

sin θ_{1}

, not

θ_{1}

itself; since

sin

is not linear, the angle grows faster than proportionally and can even exceed the physical limit.

True or false: "If light hits the grating at an angle instead of normally, the equation

d sin θ = mλ

is unchanged."

False — oblique incidence adds an incoming path difference, giving

d (sin θ_{out} \pm sin θ_{in}) = mλ

; the normal-incidence form is a special case where the incoming term is zero.

True or false: "A grating and a single wide slit produce maxima by the same physical rule."

False — a grating gives maxima where all slits add in phase (

d sin θ = mλ

); a single slit's famous condition

a sin θ = mλ

marks its dark minima instead. See Single-slit diffraction.

True or false: "Increasing

N

(slits) increases the total number of orders you can see."

False — the number of orders is capped by

m_{m a x} = ⌊ d / λ ⌋

, which depends on

d

and

λ

only; more slits sharpen peaks but add no new orders.

True or false: "The

m

and

- m

maxima are mirror images at equal angles on either side of the centre."

True —

sin (- θ) = - sin θ

, so

m

and

- m

solve the equation at

\pm θ_{m}

; the pattern is symmetric about the normal.

True or false: "Red light bends more than blue at a given order in a grating."

True — larger

λ

gives larger

sin θ_{m}

, so red (longer wavelength) sits at a wider angle than blue in every non-zero order. See Wavelength and the visible spectrum.

Spot the error

"For bright fringes on a grating we need

d sin θ = (m + \frac{1}{2}) λ

Wrong condition — half-integer path differences make neighbouring waves cancel; grating maxima need whole wavelengths,

d sin θ = mλ

d = 500

because the grating has 500 lines per mm."

d

is the spacing, not the count; you must invert,

d = 1/500 mm = 2000 nm

, and convert to metres before using it.

"Since

m

can be any integer, there are infinitely many orders."

sin θ \leq 1

physically caps the order at

m_{m a x} = ⌊ d / λ ⌋

; beyond that no real angle solves the equation.

"Small-angle approximation

sin θ \approx θ

is fine here, like in Young's experiment."

Grating angles are often tens of degrees, where

sin θ

and

θ

differ a lot; keep the full

sin θ

"The peak of the maximum is intense because a lot of light passes through the wide grating."

The intensity

\propto N^{2}

comes from

N

slits adding in phase (coherent addition), not merely from more area passing more light.

"Order

m = 4

exists because

4

is an integer, so just solve

sin θ = 4 λ / d

4 λ / d > 1

there is no real angle; an integer

m

is necessary but not sufficient — it must also keep

sin θ \leq 1

"Two gratings with the same

d

but different

N

give differently coloured spectra."

The colours land at the same angles (set by

d

); differing

N

only changes sharpness and resolving power, not the colour positions. See Resolving power and spectrometers.

"At

m = 0

the colours are spread out into a rainbow."

The zero order is where all wavelengths satisfy

θ = 0

together, so it is white and undispersed; dispersion only appears for

m \neq = 0

Why questions

Why does making neighbouring slits agree in phase guarantee all

N

slits agree?

If neighbours differ by exactly

mλ

, then slit

k

leads slit 1 by

(k - 1) mλ

— always a whole number of wavelengths — so every slit's crest lines up.

Why are grating maxima far sharper than two-slit fringes?

With

N

slits, moving slightly off the exact angle lets the many slightly-out-of-phase contributions cancel over a large group, so the bright peak collapses quickly — sharper as

N

grows.

Why does the maximum order fall when

λ

increases (for fixed

d

m_{m a x} = ⌊ d / λ ⌋

; a larger

λ

shrinks

d / λ

, so fewer whole-wavelength path differences fit before

sin θ

would exceed 1.

Why does a grating disperse light into a spectrum but a plain mirror does not?

The maximum angle depends on

λ

through

sin θ_{m} = mλ / d

, so each colour leaves in a slightly different direction; a mirror sends all colours the same way. See Interference and path difference.

Why does packing slits closer together (smaller

d

) spread the colours more?

Smaller

d

makes

sin θ_{m} = mλ / d

larger for each colour, pushing orders to wider angles and increasing the angular gap between wavelengths.

Why is the path difference

d sin θ

and not

d cos θ

or just

d

The extra path is the side of the right triangle opposite the angle

θ

between the beam and the normal; that side is

d sin θ

by the definition of sine on that triangle.

Edge cases

What is the outgoing angle for the

m = 0

order, and why?

θ = 0

— the only solution of

d sin θ = 0

; the beam goes straight through undeviated, so the centre acts like there were no grating for direction.

What happens when

d / λ

is exactly an integer, say

d / λ = 3

The highest order

m = 3

satisfies

sin θ = 1

, i.e.

θ = 9 0^{\circ}

; that maximum grazes along the grating surface and is not practically observable.

What if the wavelength is longer than the slit spacing,

λ > d

Then even

m = 1

needs

sin θ = λ / d > 1

, which is impossible, so only the central

m = 0

maximum survives — no spectrum forms.

What is the limiting behaviour of the pattern as the number of slits

N \to \infty

(spacing fixed)?

The maxima stay at the same angles but become infinitely narrow and infinitely intense relative to the background — the idealised perfectly-sharp spectral line.

What happens to the maxima if the spacing

d

is not perfectly uniform across the grating?

The condition

d sin θ = mλ

no longer holds at one clean angle for all slits, so peaks broaden and blur, degrading resolving power. See Resolving power and spectrometers.

Can a maximum exist for which

sin θ_{m}

is negative, and what does it mean?

Yes — negative

m

gives negative

sin θ

, i.e. a maximum on the opposite side of the normal at angle

∣ θ_{m} ∣

; the sign just labels which side.

What is special about the highest visible order when

d / λ

has a small fractional part, like

3.05

m_{m a x} = 3

sits at

sin θ = 3 λ / d \approx 0.98

, very close to

9 0^{\circ}

— a faint, extremely oblique maximum that is easy to miss experimentally.

Connections

Diffraction grating — condition for maxima — the parent derivation these traps stress-test.
Young's double-slit experiment — the $N = 2$ limit; same phase condition, broader fringes.
Single-slit diffraction — contrasting rule: its $a sin θ = mλ$ marks minima, not maxima.
Resolving power and spectrometers — why sharper (larger $N$ ) peaks matter.
Wavelength and the visible spectrum — why colours fan out by wavelength.