Diffraction — single slit intensity pattern derivation: Question bank

Reminder of the two anchors you will keep needing: $I (θ) = I_{0} (\frac{s i n β}{β})^{2}, β = \frac{π a s i n θ}{λ}, minima: a sin θ = mλ (m = \pm 1, \pm 2, \dots) .$

Here $a$ is the slit width, $λ$ the wavelength, $θ$ the diffraction angle, and $β$ is half the total phase spread across the slit.

True or false — justify

At a single-slit minimum the condition is

a sin θ = mλ

, so

m = 0

gives the first dark fringe

False.

m = 0

means

β = 0

, where

sin β / β \to 1

— that is the central bright maximum. Minima require

m = \pm 1, \pm 2, \dots

only.

The formula

a sin θ = mλ

marks bright fringes, just like Young's double slit

False. For a single slit this marks dark fringes: when the full-width path difference is a whole

λ

, wavelets cancel in top-bottom pairs. Same-looking formula, opposite physics.

Making the slit narrower makes the diffraction pattern narrower

False. The half-width is

θ_{1} \approx λ / a

, so smaller

a

gives a wider spread. Confining the beam more forces it to fan out more.

The secondary maxima sit exactly halfway between adjacent minima

False. They satisfy

tan β = β

, giving

β \approx 1.43 π, 2.46 π, \dots

— pulled slightly toward the centre by the

1/ β

envelope, not at the tidy

1.5 π, 2.5 π

The central maximum is exactly twice as wide (in angle) as each side maximum

True in linear width, roughly. The central peak spans

m = - 1

m = + 1

(width

\sim 2 λ / a

), while a side maximum spans only one

π

-interval of

β

(width

\sim λ / a

). So the centre is about twice as broad.

All the secondary maxima have the same brightness as each other

False. Their heights fall off roughly as

1/ (m + \frac{1}{2})^{2} π^{2}

: the first is

\approx 4.5%

I_{0}

, the second

\approx 1.6%

, and so on. The envelope decays.

If you double both

a

and

λ

, the minima appear at the same angles

True. Minima obey

sin θ = mλ / a

, which depends only on the ratio

λ / a

. Scaling both by the same factor leaves every angle unchanged.

Increasing the intensity of the incoming light shifts the fringe positions outward

False. Fringe positions depend only on

λ

a

, and geometry. Brighter light raises

I_{0}

(all peaks scale together) but moves nothing.

Spot the error

"At

β = π

we have

sin β = 0

, and since

sin β / β = 0/ π = 0

, the pattern is dark — this is the first minimum." Find the flaw

There is no flaw in this one — it is correct.

β = π

is exactly

a sin θ = λ

, the genuine first minimum. (The trap is doubting a right statement.)

"Because

sin β / β

has a

β

in the denominator, the intensity blows up at

β = 0

Wrong. As

β \to 0

sin β \approx β

, so the ratio

\to 1

, not infinity. The numerator vanishes just as fast as the denominator — the peak is finite at

I_{0}

"The first secondary maximum is at

β = 1.5 π

, so its intensity is

(sin (1.5 π) /1.5 π)^{2} = 1/ (1.5 π)^{2}

Two slips. The true peak is at

β \approx 1.43 π

, and

sin (1.5 π) = - 1

(not the max of the sinc). The

\approx 4.5%

value is a coincidentally-close estimate, not the exact peak.

"Since a single slit is 'like two slits pushed together,' the missing-order rule of the double slit applies here too."

A single slit has no separate 'orders' to miss; it is one continuous source, not two point sources. The double-slit missing-order effect comes from the diffraction envelope modulating interference fringes — a two-scale story that needs both

a

and slit separation

d

"The phase difference across the slit is

β = π a sin θ / λ

The total phase spread edge-to-edge is

2 β = 2 π a sin θ / λ

β

is only half of it — the natural half-angle variable that makes the chord formula

sin β / β

come out clean.

"At the first minimum the top edge and bottom edge of the slit are

λ /2

out of phase, which is why they cancel."

The edges are a full

λ

(i.e.

2 π

) apart at the first minimum, not

λ /2

. Cancellation happens because each strip in the top half pairs with a strip in the bottom half that is

λ /2

away — pairwise cancellation across the halves.

Why questions

Why is the centre of the pattern bright and not dark?

θ = 0

every wavelet travels the same distance, so all phases are zero and they add perfectly in step (

sin β / β \to 1

). Maximum constructive interference lives at the centre.

Why does the single-slit rule give darkness where the double-slit rule gives brightness for the same

a sin θ = mλ

A double slit has two point sources adding in step at

mλ

; a single slit is a continuum whose wavelets cancel in pairs at

mλ

. The distributed nature of the source flips the outcome.

Why does the factor

e^{i β}

in the amplitude not affect the intensity?

It has magnitude 1 (a pure phase), and intensity is

∣ E ∣^{2}

. Multiplying by a unit-magnitude phase changes direction on the complex plane but not the length that gets squared.

Why is

β

(half the phase spread) the "natural variable" rather than the full spread

2 β

The resultant of many equal phasors fanning through total angle

2 β

is the chord of that arc, and a chord's length involves the half-angle

sin β

. So

β

is what geometry hands you.

Why does the intensity fall off as we go out to larger angles even between minima?

The

1/ β

envelope: larger

β

means the phasors curl into a tighter, smaller-radius arc, so even the best partial cancellation leaves less resultant. The whole pattern rides under a shrinking

1/ β^{2}

ceiling.

Why can we treat all rays reaching one screen point as parallel?

The screen is far compared with the slit width (the Fraunhofer condition, contrasted in Fraunhofer vs Fresnel Diffraction). Rays to a distant point differ in direction by a negligible amount, so path difference is cleanly

y sin θ

Why is single-slit spreading called a "baby version" of the uncertainty principle?

Confining a photon's transverse position to width

Δ y \sim a

forces a transverse momentum spread

\sim λ / a

in angle — smaller

a

, wider angular fan. This is exactly the position-momentum trade-off of Heisenberg Uncertainty Principle.

Edge cases

What does the pattern look like as

a \to λ

(slit width shrinks to one wavelength)?

The first minimum needs

sin θ = λ / a = 1

, i.e.

θ = 9 0^{\circ}

— it slides to the edge. The central maximum fills the whole forward hemisphere; the slit acts almost like a single Huygens point source.

What happens when

a < λ

No minimum exists, since

sin θ = mλ / a

would need

sin θ > 1

. Only the central maximum survives, spread across all forward angles — essentially pure spherical-wavelet emission.

What is

I

in the limit

a \to \infty

(very wide slit)?

Minima crowd toward

θ = 0

(

sin θ = mλ / a \to 0

), so the central peak collapses to a razor-sharp spike straight ahead. This is the ray-optics limit: a wide aperture makes a sharp geometric beam, no visible diffraction.

At exactly

θ = 9 0^{\circ}

with

a sin θ

equal to some

mλ

, is that direction physically reachable?

θ = 9 0^{\circ}

means light travelling along the slit plane — a grazing, physically limiting edge. Orders that would require

sin θ > 1

simply don't exist;

sin θ = 1

is the hard ceiling.

If we shine two wavelengths through the same slit, do their central maxima overlap?

Yes — both central maxima sit at

θ = 0

regardless of

λ

, so their centres coincide. Their first minima diverge (

θ_{1} \approx λ / a

), so the longer wavelength's pattern is wider around the shared centre.

What is the intensity precisely at a secondary-maximum's neighbouring minimum, and does the peak reach

I_{0}

At any minimum

I = 0

exactly. No secondary peak ever reaches

I_{0}

; the first is only

\approx 4.5%

of it, because the

1/ β^{2}

envelope caps every later peak well below the centre.

Recall Fast self-test

Single-slit $a sin θ = mλ$ means what, and for which $m$ ? ::: DARK fringes, $m = \pm 1, \pm 2, \dots$ (never $m = 0$ ). Which way does the pattern move as the slit narrows? ::: It gets wider ( $θ_{1} \approx λ / a$ ). Where are the true secondary maxima? ::: At $tan β = β$ , i.e. $β \approx 1.43 π, 2.46 π, \dots$ , not the half-integer multiples of $π$ .

Related builds: Phasor Addition of Waves (why the chord gives $sin β / β$ ), Huygens Principle (why a slit is many sources), and Rayleigh Criterion and Resolution (why $λ / a$ sets the limit of what optics can separate).