Single Slit Diffraction

Equivalent forms

\theta_m = \arcsin\left(\frac{m\lambda}{a}\right)

y_m = \frac{m\lambda L}{a}

The same path-difference logic as interference, but here you integrate across a continuous aperture instead of summing two sources.

Symbol	Name	SI unit	Dimension	Range
$a$	Slit width Width of the single slit aperture	m	L	0.00001 – 0.001
$theta$	Angle to minimumoutput Angle from the central axis to the m-th dark fringe	rad	1	0 – 0.5
$m$	Order number Integer order of the dark fringe (1, 2, 3, ...)	dimensionless	1	1 – 10
$lambda$	Wavelength Wavelength of the incident light	m	L	3.8e-7 – 7.5e-7

Unit systems

Symbol	SI	CGS	Imperial
$a$	m	cm	in
$theta$	rad	rad	deg
$lambda$	m	cm	in

Where it holds

Valid in the Fraunhofer (far-field) regime where screen distance L >>

a^{2}/\lambda

. For near-field, use Fresnel diffraction theory.

Dimensional analysis

a\cdot \sin (\theta ) \to [L]\cdot [1] = [L]

m\cdot \lambda \to [1]\cdot [L] = [L]

Both sides are

[L] =

meters.

\checkmark

Discovery

Francesco Maria Grimaldi · 1665

Grimaldi first observed diffraction fringes. Fraunhofer later provided the mathematical treatment for far-field single-slit patterns.

Try this

Why does a laser beam spread out after passing through a narrow slit?

A 633 nm laser shines through a 0.05 mm slit. Find the angle to the first dark fringe using a*sin(theta) = m*lambda.

Research status: stable

Real-world applications

Telescope resolution: the Airy disk is the 2D analog, setting the diffraction limit.
CD/DVD reading: laser diffraction from pit tracks encodes data.
X-ray crystallography: diffraction from atomic-scale 'slits' reveals molecular structure.
Acoustic diffraction: sound bending around doorways and obstacles.

Common misconceptions

The formula gives bright fringes — it gives dark fringes (minima). Bright maxima are approximately halfway between.
Narrower slit means narrower pattern — the opposite: narrower slit produces wider diffraction spread.
Diffraction and interference are different phenomena — diffraction IS interference from a continuous aperture.

Experimental verification

Grimaldi first observed (1665). Fraunhofer provided precise measurements using his diffraction grating. Modern: single-slit patterns are reproduced routinely with lasers and adjustable slits in undergraduate labs.

Derivation

Divide the slit of width a into infinitesimal strips.

Each strip acts as a Huygens wavelet source.

At angle

\theta

, the path difference between the top and bottom of the slit is

a\cdot \sin (\theta )

.

Dark fringes occur when this equals

m\lambda (m = \pm 1

,

\pm 2

, ...) because the slit can be divided into m pairs of strips that cancel pairwise.

Formally, integrating the electric field contributions:

E(\theta ) = E_{0}\cdot

sinc

(\pi a\cdot \sin \theta /\lambda )

, giving minima at

a\cdot \sin (\theta ) = m\lambda

.

Limiting cases

a \to \infty

⟶

\theta \to 0

Wide slit means negligible diffraction — geometric optics limit.

a \to \lambda

⟶

\theta \to 90^{\circ}

Slit width equals wavelength: light spreads into a hemisphere.

\lambda \to 0

⟶

\theta \to 0

Short wavelength means sharp shadows — ray optics regime.

What if…

What if a >>

\lambda

?

Diffraction becomes negligible $(\theta \to 0)$ . Light passes in a straight beam — the geometric optics limit.

What if

a \approx \lambda

?

First minimum $at \theta \approx 90^{\circ}$ . Light spreads into a hemisphere — maximum diffraction.

What if you use a circular aperture instead?

The pattern becomes an Airy disk: concentric bright and dark rings. The first dark ring is at $\sin (\theta ) = 1.22\lambda /D$ .

1

First dark fringe of a laser through a slit

Given ·

a:: 0.00005
lambda:: 6.33e-7
m:: 1

Find · theta

Steps

Apply $a\cdot \sin (\theta ) = m\lambda for$ the first minimum $(m=1)$
$\sin (\theta ) = m\lambda$ / $a = 6.33\times 10^{-7} / 5\times 10^{-5} = 0.01266$
$\theta = \arcsin (0.01266) \approx 0.01266\,\mathrm{rad}$
Convert: $0.01266 \times 180/\pi \approx 0.725^{\circ}$

Result ·

\theta = \arcsin (m\lambda

/a

) = \arcsin (1 \times 6.33\times 10^{-7} / 5\times 10^{-5}) = \arcsin (0.01266) = 0.01266\,\mathrm{rad} \approx 0.725^{\circ}

2

Central maximum width on a screen

Given ·

a:: 0.0001
lambda:: 5e-7
L:: 2

Find · width of central maximum

Steps

Central maximum extends between first minima on both sides
Position of first minimum: $y_{1} = \lambda L$ /a
Full width $= 2y_{1} = 2\lambda L$ /a
$= 2 \times 5\times 10^{-7} \times 2.0 / 1\times 10^{-4} = 0.02\,\mathrm{m} = 20\,\mathrm{mm}$

Result · Width

= 2\lambda L

/

a = 2 \times 5\times 10^{-7} \times 2.0 / 1\times 10^{-4} = 0.02\,\mathrm{m} = 20\,\mathrm{mm}

Young's Double Slit Interference→Wave Speed Equation→