Young's Double Slit Interference

Equivalent forms

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

\Delta y = \frac{\lambda L}{d}

A simple geometry of path difference produces the entire interference pattern — the fingerprint of wave behavior.

Symbol	Name	SI unit	Dimension	Range
$d$	Slit separation Distance between the centers of the two slits	m	L	0.00001 – 0.001
$theta$	Angle to maximumoutput Angle from the central axis to the m-th bright fringe	rad	1	0 – 0.1
$m$	Order number Integer order of the bright fringe (0, 1, 2, ...)	dimensionless	1	0 – 10
$lambda$	Wavelength Wavelength of the incident light	m	L	3.8e-7 – 7.5e-7

Unit systems

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

Where it holds

Valid for coherent, monochromatic light with slit width << slit separation. Assumes far-field (Fraunhofer) regime where L >> d.

Dimensional analysis

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

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

Both sides are

[L] =

meters.

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Discovery

Thomas Young · 1801

Young's experiment provided the first definitive proof of the wave nature of light, overturning Newton's corpuscular theory.

Try this

How can two slits in a card prove that light is a wave?

Monochromatic light (lambda = 550 nm) hits two slits separated by 0.1 mm. Find the fringe spacing on a screen 1 m away using y = m*lambda*L/d.

Research status: stable

Real-world applications

Measuring wavelength: fringe spacing directly gives $\lambda if d$ and L are known.
Anti-reflective coatings: thin film interference cancels reflected light.
Holography: interference patterns encode 3D information on a 2D surface.
Quantum mechanics foundations: single-particle interference proves wave-particle duality.

Common misconceptions

Dark fringes mean light is destroyed — energy is redistributed, not lost. Bright fringes get extra energy.
Interference only works with lasers — any coherent source works; Young used filtered sunlight.
Wider slits give sharper fringes — wider slits reduce coherence and blur the pattern.

Experimental verification

Thomas Young's 1801 experiment using sunlight through two pinholes. Modern: laser double-slit experiments reproduce the pattern precisely. Single-photon double-slit experiments confirm wave-particle duality at the quantum level.

Derivation

Two coherent sources separated by distance d emit waves.

At angle

\theta

, the path difference

is \Delta = d\cdot \sin (\theta )

.

Constructive interference (bright fringe) occurs when

\Delta = m\lambda

, giving

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

.

For small angles

(\sin \theta \approx \tan \theta = y/L)

, the fringe position is

y_m = m\lambda L/d

.

Limiting cases

d \to \infty

⟶

\theta \to 0

Very wide slit separation compresses fringes to the center.

\lambda \to 0

⟶

\theta \to 0

Short wavelength means fringes crowd near the center — approaching ray optics.

m = 0

⟶

\theta = 0

Central maximum is always straight ahead regardless of wavelength.

What if…

What if you use white light instead of monochromatic?

Each wavelength produces its own fringe pattern at different spacings. The central maximum is white, but higher orders show rainbow-colored fringes with overlapping colors.

What if you close one slit?

The interference pattern disappears. You get a single-slit diffraction pattern instead — broader central maximum, no sharp fringes.

What if d decreases?

Fringes spread apart $(\Delta y = \lambda L/d)$ . In the limit $d \to 0$ , fringes become infinitely wide — effectively one source.

1

Fringe spacing on a screen

Given ·

lambda:: 5.5e-7
d:: 0.0001
L:: 1
m:: 1

Find · y_1 (first bright fringe position)

Steps

Use small-angle formula: $y_m = m\lambda L/d$
Substitute: $y_1 = 1 \times 5.5\times 10^{-7} \times 1.0 / 1\times 10^{-4}$
$y_1 = 5.5\times 10^{-3} m = 5.5\,\mathrm{mm}$
Fringe spacing $\Delta y = \lambda L/d = 5.5\,\mathrm{mm}$ between consecutive bright fringes

Result ·

y_1 = m\lambda L/d = 1 \times 5.5\times 10^{-7} \times 1.0 / 1\times 10^{-4} = 5.5\times 10^{-3} m = 5.5\,\mathrm{mm}

2

Finding wavelength from fringe pattern

Given ·

d:: 0.0002
L:: 1.5
y spacing:: 0.004

Find · lambda

Steps

From $\Delta y = \lambda L/d$ , rearrange: $\lambda = \Delta y\cdot d/L$
$\lambda = 0.004 \times 2\times 10^{-4} / 1.5$
$\lambda = 5.33\times 10^{-7} m \approx 533\,\mathrm{nm}$
This falls in the green part of the visible spectrum

Result ·

\lambda = \Delta y \times d / L = 0.004 \times 2\times 10^{-4} / 1.5 = 5.33\times 10^{-7} m \approx 533\,\mathrm{nm}

(green)

Single Slit Diffraction→Wave Speed Equation→