Interference and Diffraction

Concise answer

All interference problems reduce to counting path difference in wavelengths: $d\sin\theta = m\lambda$ gives double-slit and grating maxima, $a\sin\theta = m\lambda$ gives single-slit minima, and thin films add the wavelength-in-film and reflection phase shifts to the same bookkeeping.

Definitions

Path difference: The difference in distance traveled by two interfering waves; measured in wavelengths, it decides constructive versus destructive interference.
Constructive interference: Superposition in phase — path difference a whole number of wavelengths (plus any reflection shifts) — giving maximum intensity.
Destructive interference: Superposition out of phase — path difference an odd number of half-wavelengths — giving minimum intensity.
Diffraction grating: Many equally spaced slits; spacing $d$ is the inverse of the line density, and the many slits make the maxima sharp and bright.
Reflection phase shift: The half-wavelength ($\pi$) phase change a wave acquires when reflecting off a medium of higher refractive index; reflection off a lower index gives no shift.

Intuition

Every setup is the same question: by how many wavelengths do the two paths differ? Whole wavelengths reinforce; half-integer offsets cancel. Geometry (slits, films, gratings) only changes how the path difference is computed.

The double-slit condition marks maxima, but the single-slit condition marks minima. The reason: a single slit is a continuum of sources, and at $a\sin\theta = \lambda$ the slit splits into halves that cancel pairwise — the same algebra, opposite meaning.

Concept walkthrough

Two slits separated by $d$ produce bright fringes where $d\sin\theta = m\lambda$ for $m = 0, \pm 1, \pm 2, \ldots$ . For a distant screen and small angles the fringes are equally spaced with $\Delta y = \lambda L/d$ . A grating obeys the same maxima condition with $d$ the line spacing (the inverse of lines per meter); with many slits the maxima sharpen into bright, narrow lines, which is what makes gratings good spectrometers.

A single slit of width $a$ produces dark fringes where $a\sin\theta = m\lambda$ for $m = \pm 1, \pm 2, \ldots$ (no $m = 0$ minimum — that is the central maximum, twice the width of the side lobes). In a real double-slit experiment the single-slit envelope modulates the double-slit fringes: separation $d$ sets the fine fringe spacing, width $a$ sets the envelope.

Thin films add two wrinkles to the path-difference bookkeeping. First, inside a film of index $n$ the wavelength is $\lambda/n$ , so the extra path $2t$ is counted in film wavelengths: the interference is governed by $2nt$ . Second, each reflection off a higher-index medium adds a half-wavelength shift. With exactly one such shift (a soap film in air, an anti-reflective coating between air and higher-index glass), constructive reflection requires $2nt = (m + \tfrac{1}{2})\lambda$ ; with zero or two shifts the roles of the conditions swap.

After this page, you should be able to

Locate double-slit and grating maxima with $d\sin\theta = m\lambda$ and convert to screen positions in the small-angle limit.
Locate single-slit minima with $a\sin\theta = m\lambda$ and explain why the central maximum is twice as wide as the others.
Account for the wavelength inside a film ($\lambda/n$) and reflection phase shifts in thin-film problems.
Distinguish the roles of slit separation, slit width, and slit count in a fringe pattern.

Formulas and assumptions

Double-slit maxima

d\sin\theta = m\lambda

Variables

d: slit separation in meters
theta: angle from the central axis
m: order of the bright fringe
lambda: wavelength in meters

Assumptions

The screen is far compared with the slit separation.
The same condition with half-integer m locates the dark fringes.

Fringe spacing (small angles)

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

Variables

Delta y: distance between adjacent bright fringes in meters
L: slit-to-screen distance in meters
d: slit separation in meters
lambda: wavelength in meters

Assumptions

Small angles: sin(theta) is approximately tan(theta).
Fringes are equally spaced only in this limit.

Diffraction grating maxima

d\sin\theta = m\lambda,\quad d = \dfrac{1}{N}

Variables

d: spacing between adjacent grating lines in meters
N: line density in lines per meter
m: diffraction order
lambda: wavelength in meters

Assumptions

Same maxima condition as the double slit; many slits make the maxima far sharper.
Orders exist only while |m| lambda / d is at most 1.

Single-slit minima

a\sin\theta = m\lambda \quad (m \ne 0)

Variables

a: slit width in meters
theta: angle from the central axis
m: order of the dark fringe
lambda: wavelength in meters

Assumptions

This condition locates minima, not maxima.
m = 0 is excluded; the center of the pattern is the brightest point.

Thin-film constructive reflection (one phase shift)

2nt = \left(m + \tfrac{1}{2}\right)\lambda

Variables

n: refractive index of the film
t: film thickness in meters
lambda: vacuum wavelength in meters
m: order 0, 1, 2, ...

Assumptions

Near-normal incidence.
Exactly one of the two reflections has a half-wavelength phase shift (e.g. a soap film in air); with zero or two shifts, this condition gives destructive reflection instead.

Worked example

Fringe spacing in a double-slit experiment

Light of wavelength $600\,\text{nm}$ passes through two slits separated by $d = 0.30\,\text{mm}$ , and fringes form on a screen $L = 2.0\,\text{m}$ away. Find the fringe spacing and the distance of the third-order bright fringe from the center.

1Check the small-angle limit: $\sin\theta_1 = \lambda/d = (6.0\times 10^{-7})/(3.0\times 10^{-4}) = 2.0\times 10^{-3}$ , tiny, so $\sin\theta \approx \tan\theta$ is safe.
2Fringe spacing: $\Delta y = \dfrac{\lambda L}{d} = \dfrac{(6.0\times 10^{-7})(2.0)}{3.0\times 10^{-4}} = \dfrac{1.2\times 10^{-6}}{3.0\times 10^{-4}} = 4.0\times 10^{-3}\,\text{m}$ .
3So adjacent bright fringes are $4.0\,\text{mm}$ apart.
4Third-order fringe: $y_3 = 3\Delta y = 3(4.0\,\text{mm}) = 12\,\text{mm}$ from the central maximum.

The fringe spacing is $4.0\,\text{mm}$ , and the $m = 3$ bright fringe sits $12\,\text{mm}$ from the center of the pattern.

Common traps

Using $a\sin\theta = m\lambda$ to find single-slit maxima; for a single slit that condition locates the dark fringes.
Confusing slit separation $d$ with slit width $a$ ; $d$ sets the fine fringe spacing while $a$ sets the diffraction envelope.
Forgetting the wavelength shortens to $\lambda/n$ inside a film, so the film condition uses $2nt$ , not $2t$ .
Dropping a reflection phase shift: each reflection off a higher-index medium adds half a wavelength, and only the net number of shifts matters.
Expecting arbitrarily high grating orders; once $m\lambda/d$ exceeds 1 the order simply does not exist.

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Sources

GRE Subject Test Content and Structure — ETS. Accessed 2026-07-06. Use as a cited source for exam facts; do not imply affiliation or reproduce protected test material.
University Physics Volume 3, Section 3.1: Young's Double-Slit Interference — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 3, Section 3.4: Interference in Thin Films — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 3, Section 4.1: Single-Slit Diffraction — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 3, Section 4.4: Diffraction Gratings — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.

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Last source check: 2026-08-02
Next scheduled review: 2026-11-02

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Interference and Diffraction

Concise answer

Definitions

Intuition

Concept walkthrough

After this page, you should be able to

Formulas and assumptions

Double-slit maxima

Fringe spacing (small angles)

Diffraction grating maxima

Single-slit minima

Thin-film constructive reflection (one phase shift)

Worked example

Fringe spacing in a double-slit experiment

Common traps

Related pages and practice

Sources

Review and maintenance