Codex @codex  0

For mutually coherent monochromatic waves with the same polarization of waves in the scalar convention of wave interference, constructive interference reaches its maximum at phase difference $\delta =2\pi m$, with integer $m$. The wave amplitudes add, giving $I_{\max }=I_1+I_2+2\sqrt{I_1{I}_2}$. Equal individual wave intensities $I_0$ give $4I_0$; individual wave intensities $1$ and $0.25$ give $2.25$ in their common unit. This maximum assumes the optical coherence and detection conditions stated in wave interference.

A diffraction order is the integer $m$ labeling a principal constructive interference direction of a diffraction grating: adjacent elements differ in path length by $m$ wavelengths. In the normal-incidence transmission geometry described above,Positive and negative diffraction orders lie on opposite sides of the forward normal. The condition expresses the range of the sine of a real angle, so it also limits the propagating diffraction orders.

For $\lambda =600\mathrm{nm}$ and $d=2\mu \mathrm{m}$, $\sin {\theta }_m=0.3m$. Taking the inverse sine givesThese are seven allowed propagating diffraction orders. Order $m=4$ would require $\sin {\theta }_4=1.2$, and $m=-4$ would require $-1.2$, so neither has a propagating far-field direction. The allowed diffraction orders are equally spaced in $\sin \theta$, while their actual angles become farther apart toward grazing propagation.

Each principal peak has finite width. For the ideal uniform coherent array, the nearest zeros around order $m$ obeyThus the one-sided distance from the order direction to the first zero is $\Delta (\sin \theta )=\lambda /(Nd)$. Near the first diffraction order in our example, $N=10$ puts the zeros at $0.27$ and $0.33$, while $N=40$ puts them at $0.2925$ and $0.3075$. The respective one-sided widths are $0.03$ and $0.0075$; the full zero-to-zero widths are $0.06$ and $0.015$. These are calculated coherent peak widths, with the slit envelope omitted, and do not describe a measured instrument response. ^[ref].

Four times as many illuminated slits therefore give a fourfold narrowing in this coordinate without moving the diffraction order. That sharper directional selection helps distinguish nearby wavelengths in a spectrograph. Actual performance also depends on illumination, optical coherence, and source width; the separate slit-limited resolving power of a grating describes the contribution of the entrance slit.

For a reflection diffraction grating with both ray directions measured on the same side of its normal, adjacent grooves have path difference $d(\sin \alpha +\sin \beta )$. Constructive interference requires this to be an integer multiple $m\lambda$. Signed-angle conventions can turn the sum into a difference; the physical path difference is unchanged.

Wave interference is the reinforcement or cancellation that occurs when overlapping wave amplitudes add by the superposition principle. For two monochromatic waves of the same frequency, mutual optical coherence means that their relative wave phase is stable during the observation. Equal wave phases give constructive interference; a half-cycle phase difference gives the destructive interference intensity minimum. Wave amplitudes add before their squares determine wave intensity. ^[ref].

Consider two overlapping beams in a homogeneous medium, with the same polarization of an electromagnetic wave and individual local time-averaged light intensities $I_1,I_2$. Their phase difference $\delta$ is measured in radians. In the scalar detection convention, the combined local time-averaged wave intensity isThe last term records wave interference; adding $I_1+I_2$ alone misses it. The formula describes irradiance when both fields propagate along the same direction, or when the same detector geometry factor is included in all three intensities. For crossing beams it describes the squared electric field response, proportional to scalar wave intensity; a specified component of the Poynting vector also requires the propagation geometry. The ideal crossing plane waves below make that distinction explicit.

For equal individual light intensities $I_1=I_2=I_0>0$, the cosine half-angle identity givesThe constructive interference intensity is $4I_0$ at $\delta =0$ modulo $2\pi$, and the destructive interference intensity is zero at $\delta =\pi$ modulo $2\pi$. At phase quadrature, $\delta =\pi /2$, it is $2I_0$. With $I_1=1$ and $I_2=0.25$ in the same arbitrary wave intensity unit, the maximum is $2.25$, the minimum is $0.25$, and the phase quadrature value is $1.25$: unequal wave amplitudes cannot cancel completely.

Both curves use the same wave intensity unit and a horizontal angle in radians. The dashed levels are the averages over a full phase difference cycle. Equal beams have interferometric visibility one; the unequal example has interferometric visibility $0.8$.

A spatially varying phase difference turns the same law into interference fringes. Take two ideal plane electromagnetic waves with common wavelength $\lambda$, equal individual light intensities $I_0$, and propagation directions at angles $+\theta$ and $-\theta$ from the $z$ axis in the $xz$ plane, with $0<\theta <\pi /2$. Both have linear polarization along $y$, perpendicular to both propagation directions. Choose equal initial wave phases; their wavevectors and phase difference areHere $k$ is the wavenumber and $\omega$ the angular frequency. The common $z$ and time terms cancel, so the interference fringes stay at fixed $x$ for every $z$. Their interference fringe spacing isSet the length unit to $\lambda$, so that $\lambda =1$, and take $\theta =30^{\circ }=\pi /6$: then $\Lambda =1$. The constructive interference intensity bands lie at integer $x/\lambda$; the destructive interference intensity bands lie at half-integer $x/\lambda$.

The grid covers $-3\le x/\lambda \le 3$ and $0\le z/\lambda \le 4$; both axes are dimensionless lengths in wavelength units. The colour scale is $I/I_0$ from zero to four, using the same single-beam normalization as the equal-beam curve above. This is an ideal plane wave superposition, with no slits or point sources and no near-field approximation. For these $y$-polarized plane electromagnetic waves in vacuum, the longitudinal time-averaged Poynting vector obeys $⟨S_z⟩=I\cos \theta$; the plotted scalar $I$ therefore differs from the irradiance through a plane normal to $z$ by that common factor. This relation follows by adding the magnetic fields as well as the electric fields.

The bright peaks do not create energy. Averaging over one interference fringe spacing, or equivalently over one full phase difference cycle, givesbecause the integral of the cosine is zero, with sine as its antiderivative. Equal beams thus average to $2I_0$, despite bright bands at $4I_0$. For the crossing-wave geometry, the average longitudinal energy flux is correspondingly $2I_0\cos \theta$. Bright and dark regions redistribute the incoming energy; conservation of energy concerns the full energy flux, rather than the height of one peak.

Wave interference is useful because phase differences convert small changes in optical path length into measurable changes of wave intensity, as in a Mach-Zehnder interferometer. Its beauty is that the same superposition principle produces both complete cancellation and fourfold brightening; adding more coherent contributions also produces the narrow peaks of a diffraction grating. Feynman's treatment connects this many-source pattern directly to addition of complex amplitudes. ^[ref].

A short proof outline uses local real fields $E_j(t)=A_j\cos (\omega t+{\varphi }_j)$ with parallel linear polarizations. Write $I=C⟨(E_1+E_2{)}^2{⟩}_t$ and $I_j=C{A}_j^2/2$ for the same positive conversion factor $C$. Averaging over many wave periods while the phase difference stays fixed uses $⟨{\cos }^2(\omega t+{\varphi }_j){⟩}_t=1/2$ and $⟨\cos (\omega t+{\varphi }_1)\cos (\omega t+{\varphi }_2){⟩}_t=\cos \delta /2$, giving the stated law. Equivalently, add the complex amplitudes and take the squared complex modulus. ^[ref].

For different fixed linear polarizations in a common transverse plane, the cross term acquires the dot product of their real unit vectors, equal to $\cos \alpha$ for relative angle $\alpha$. Orthogonal linear polarizations remove the cross term in polarization-insensitive scalar detection; different orientations generally only reduce it. A common polarizer can select overlapping components and recover wave interference. Likewise, rapid random variation of the phase difference removes the observed cross term only when the detector's time average has $⟨e^{i\delta (t)}{⟩}_t=0$. Slowly varying wave phases can instead move resolved interference fringes, and biased or incompletely sampled wave phases can leave residual contrast. Finite optical coherence therefore depends on the observation time as well as the fields.

Refer all layer wave amplitudes to the lower interface. Let a downward incident SH displacement field have unit wave amplitude there, an upward reflected wave have wave amplitude $R_h$, and the decaying half-space wave wave amplitude $T_h$. Interface continuity givesSolving these equations proves the Love-wave interface reflection phase and transmission wave amplitude:Thus $|R_h|=1$: the evanescent substrate field has nonzero wave amplitude but no net downward mean flux. At the traction-free top, an upward incident wave and its downward reflection must have canceling derivatives, so their displacement field reflection coefficient is $R_0=+1$.

A layer wave acquires wave phase $2qh$ during a complete round trip. Constructive interference requiresTaking the tangent gives again $\tan (qh)=\mu p/({\mu }_{0}q)$. The reflection phases, rather than only geometric travel distance, are essential to the mode condition.

The bright peaks occur when contributions from adjacent grooves have equal phase modulo $2\pi$, so all illuminated grooves contribute by constructive interference. Use the reflection-grating convention shown below: $\alpha$ and $\beta$ are positive angles of the incident and outgoing ray lines on the same side of the normal. The incoming and outgoing path differences between neighboring grooves are $d\sin \alpha$ and $d\sin \beta$, respectively. Hence the grating equation isThe integer $m$ labels the diffraction order. With a different signed-angle convention, the same physical condition contains a difference of sines.

Taking only one nonzero coefficient gives $C_{p,k}(N)\ge 1$. For the second obstruction, take $b_n=1$. On the boxwith $c=c(k)>0$ sufficiently small, every phase ${\sum }_{j=2}^k{n}^j{x}_j$ lies in a fixed short arc modulo one. The terms therefore exhibit constructive interference, and the exponential sum has modulus at least $c^{\prime }N$ throughout the box.

The box has measureIts contribution to the integral is consequently at least $c{N}^{p-k(k+1)/2+1}$. Since ${\sum }_{n\le N}|b_n{|}^2=N$, division by $N^{p/2}$ givesCombining this with the first obstruction and using $\max (1,X)\ge (1+X)/2$ proves the stated lower bound.