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The angle of incidence is the angle between the incident light ray and the line of the surface normal vector at the point of incidence. Use the acute angle on the incident side, with $0\le {\theta }_i<90^{\circ }$ for a nongrazing incident light ray. An angle measured from the surface itself is its complement $90^{\circ }-{\theta }_i$. The same convention applies to wavevectors in the transparent media with isotropy considered here.

A beam splitter is an optical instrument that routes an incident beam of light into transmitted and reflected beams. It also combines incident complex amplitudes: two inputs can contribute to the same output. A common geometry has four optical ports, with two chosen as inputs and two as outputs. A beam splitter changes the field; a photodetector supplies a measurement.^[ref]

For one illuminated input of a passive beam splitter, at specified wavelength, angle of incidence and electromagnetic polarization, define power transmittance $T=P_{\mathrm{transmitted}}/P_{\mathrm{in}}$, reflectance $R=P_{\mathrm{reflected}}/P_{\mathrm{in}}$ and radiometric absorptance $A=P_{\mathrm{absorbed}}/P_{\mathrm{in}}$. In steady operation, with no gain or other source, no net stored energy change, and all escaping light included in the transmitted or reflected collection,This is conservation of energy divided by incident power. If $T$ and $R$ count only the intended output beams, include a separate scattered or otherwise uncollected fraction $L$: then $T+R+A+L=1$. Missing output power need not be wave absorption. For an original specified example, $P_{\mathrm{in}}=10\mathrm{mW}$, $T=0.45$, $R=0.50$ and $A=0.05$ give $4.5\mathrm{mW}$ transmitted, $5.0\mathrm{mW}$ reflected and $0.5\mathrm{mW}$ absorbed; these are assumed values, not measurements.

A 50:50 beam splitter specifies equal output powers for one input, not equal complex amplitudes without a relative wave phase. Recombination needs the beam splitter scattering matrix, including its wave phases. A nonpolarizing beam splitter approximately divides both orthogonal linear polarizations in the same power ratio over its specified operating range; a polarizing beam splitter deliberately routes the two differently.

The useful surprise is that the same component divides power with one input and steers power by wave interference with two inputs. In a Michelson interferometer or Mach-Zehnder interferometer, mirror motion changes relative wave phase, turning a small change in optical path length into a measurable change in output power.

For refractive indices $n_{-}>n_{+}$, the critical angle for total internal reflection satisfies $\sin {\theta }_c=n_{+}/n_{-}$. Above it, the transmitted normal wavenumber is imaginary and the transmitted field is an evanescent wave.

The critical angle for total internal reflection is the boundary between outgoing refraction and an evanescent wave, rather than an angle at which the external electric field abruptly disappears. Write $n_1=n_{-}$ and $n_2=n_{+}$. At the threshold, Snell law for electromagnetic waves has ${\theta }_t=90^{\circ }$, henceHere inverse sine selects $0<{\theta }_c<90^{\circ }$. For $n_1=1.5$ and $n_2=1$, $\sin {\theta }_c=2/3$ and ${\theta }_c\approx 41.8103149^{\circ }$. The threshold is independent of polarization of an electromagnetic wave under the stated assumptions.

A short proof using phase matching at a planar wave interface reveals all three cases at once. Choose a Cartesian coordinate system with boundary $z=0$, incident medium $z<0$, and transmitted medium $z>0$. With vacuum wavenumber $k_0=2\pi /{\lambda }_0$, the shared tangential wavevector component is $q=n_1{k}_0\sin \theta$. The dispersion relation in medium two givesBelow ${\theta }_c$, choose the positive real square root for an outgoing plane electromagnetic wave. At ${\theta }_c$, $k_{2z}=0$: there is a grazing wave, with zero normal average power but no exponential confinement in $z$. Above ${\theta }_c$, choose $k_{2z}=i\kappa$ with $\kappa >0$, so the wave amplitude decays into the second half-space.

If $n_1\le n_2$, no nongrazing angle of incidence produces this threshold. As $n_2/n_1\to 1$ from below, ${\theta }_c\to 90^{\circ }$; as that ratio tends to zero, ${\theta }_c\to 0$. These limits explain why a larger refractive index contrast permits confinement over a wider range of internal angles.

Define the amplitude evanescent penetration depth as the distance at which the electric field wave amplitude falls to $1/e$ of its surface value:The squared-field envelope decays twice as fast:The same normalized exponential decay describes the tangential average Poynting vector for the TE polarization above. An intensity penetration depth convention often means $d_{1/e}$; it must not be confused with amplitude depth $\delta$ or interpreted as outgoing normal light intensity.

For the stipulated $n_1=1.5$, $n_2=1$, ${\lambda }_0=633\mathrm{nm}$ and $\theta =50^{\circ }$,At $z=200\mathrm{nm}$ the normalized wave amplitude is approximately $0.325099$, and its squared envelope is approximately $0.105689$. At $z=\delta$, these become $e^{-1}$ and $e^{-2}$, respectively. The quantity inside the square root is dimensionless, so $\kappa$ has inverse-length units and both depths have length units.

The sharp angular dependence has a simple explanation. For $\theta ={\theta }_c+\epsilon$ with $\epsilon >0$ small and measured in radians, a first-order Taylor expansion givesThus $\delta \to \infty$ as $\theta \downarrow {\theta }_c$: exponential confinement disappears continuously. With fixed refractive indices, $\delta$ decreases as the angle of incidence increases and approaches ${\lambda }_0/(2\pi \sqrt{n_1^2-n_2^2})\approx 90.1091\mathrm{nm}$ as $\theta \to 90^{\circ }$. This is a limit of the normalized depth; exactly grazing incidence supplies zero incident normal power. Real beam widths, angular spreads and optical dispersion matter when applying these ideal plane electromagnetic wave predictions near the threshold.

A Fabry–Pérot interferometer uses repeated specular reflections between two nearly parallel, partially transmitting surfaces. Its transmitted light is the sum of contributions that have made zero, one, two, or more additional round trips. Their complex amplitudes reinforce only near particular wave phases, producing narrow transmission peaks that select wavelength intervals. Scanning the spacing or angle of incidence can build an imaging spectral cube: an image at each selected wavelength.

The surprising feature is that highly reflecting mirrors can together transmit all the incident power at resonance in the ideal symmetric, lossless model. Each individual contribution is weak, but many contributions arrive in phase. Between resonances the same contributions cancel strongly. Increasing mirror reflectivity sharpens the peaks instead of simply suppressing every transmitted signal; finesse quantifies that selectivity. An optical spectrum with nearby spectral lines can therefore produce distinguishable peaks, provided their separation is large enough relative to the instrument's peak width and their orders are identified.

Charles Fabry and Alfred Perot developed the instrument together; Optica dates its development from 1896 to 1898.^[ref]

At normal incidence, with no optical dispersion and zero round-trip reflection wave phase modulo $2\pi$, the resonance condition in Fabry–Pérot transmission from multiple reflections isHere $\nu$ is frequency and $c$ is the speed of light in vacuum. A change of $\nu$ by one free spectral range advances the round-trip wave phase by $2\pi$. At fixed internal angle of incidence $\theta$, the same nondispersive model gives $c/(2nd\cos \theta )$. Holding the external angle of incidence fixed in a dispersive medium need not hold the internal $\theta$ fixed, because of Snell's law.

For stipulated $n=1$ and $d=10\mathrm{mm}$, the free spectral range is $14.9896229\mathrm{GHz}$ using $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. A frequency difference of $0.10\Delta {\nu }_{\mathrm{FSR}}$ is therefore $1.49896229\mathrm{GHz}$. Transmission repeats every free spectral range: one isolated scan cannot distinguish that separation from $1.10\Delta {\nu }_{\mathrm{FSR}}$. Independent knowledge of the optical spectrum, an additional filter, or a changed mirror separation is needed to assign orders.

Equal spacing in frequency does not imply equal spacing in wavelength. At normal incidence the exact adjacent resonance wavelengths obeyFor example, choosing $2nd=10{\lambda }_{10}$ makes the exact gap to order 11 equal to ${\lambda }_{10}/11$, while the high-order estimate gives ${\lambda }_{10}/10$. The approximation is already ten percent larger at order 10. With optical dispersion or frequency-dependent reflection wave phase, obtain local spacings from the total phase difference instead: $\Delta \nu \simeq 2\pi /(d\delta /d\nu )$ when its derivative varies little over one order.

Consider two identical, lossless parallel mirrors with intensity reflectivity $0\le R<1$, separation $d$, and a uniform transparent medium of refractive index $n$. Fix one polarization of an electromagnetic wave and an internal angle of incidence $\theta$ measured from the mirror normal vector. Illuminate with a monochromatic wave having planar wavefronts; the emerging contributions must overlap coherently in the detected direction. Define $T=I_{\mathrm{transmitted}}/I_{\mathrm{incident}}$ using the same incident and exit media. Each mirror transmits a fraction $1-R$ of incident power. Its reflected wave amplitude has magnitude $\sqrt{R}$, so two internal specular reflections multiply the circulating complex amplitude by magnitude $R$, not $R^2$.

Let $\lambda$ be the vacuum wavelength and choose the reflection wave phase convention in which the product of the two internal amplitude reflection coefficients is real and positive. The phase difference between successive transmitted contributions, evaluated on a common outgoing wavefront, isWith other mirror coatings or phase conventions, use the total $\delta =4\pi nd\cos \theta /\lambda +{\varphi }_{\mathrm{rt}}$, where ${\varphi }_{\mathrm{rt}}$ is the sum of the two internal reflection wave phases. The stated resonance positions presume ${\varphi }_{\mathrm{rt}}=0$ modulo $2\pi$; a constant nonzero ${\varphi }_{\mathrm{rt}}$ shifts them, while a frequency-dependent one can also change their spacing.

Why does the cosine appear rather than its reciprocal? Inside the gap, the extra two legs have optical path length $2nd/\cos \theta$. But the later emerging contribution starts $2d\tan \theta$ farther along the mirror. Comparing both on one outgoing wavefront subtracts $2nd\tan \theta \sin \theta$ from that path difference, leaving $2nd\cos \theta$. Equivalently, a plane wave's mirror-normal wavevector component is $k_z=(2\pi n/\lambda )\cos \theta$, so the round-trip wave phase is $2k_{z}d$. The diagram uses the same medium outside and inside to show the common-wavefront subtraction directly; at a refracting exit interface, tangential wavevector matching in Snell's law supplies the same result.

Absorb the common first-pass wave phase into the incident complex amplitude $E_{\mathrm{in}}$. Successive transmitted contributions are $(1-R)E_{\mathrm{in}}$, $(1-R)E_{\mathrm{in}}R{e}^{i\delta }$, and so on. The superposition principle and the convergent geometric series giveThe last step expands the squared complex modulus as $1+R^2-2R\cos \delta$. This transmission profile is often called the Airy transmission function; it is distinct from the Airy disk, a diffraction image of a circular aperture. The University of Washington's lab handout gives the same transmission profile and distinguishes mirror reflectivity from other sources of peak broadening.^[ref]

For an original example, set $R=3/4$. At $\delta =0$, the amplitudes add to $(1/4)/(1-3/4)=1$, so $T=1$. At $\delta =\pi$, they alternate in sign and add to $(1/4)/(1+3/4)=1/7$, so $T=1/49$. At $\delta =\pi /2$, the amplitude is $4/25+3i/25$ and $T=1/25$. Adding the individual intensities instead would give $(1-R{)}^2{\sum }_j{R}^{2j}=(1-R)/(1+R)=1/7$, independent of wave phase; that discards the wave interference responsible for the peaks.

In general the maximum is $T_{\max }=1$ at $\delta =2\pi m$, with integer $m$, and the minimum is $T_{\min }=[(1-R)/(1+R){]}^2$ halfway between peaks. Perfect resonant transmission requires identical lossless mirrors, coherent illumination, and sufficient time to establish the steady field. Absorption, unequal mirror transmission, finite beams that fail to overlap, and fluctuations of $d$ can reduce it. The exactly reflecting case $R=1$ is excluded: no incident power enters through a perfectly reflecting mirror, and the resonant limit $R\to 1^{-}$ is singular.

For a light ray launched on the centreline in the plane of a circular optical fiber bend, let $R$ be centreline radius, $r$ core radius and $\beta$ the internal launch angle from the tangent. The straight segment's distance from the bend centre is $R\cos \beta$. Thus its outer-wall angle of incidence satisfies $\sin i=R\cos \beta /(R+r)$. Equating this to $n_2/n_1$ gives the limiting total internal reflection threshold in the title, provided $n_1\cos \beta >n_2$. Radially displaced or nonplanar launches need different geometry; this is not a universal wave-optical loss cutoff.

A nonpolarizing beam splitter is designed for approximately equal power splitting ratios for the two orthogonal linear polarizations, at specified wavelengths and angle of incidence. It need not be 50:50, lossless, or have equal transmission and reflection wave phases for both linear polarizations. A scalar beam splitter scattering matrix therefore still requires a specified electromagnetic polarization and operating range. Polarizing beam splitters instead aim to separate those linear polarizations into different paths.

Snell's law is $n_a\sin {\theta }_a=n_b\sin {\theta }_b$, with both angles measured from the interface normal vector. Let $\psi$ be the meridional ray's angle to the optical axis inside the core and let $n_0$ be the incident medium's refractive index. At the entrance face,At the cylindrical wall the angle of incidence is $i=\pi /2-\psi$. For $n_{\mathrm{inner}}>n_{\mathrm{outer}}$, the critical angle for total internal reflection satisfies $\sin i_c=n_{\mathrm{outer}}/n_{\mathrm{inner}}$. At the limiting accepted ray,Consequently the optical fiber acceptance angle obeysThe requested expression follows for incidence from air, $n_0\simeq 1$, provided the right-hand side is at most one. Strict total internal reflection occurs inside this limiting cone. The derivation uses the standard meridional ray model; skew launches need additional geometry.

If $\sqrt{n_{\mathrm{inner}}^2-n_{\mathrm{outer}}^2}>n_0$, every forward entrance angle allowed by this ideal model meets the side-wall condition, and the limiting half-angle is $\pi /2$, rather than an undefined inverse sine. For example, $n_{\mathrm{inner}}=1.6$, $n_{\mathrm{outer}}=1.1$, $n_0=1$ gives $\sqrt{1.35}>1$. Thus the unsaturated printed formula is not valid for every possible pair of refractive indices.

Regard the refraction angle as a function of the angle of incidence, $\beta =\arcsin (\sin \alpha /n)$. Implicit differentiation of Snell's law gives ${\beta }^{\prime }=\cos \alpha /(n\cos \beta )$. Differentiating the deviation angle gives ${\delta }^{\prime }=2-4{\beta }^{\prime }$. An interior stationary point therefore satisfiesThe stationary deviation of a primary rainbow ray is a minimum: differentiating $\cos \alpha =n\cos \beta {\beta }^{\prime }$ again and using Snell's law gives ${\beta }^{\prime \prime }=\tan \beta (({\beta }^{\prime }{)}^2-1)$. At the stationary point, ${\beta }^{\prime }=1/2$, and hence ${\delta }^{\prime \prime }=-4{\beta }^{\prime \prime }=3\tan \beta >0$ for a positive interior refraction angle. This minimum is responsible for the concentration of emerging light rays near the primary rainbow direction.

Square the stationary condition and eliminate the angle of incidence using Snell's law:Rearranging yields $3n^2{\sin }^2\beta =4-n^2$. Taking the positive root for a physical incident light ray,An interior minimum exists for $1<n<2$, which includes water at visible wavelengths. For illustration, $n=4/3$ gives $\beta =40.20^{\circ }$, $\alpha =59.39^{\circ }$ and ${\delta }_{\min }=137.97^{\circ }$. The corresponding angle from the antisolar direction is $180^{\circ }-{\delta }_{\min }=42.03^{\circ }$. At $n=2$ the stationary solution reaches the endpoint, and for $n>2$ the displayed square root is not real; the water result is not a formula for every possible refractive index.

With the Sun behind the observer and raindrops ahead, the one-reflection branch sends many light rays toward directions close to the minimum deviation angle. Nearby values of the angle of incidence produce nearly the same outgoing direction there, forming a wave caustic. For water the resulting cone is approximately $42^{\circ }$ from the antisolar direction. The observer sees the intersecting drops as a circular rainbow, or the part of that circle visible above the ground.

Normal dispersion makes the refractive index larger for violet than for red light. At the minimum, increasing the refractive index increases the deviation angle, and therefore decreases the viewing angle $\pi -{\delta }_{\min }$. Red is on the outer edge of the primary rainbow, violet on the inner edge. The internal specular reflection is partial: with $n=4/3$, the relevant $\beta \simeq 40.2^{\circ }$ is below the critical angle $\arcsin (1/n)\simeq 48.6^{\circ }$. Diffraction and the finite angular size of the Sun smooth the geometrical concentration; the physical intensity does not actually become infinite.

In the Huygens-Fresnel principle construction, after a time $\Delta t$ the incident disturbance advances $v_1\Delta t$ while the secondary disturbance in the second medium advances $v_2\Delta t$. The two right triangles share the segment along the interface. HenceHere both angles are measured from the normal vector, and $v_j=c/n_j$ are phase velocities. This is Snell's law. For $n_1>n_2$, no real transmitted angle exists whenThis is total internal reflection; equality is the limiting grazing geometry. The statement concerns an ideal transparent interface, not material absorption or losses at a nearby second boundary.

Let $\beta$ be the internal angle to the optical axis in the straight step-index optical fiber. At its flat entrance, Snell's law gives $\sin {\theta }_i=n_1\sin \beta$. At the side wall a meridional ray has angle of incidence $\pi /2-\beta$. The limiting total internal reflection condition is $\cos \beta =n_2/n_1$. Thus the optical fiber acceptance angle satisfiesThe numerical aperture expression assumes the square root is at most one for air; otherwise all forward launch angles are admitted by this ideal ray test. Strictly confined meridional rays lie inside the limiting cone.

For $n_1=1.48$ and $n_2=1.46$, $\mathrm{NA}=0.242487$ and ${\theta }_{i,\max }=14.0334^{\circ }$. A light ray at internal angle $\beta$ travels distance $L/\cos \beta$ along a guide of axial length $L$. In the nondispersive approximation its travel time is $t=n_{1}L/(c\cos \beta )$. ThereforeHere $L=1000\mathrm{m}$ and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$; using $3.00\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$ gives $5.001\mu \mathrm{s}$ and $4.950\mu \mathrm{s}$. The halving applies to the external launch angle, not to $\beta$. For actual short pulses the group refractive index $n_{\mathrm{g}}$ replaces the prefactor $n_1$ in $t=n_{\mathrm{g}}L/(c\cos \beta )$; the source supplies no optical dispersion data. The difference calculated here is geometrical intermodal dispersion in a nondispersive geometrical optics model.

Bending lowers the outer-wall angle of incidence, allowing a previously guided light ray to escape into the cladding. To calculate a definite meridional ray bend threshold, continue the on-axis launch assumption and take a meridional ray in the plane of the circular bend. Its initial internal angle to the centreline tangent obeys $\sin \beta =\sin 10^{\circ }/n_1$. Its straight path has distance $b_{\mathrm{ray}}=R\cos \beta$ from the bend centre. At the outer wall of radius $R+r$, elementary right-triangle geometry givesAt the loss threshold, this equals $n_2/n_1$. Solving for the centreline radius givesFor this launch, $R<R_c$ violates the total internal reflection condition at the outer wall; $R=R_c$ is the limiting case. A real optical fiber has gradual bend loss governed by its electromagnetic wave modes, rather than a universal abrupt cutoff.

Launch dependence. The final sentence does not separately specify the entry position or orientation of the bend relative to the incident light ray. For a planar launch displaced by $u$ from the centreline toward the outer wall, replace the numerator of $\sin i_{\mathrm{out}}$ by $(R+u)\cos \beta$. For example, a launch at the inner edge, $u=-r$, instead givesThus $14.93\mathrm{mm}$ is the result for the stated on-axis continuation; it is not an entry-independent threshold. No missing figure is needed for these explicitly specified ray models.

For a sinusoidal plane electromagnetic wave in vacuum, write the maximum electric field as $E_0$. The magnetic field maximum is $B_0=E_0/c$, so the cycle-averaged Poynting vector magnitude, or light intensity, isEach photon carries energy $hf$, where $h$ is the Planck constant. Therefore the photon number flux density, namely the number per unit normal area per unit time, isMultiplying by an illuminated normal area gives the photon arrival rate. Each absorbed photon transfers momentum $hf/c$ at normal incidence, giving the radiation pressure on a perfectly absorbing surfaceThese are time-averaged quantities. At oblique angle of incidence $\theta$, the normal radiation pressure is $I{\cos }^2\theta /c$; the simple displayed expression assumes normal incidence.

The Huygens-Fresnel principle treats every point of an advancing wavefront as a source of secondary spherical waves; their forward envelope constructs the next wavefront. Consider two points where an oblique plane wavefront meets a plane interface a distance $AB$ apart. After its first point reaches $A$, let time $t$ elapse until the second reaches $B$. In medium 1, the incident travel distance is $v_{1}t=AB\sin i$. Meanwhile, a secondary wave from $A$ has travelled distance $v_{2}t$ in medium 2. The tangent from $B$ to that secondary wavefront gives $v_{2}t=AB\sin r$. ThereforeHere $i,r$ are the angle of incidence and angle of refraction, measured from the interface normal vector, and $n_j=c/v_j$ are the refractive indices at the common frequency. This is Snell's law.

Because optical dispersion makes the refractive index depend on wavelength, no single angle of incidence makes every component of white light travel parallel to the base. Choose the supplied central visible wavelength $535\mathrm{nm}$, with $n_0=1.86$, as the design component. Snell's law givesKeep this angle of incidence fixed for all colours. The other components haveSubstitution givesAll three emerge: the largest value of $n\sin r_2$ is about $0.9323<1$. The angular spread between the extreme supplied wavelengths isThe respective deviations are $72.988^{\circ }$, $63.376^{\circ }$ and $60.398^{\circ }$. Applying the symmetric formula independently to the two extreme refractive indices would instead give $72.221^{\circ }-60.344^{\circ }=11.877^{\circ }$; that compares differently oriented optical prisms, not the emerging components of one fixed white-light beam. It can be a rough approximation but misses the stated fixed-incidence geometry. The supplied $370\mathrm{nm}$ value is near ultraviolet, so the computed spread refers to the entire supplied band, not a universally defined visible-white spectrum.

The distance from the origin to the incident straight light ray is $d_a$ (taking $d_a\ge 0$). At its intersection with the sphere, let $i$ and $t$ be the angle of incidence and angle of refraction, measured from the radial normal vector. Elementary right triangle geometry gives $d_a=r_b\sin i$. By Snell's law,The refracted straight light ray inside the sphere has distance $d_b=r_b\sin t$ from the origin. ThereforeThe angle ${\theta }_b$ describes the normal vector of the new line; it need not equal the propagation angle. The existence of a crossing requires $|d_a|<r_b$. A signed convention for $d_a$ gives the same line equation after choosing the orientation of its normal vector consistently.

A 2013 experiment by Paul A. Williams and collaborators mounted a 25-mm dielectric mirror on a commercial force-restoration balance. Removing its pan-weight compensation spring allowed horizontal force sensing with a horizontal laser path. They measured mirror optical reflectance and varied incidence from about $7^{\circ }$ to $45^{\circ }$. A 1071-nm laser, capable of 530 W, was switched at 0.1 Hz. Digitized mass-equivalent readings were converted to force using $g$; quadrature demodulation isolated the switched signal from drift. ^[ref]

Measurement calibration used a 20-g mass, with checks down to 1 mg. The optical forces lay below that checked range, limiting the accuracy claim. An angle scan tested the predicted cosine dependence at fixed intercepted power. Replacing the mirror with a carbon nanotube coated absorber increased absorbed power while reducing the predicted optical force; no modulation peak was detected. This bounded gas-heating artifacts under those conditions, rather than proving their universal absence. The fitted slope was about 7% below $2/c$. Inferring laser power requires calibrated force, optical reflectance, radiometric absorptance and angle of incidence, with measurement uncertainty for each. ^[ref]

For normal incidence in vacuum, a uniform beam delivers energy $IA\Delta t$ to an illuminated surface area $A$ during time $\Delta t$. A perfect absorber receives momentum $IA\Delta t/c$, so force divided by surface area gives $p_{\mathrm{abs}}=I/c$. A perfect stationary mirror sends the photons back with reversed momentum, so momentum conservation gives twice the normal transfer, $p_{\mathrm{refl}}=2I/c$. These are time-averaged pressures; the microscopic arrivals need not be uniform.

For a stipulated model light intensity $I=1000\mathrm{W}{\mathrm{m}}^{-2}$ and the exact SI speed of light $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$,These are computed predictions, not measured data or a universal sunlight value. If the beam fully covers a normally oriented surface area $A=0.010{\mathrm{m}}^2$, the intercepted radiant flux is $\Phi =IA=10\mathrm{W}$ and the absorber's force is $\Phi /c\simeq 33.356\mathrm{nN}$. Using the total source power would be wrong if only part of its beam reaches the receiver. The absorption prediction assumes complete photon absorption, while the reflection prediction assumes complete specular reflection; transmission, scattering of light and directed thermal photon emission require their own outgoing momentum accounting.

Measure the angle of incidence $\theta$ between the incoming path and the surface's normal line, with $0\le \theta <\pi /2$. There are two separate cosine factors. A fixed actual surface area $A$ intercepts only its projected area $A\cos \theta$, so $\Phi =IA\cos \theta$. Each incoming photon supplies normal momentum $(E_{\gamma }/c)\cos \theta$ on absorption, or twice that on specular reflection. Thus, for radiometric absorptance $a$ and reflected power fraction $R$,This model assumes transmitted light continues straight in the same medium and carries away its original momentum. It omits thermal recoil and assumes a stationary receiver. At $\theta =60^{\circ }$, fixed beam light intensity therefore gives one quarter of the normal-incidence pressure. For fixed intercepted radiant flux $\Phi$, however, the normal force has only one cosine factor. Equivalently, if the given quantity is receiving-surface irradiance $E_{\mathrm{inc}}=I\cos \theta$, then $p_{\perp }=(a+2R)E_{\mathrm{inc}}\cos \theta /c$. The distinction matters when interpreting an experiment's angle scan. ^[ref]

Absorption also transfers a tangential momentum component at oblique incidence. It contributes a tangential force, rather than normal pressure. Ideal specular reflection preserves the tangential momentum component, so it transfers only normal momentum.