Snell’s law of refraction

The normal is perpendicular to the interface; replace an angle measured from the surface by its complement to $90^\circ$ before substituting it. You need sine, inverse sine and the distinction between angles from the surface and from its normal. The normal is an imaginary perpendicular through the crossing point. Draw it first so both angles represent the quantities used in the formula. A sketch prevents more errors here than extra decimal places.[1]

Consider a stationary, locally planar interface between homogeneous isotropic media with positive real indices. A ray approximates light when wavelength is small compared with the relevant dimensions. “Optically denser” means larger refractive index here, not necessarily larger mass density. Use both indices at the same frequency.[1][4]

Snell’s law equates refractive index times the sine of the angle from the normal in the two media. Draw the interface normal through the point of incidence. α₁ belongs to the incoming ray in medium 1, α₂ to the transmitted ray in medium 2. Measure each from the normal inside its own medium. For ordinary transmission considered here, the angles lie between 0° and 90°. An angle measured from the surface is complementary and cannot be substituted unchanged.[1]

Number the media in actual propagation order rather than assigning fixed numbers to materials. On reversal, both media and angles exchange roles. For the ordinary reciprocal isotropic media used here, the geometric path reverses. A limiting ray running along the interface requires separate treatment of the grazing case. By contrast, the thin-lens equation describes image formation in the paraxial thin-lens model; it does not replace local refraction at an individual interface.[1][5]

A monochromatic plane wave has phase $\vec k\cdot\vec r-\omega t$. At a stationary interface, temporal and tangential spatial phases must agree so boundary conditions hold at every point and time. Frequency ω and the tangential component of the wavevector therefore match. This is a wave argument, not a claim that photons lose energy when crossing the boundary.[2]

An alternative varies optical path length until it is stationary. Fermat’s principle requires stationary travel time, not a global minimum in every optical system. At a simple planar boundary with positive indices, varying the crossing point gives the same relation. Historically, discovery of the relation and its later theoretical explanations should be distinguished.[4]

The dimensionless index $n=\frac{c}{v_{\mathrm{ph}}}$ refers to phase velocity. In dispersive media it depends on frequency; pulse group velocity is generally different. Frequency f stays unchanged at a stationary boundary, so wavelength becomes $\lambda=\frac{v_{\mathrm{ph}}}{f}=\frac{\lambda_0}{n}$. Use indices for the same vacuum wavelength or frequency and for the relevant material conditions.[2]

First isolate $\sin\alpha_2$ in Snell’s law, then apply the inverse sine. The example uses degrees; use the same degree mode for sine and inverse sine. A sine value above one allows no propagating refracted ray and must not be rounded down to one.[1]

For parallel plate faces surrounded by the same external medium, the two angular changes cancel. The exiting ray is parallel to the incident ray but lies on a displaced line. Inside the plate, the second crossing point lies at $d\tan\alpha_2$. The perpendicular distance between incident and emerging ray lines is a different quantity.[4]

For known n₁, the ratio $n_2=n_1\frac{\sin\alpha_1}{\sin\alpha_2}$ infers index from a ray pair. This is a measurement result, not a material property independent of color and temperature. Near-normal incidence is sensitive because two small sine values can amplify relative angle errors. Several incidence angles at fixed frequency should yield the same index; systematic variation suggests geometric or model errors.[1]

Snell’s law predicts propagation direction, not the transmitted power fraction. Reflection and transmission require Fresnel coefficients, polarization and the appropriate energy-flux factor. Partial reflection can occur even when refraction is allowed. Above the critical angle from higher to lower index there is no propagating refracted ray, but an evanescent field remains. Total internal reflection requires $n_1>n_2$ and incidence above the critical angle for total internal reflection; equality is the limiting case. The Brewster angle instead describes vanishing p-polarised reflection in the appropriate two-medium model and is not a criterion for total internal reflection.[3][6][7]

The scalar form assumes homogeneous isotropic media and a locally smooth stationary boundary. In birefringent crystals, energy flow may differ from wavevector direction. Strong absorption requires complex indices; wavelength-scale structures require diffraction rather than one refracted ray. For lenses, apply the law locally at each boundary, not once to the complete objective.[4]

Draw the normal, number media in propagation order and check the sine value before taking inverse sine. A higher index reduces the angle from the normal. At a stationary boundary frequency stays unchanged while wavelength changes. A parallel slab can restore the external direction while still displacing the ray. Power fractions additionally require a reflection and transmission model.[1][3]