Connect refraction to wave speed and wavelength

Compare three media with one incident ray

Keep incidence at 30 degrees. Medium 1 has index 1.50. Capture second indices 1.00, 1.50 and 2.00. Transmitted angles are approximately 48.590378°, 30° and 22.024313°, respectively. The smaller angle in the last case means a direction closer to the downward normal; the normal does not rotate with the ray.

Read the properties table beside each capture. With rounded vacuum speed c = 300,000,000 m/s and frequency f = 500,000,000,000,000 Hz, medium 1 has speed 200,000,000 m/s and wavelength 400 nm. Medium 2 at index 1.00 has speed 300,000,000 m/s and wavelength 600 nm. At index 2.00 these become 150,000,000 m/s and 300 nm. A nanometre, nm, is one billionth of a metre.

Make index and wavelength meaningful

Refractive index n = c/v is dimensionless: c is vacuum light speed and v is wave speed in the material, both in metres per second. Thus v = c/n. A larger positive index means a smaller speed in this model. These are ideal index values at one frequency, not measured material identifications.

At a stationary boundary, transmitted light keeps the incoming frequency. Wave speed is v = fλ, where λ is wavelength in metres and f is frequency in hertz. Hence λ = c/(nf). For n = 2.00, this gives 3 × 10⁸ / (2 × 5 × 10¹⁴) = 3 × 10⁻⁷ m = 300 nm. Unit conversion changes the number, not the physical length.

Use a short trigonometry refresher

For an acute angle in a right triangle, sine is opposite-side length divided by hypotenuse length. It is dimensionless and lies between zero and one. For a unit-length ray tilted by θ from the vertical normal, the horizontal component has magnitude sin θ. For example, sin 30° = 0.5. The browser displays the needed sine calculation; no coding is required.

Snell’s law is n₁ sin θ₁ = n₂ sin θ₂. Subscripts identify incident and transmitted media; angles are measured from the respective normal directions. Solve for sin θ₂ = (n₁/n₂) sin θ₁, then use inverse sine in degree mode to recover θ₂. Inverse sine asks which angle has that sine; it is not the reciprocal of sine.

At 30 degrees from index 1.50 into index 1.00, the required sine is 1.50 × 0.5 / 1.00 = 0.75, giving 48.590378 degrees. Into index 2.00 the required sine is 0.375, giving 22.024313 degrees. It is the sine relationship, not a direct ratio of angles, that determines the transmitted direction.

Test normal incidence

Set incidence to zero and second index to 2.00. Both ray angles are zero, but speed still falls from 200,000,000 to 150,000,000 m/s and wavelength falls from 400 to 300 nm. A speed change need not change direction at normal incidence. At equal indices, direction, speed and wavelength all remain unchanged.

Wave cycles stay synchronized along a stationary interface. Changing wavelength while maintaining this boundary matching accounts for bending at oblique incidence. The diagram represents ray geometry, not individual photon trajectories or a microscopic model of the material.

Infer an unknown index

In a separate exact synthetic case, n₁ = 1.50, incidence is 30 degrees and the transmitted angle is 40 degrees. Find n₂, decide whether light speeds up or slows down, and calculate wavelength at the same frequency of 5 × 10¹⁴ Hz. These exact angles are model data, not physical measurements.

Check the refraction transfer

n₂ = 1.50 sin 30° / sin 40° = approximately 1.166793. It is below 1.50, so the wave speeds up and bends away from the normal. Wavelength is 600/n₂ nm, approximately 514.230088 nm. Using the raw angle ratio would give the wrong index.

OpenStax: refraction develops the macroscopic ray law. Continue to total internal reflection for a required sine too large for a real angle.