C.3.10—Reflection, refraction and transmission at boundaries
- Syllabus
- First assessment 2025
- Objective
- —
- Level
- HL
A boundary can split wave energy
When a travelling wave reaches a boundary, part of its energy may be reflected back into the first medium and part may be transmitted into the second. The transmitted wave is refracted when its speed changes and it crosses the boundary at a non-zero angle.
| Behaviour | What happens | Direction rule |
|---|---|---|
| Reflection | wave remains in medium 1 | angle of reflection equals angle of incidence |
| Transmission | wave enters medium 2 | continues across the boundary |
| Refraction | transmitted wave changes direction because speed changes | toward the normal if speed decreases; away if speed increases |
Read rays and wavefronts together
Angles are measured from the normal. Rays show energy-transfer direction and remain perpendicular to wavefronts. Frequency stays fixed at a stationary boundary; a change in speed therefore changes wavelength and wavefront spacing.
Boundary cases
At normal incidence the transmitted ray does not bend, although its speed and wavelength may change. Unless absorption is stated, reflected and transmitted energy together account for the incident energy; their amplitudes do not generally add directly.
Questions ask for a missing reflected wavelength or the minimum film thickness for constructive reflection. The evidence rewards including refractive index and the correct phase-shift condition.
Determine / What is
Draw or identify the top and bottom reflected rays, calculate the optical path contribution 2nt at normal incidence, and count phase reversals before selecting the constructive condition. For the air–film–air minimum-thickness case, use t = λ/(4n).
Using 2t instead of 2nt, or applying the air–film–air quarter-wave result without checking the phase changes at the two surfaces.
Representative question
The refractive index of the coating is 1.63 and the refractive index of the glass is 1.52 .
The thickness of the coating is 143 nm .
Determine the wavelength, in nm , that is missing in the light reflected to the girl assuming that the light is incident normally on the window.
Use of 2dn=mλ
Use of n=1.63
470 «nm»