Wave Behaviour
Orientation
Lesson goal: explain reflection, refraction, diffraction, interference, and standing waves using the correct wave model and representation.
Students should be able to state what changes, what stays constant, and what evidence supports the explanation.
Core Content
Wave behaviour is controlled by boundary conditions and superposition. Reflection, refraction, diffraction, and interference are not interchangeable terms. Each describes a different physical interaction.
| Behaviour | What changes | What remains fixed | Key evidence |
|---|---|---|---|
| Reflection | direction | speed in same medium, frequency | angle in equals angle out |
| Refraction | direction, speed, wavelength | frequency | wave bends at boundary |
| Diffraction | spreading pattern | frequency | greater spreading for wider wavelength or narrower gap |
| Interference | resultant amplitude | source frequencies | constructive or destructive pattern |
Key equations and rules:
$$\theta_i = \theta_r$$
$$n_1\sin\theta_1 = n_2\sin\theta_2$$
$$n = \frac{c}{v}$$
In refraction, the source fixes the frequency. The speed changes because the medium changes, so the wavelength must change to keep $v = f\lambda$ true.
Concept Check
-
In reflection, angles are measured relative to:
- A. the surface
- B. the normal
- C. the wave crest
- D. the source
Answer: B.
-
When a wave refracts at a boundary, which quantity remains constant?
- A. frequency
- B. speed
- C. wavelength
- D. direction
Answer: A.
-
Diffraction is most pronounced when:
- A. the gap is much larger than the wavelength
- B. the wave has zero amplitude
- C. the gap size is comparable to wavelength
- D. frequency is zero
Answer: C.
-
Short response: explain why a ray bends toward the normal when light slows down entering a higher refractive-index medium.
Applied Practice
Worked Example
Light passes from air into glass. The incident angle is $40^\circ$ to the normal. Take $n_1 = 1.00$ and $n_2 = 1.50$. Find the refracted angle.
-
State Snell's law:
$$n_1\sin\theta_1 = n_2\sin\theta_2$$
-
Substitute:
$$1.00\sin40^\circ = 1.50\sin\theta_2$$
-
Rearrange:
$$\sin\theta_2 = 0.4285$$
-
Calculate:
$$\theta_2 = 25.4^\circ$$
Final answer: the refracted angle is $25.4^\circ$ to the normal. The ray bends toward the normal because light travels more slowly in glass.
Practice Problem
A light ray moves from glass with $n = 1.50$ into air with $n = 1.00$. The incident angle is $30^\circ$ to the normal. Calculate the refracted angle and state whether the ray bends toward or away from the normal.
Deep Practice And Writing
Prompt: write a paragraph comparing reflection and refraction. Your paragraph must use the terms direction, speed, frequency, and boundary.
Strong response pattern:
- identify the boundary interaction,
- state the conserved quantity,
- state the changed quantity,
- explain the visible consequence.
Tutor Context
Use this lesson context when the student asks about:
- angle conventions,
- reflection law,
- Snell's law,
- constant frequency in refraction,
- diffraction and interference,
- nodes and antinodes in standing waves.
Tutor should first check whether the student is measuring angles from the normal or from the surface.
Useful tutor diagnostic:
In refraction, which quantity is fixed by the source, and what must change if wave speed changes?
Maintenance Loop
Fast retrieval:
- Angles in reflection/refraction are measured from the ______.
- Refraction changes speed and wavelength, but not ______.
- Diffraction increases when gap size is closer to ______.
Source Trace
This lesson is materialised from the existing textbook section, app lesson YAML, roadmap lesson, roadmap concept questions, and Module 3 notes. It is ready for conversion into a typed content manifest and panelled-course render block.