Unit 3 · Lesson 3b

3bWave Behavior: Reflection and Refraction

Master the law of reflection, apply Snell's law to refraction, and explore how waves diffract around obstacles.

Reflection and refraction are the foundation of lenses, mirrors, fiber optics, and medical imaging — understanding wave behavior at boundaries is essential for modern technology.

What happens to a wave when it encounters a boundary — and why does light bend when it enters water?

Lesson Overview

When waves encounter boundaries or obstacles, they can reflect, refract, or diffract. The law of reflection governs how waves bounce off surfaces. Snell's law describes refraction — the bending of waves as they cross from one medium to another at different speeds. Diffraction explains how waves spread around obstacles. Understanding these behaviors is essential for optics, acoustics, and telecommunications.

Key Concepts

Law of Reflection

Angle of incidence equals angle of reflection (θᵢ = θᵣ), measured from the normal

Refraction

Bending of a wave as it crosses a boundary where wave speed changes

Snell's Law

n₁ sin θ₁ = n₂ sin θ₂ — relates angles and indices of refraction at a boundary

Index of Refraction (n)

n = c/v; ratio of speed of light in vacuum to speed in the medium

Diffraction

Spreading of waves around obstacles or through openings; most pronounced when λ ≈ gap size

Wave Speed at Boundaries

Wave speed changes at a boundary; frequency stays constant; wavelength changes

Worked Examples

Example 1

A ray of light strikes a flat mirror at an angle of incidence of 35°. What is the angle of reflection?

Answer:By the law of reflection, angle of reflection = angle of incidence = 35°. Both angles are measured from the normal to the surface.
Example 2

Light travels from air (n₁ = 1.00) into glass (n₂ = 1.50) at an angle of incidence of 30°. Find the angle of refraction.

Answer:n₁ sin θ₁ = n₂ sin θ₂ → 1.00 × sin 30° = 1.50 × sin θ₂ → sin θ₂ = 0.500/1.50 = 0.333 → θ₂ = sin⁻¹(0.333) ≈ 19.5°. The ray bends toward the normal as it enters the denser medium.
Example 3

Light travels from glass (n = 1.50) into water (n = 1.33) at 25°. Find the refracted angle.

Answer:1.50 × sin 25° = 1.33 × sin θ₂ → sin θ₂ = (1.50 × 0.4226)/1.33 = 0.6339/1.33 = 0.4766 → θ₂ = sin⁻¹(0.4766) ≈ 28.4°. The ray bends away from the normal (going to less dense medium).
Example 4

Explain why a straw appears bent when placed in a glass of water.

Answer:Light from the submerged part of the straw travels from water (n = 1.33) into air (n = 1.00). It bends away from the normal (refracts), so the straw appears displaced from its actual position — the classic refraction illusion.
Example 5

Sound waves with wavelength 0.50 m pass through a doorway 0.60 m wide. Will significant diffraction occur? Explain.

Answer:Diffraction is most pronounced when the wavelength is comparable to the gap size. Here λ = 0.50 m ≈ gap = 0.60 m, so yes — significant diffraction occurs and the sound spreads around the doorway into the next room.

Guided Problems

Guided Problem 1

A light ray hits a mirror at 50° to the surface (not the normal). What is the angle of reflection measured from the normal?

Hint: The angle of incidence is measured from the normal, not the surface. If the ray is 50° from the surface, it is 90° − 50° = 40° from the normal.

Guided Problem 2

Light passes from water (n = 1.33) into air (n = 1.00) at 20°. Use Snell's law to find the refracted angle.

Hint: Apply n₁ sin θ₁ = n₂ sin θ₂. Solve for sin θ₂ then take the inverse sine.

Guided Problem 3

Why does a fish in a pond appear shallower than it actually is?

Hint: Think about how light from the fish refracts as it exits the water into air — does it bend toward or away from the normal?

Guided Problem 4

Radio waves (λ ≈ 1 m) diffract around buildings, but visible light (λ ≈ 500 nm) does not. Explain why.

Hint: Compare the wavelength to the size of the obstacle. Diffraction is significant when λ ≈ obstacle size.

Guided Problem 5

When a wave crosses a boundary from a fast medium to a slow medium, does it bend toward or away from the normal?

Hint: Use Snell's law: if v decreases, n increases, so sin θ₂ < sin θ₁, meaning θ₂ < θ₁.

Key Vocabulary

Law of Reflection

The angle of incidence equals the angle of reflection, both measured from the normal to the reflecting surface.

Example: A mirror reflects a laser beam at the same angle it arrives — θᵢ = θᵣ.

Refraction

The bending of a wave as it passes from one medium to another where its speed changes.

Example: A pencil in a glass of water appears bent because light refracts at the water–air boundary.

Snell's Law

n₁ sin θ₁ = n₂ sin θ₂ — the relationship between angles and indices of refraction at a boundary.

Example: Light entering glass from air bends toward the normal because glass has a higher index of refraction.

Diffraction

The spreading of waves around obstacles or through openings; most significant when the wavelength is comparable to the obstacle or gap size.

Example: Sound diffracts around corners, which is why you can hear someone talking in the next room even without line of sight.

Interactive Practice — 5 Questions

1

The law of reflection states that the angle of incidence equals:

2

Light travels from air (n=1.00) into diamond (n=2.42) at 30°. The refracted ray bends:

3

Snell's law is written as:

4

Diffraction is most pronounced when the wavelength is:

5

When a wave crosses a boundary, which property stays constant?

Independent Practice

1

A light ray strikes a mirror at 42° to the normal. Draw a diagram and find the angle of reflection.

2

Light passes from air (n = 1.00) into water (n = 1.33) at an angle of incidence of 45°. Calculate the angle of refraction.

3

Explain total internal reflection: what condition must be met, and give one technological application.

4

A wave travels from medium A (speed 400 m/s) to medium B (speed 200 m/s) at 30°. Calculate the angle of refraction using the wave form of Snell's law: sin θ₁/v₁ = sin θ₂/v₂.

5

★ A fiber-optic cable uses total internal reflection to transmit light. Explain how Snell's law governs this, calculate the critical angle for glass (n = 1.50) to air (n = 1.00), and describe what happens to a ray that hits the boundary at an angle greater than the critical angle.

Challenge
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Common Mistakes

Measuring the angle of incidence from the surface instead of the normal.

All angles in reflection and refraction (θᵢ, θᵣ, θ₂) are measured from the normal — the line perpendicular to the surface at the point of incidence.

Thinking refraction only applies to light.

Refraction applies to all waves — sound, water waves, and seismic waves all refract when their speed changes at a boundary.

Confusing diffraction with refraction.

Refraction is bending due to a speed change at a boundary. Diffraction is spreading around obstacles or through gaps — no boundary crossing required.

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Math Tips

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In Snell's law, always use the sine of the angle, not the angle itself. Use a calculator: sin⁻¹ to find the angle from its sine.

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The index of refraction n = c/v is always ≥ 1 (since v ≤ c). A higher n means slower wave speed in that medium.