5bProperties of Light
Investigate reflection, refraction, index of refraction, total internal reflection, and the fiber optics that power modern communications.
Reflection and refraction govern lenses, cameras, eyeglasses, and fiber-optic internet. Total internal reflection makes modern telecommunications possible — understanding these principles is essential for optics and photonics.
Lesson Overview
Light travels at c = 3 × 10⁸ m/s in a vacuum but slows when entering a denser medium. The index of refraction n = c/v quantifies this slowing. In this lesson you will apply Snell's law to refraction, derive the critical angle for total internal reflection, and explore how fiber optics exploits these principles.
Key Concepts
Speed of Light
c = 3 × 10⁸ m/s in vacuum; slows in any material medium
Index of Refraction
n = c/v; higher n means slower light and more bending
Snell's Law
n₁ sin θ₁ = n₂ sin θ₂; governs the angle of refraction
Total Internal Reflection
Occurs when light in a denser medium hits the boundary at θ ≥ θ_c
Critical Angle
sin θ_c = n₂ / n₁ (for light going from medium 1 to less-dense medium 2)
Fiber Optics
Uses total internal reflection to transmit light signals over long distances with minimal loss
Light travels from air (n = 1.00) into glass (n = 1.50) at an angle of incidence of 30°. Find the angle of refraction.
What is the speed of light in glass with n = 1.50? (c = 3 × 10⁸ m/s)
Find the critical angle for total internal reflection at a glass–air interface. (n_glass = 1.50, n_air = 1.00)
A ray of light in water (n = 1.33) hits the water–air surface at 50°. Does total internal reflection occur?
Light strikes a mirror at an angle of incidence of 35°. What is the angle of reflection, and what is the angle between the incident and reflected rays?
Light travels from water (n = 1.33) into air (n = 1.00) at 30°. Find the angle of refraction.
Hint: Apply Snell's law: n₁ sin θ₁ = n₂ sin θ₂. Since n₂ < n₁, the ray bends away from the normal.
A diamond has n = 2.42. What is the speed of light inside a diamond?
Hint: Use v = c / n.
Find the critical angle for total internal reflection at a diamond–air interface. (n_diamond = 2.42)
Hint: sin θ_c = n_air / n_diamond = 1.00 / 2.42.
Explain why a fiber-optic cable can transmit light signals around bends without losing the signal.
Hint: Think about what happens when light hits the glass–cladding boundary at a large angle.
A light ray in air hits a flat glass surface (n = 1.60) at 45°. (a) Find the refracted angle. (b) Does total internal reflection occur?
Hint: For part (b), TIR only occurs when light goes from a denser to a less-dense medium.
Key Vocabulary
Index of Refraction
The ratio of the speed of light in a vacuum to its speed in a medium: n = c/v. A higher index means slower light.
Example: Glass has n ≈ 1.5, meaning light travels at 2/3 of its vacuum speed inside glass.
Snell's Law
The law governing refraction: n₁ sin θ₁ = n₂ sin θ₂, where θ is measured from the normal.
Example: A ray entering water from air at 45° refracts to about 32° because water has a higher index of refraction.
Total Internal Reflection
Complete reflection of light back into a denser medium when the angle of incidence exceeds the critical angle.
Example: Fiber-optic cables use total internal reflection to guide light pulses over thousands of kilometers.
Critical Angle
The minimum angle of incidence (measured from the normal) at which total internal reflection occurs: sin θ_c = n₂/n₁.
Example: For glass (n = 1.5) to air, the critical angle is about 41.8°.
Interactive Practice — 5 Questions
What does the index of refraction n = c/v tell us?
Light travels from glass (n = 1.5) into air (n = 1.0) at the critical angle. What happens?
Snell's law states that n₁ sin θ₁ = n₂ sin θ₂. If n₂ > n₁, the refracted ray:
Which condition is required for total internal reflection?
Fiber-optic cables transmit data using:
Independent Practice
Light in air (n = 1.00) hits a water surface (n = 1.33) at 40°. Find the angle of refraction using Snell's law.
Calculate the speed of light in a medium with n = 1.75. (c = 3 × 10⁸ m/s)
Find the critical angle for a glass–water interface. (n_glass = 1.50, n_water = 1.33)
Explain in your own words why a straw appears bent when placed in a glass of water.
★ A fiber-optic cable has a glass core (n = 1.48) surrounded by cladding (n = 1.46). (a) Find the critical angle at the core–cladding interface. (b) Explain why the cladding index must be less than the core index for the cable to work.
ChallengeCommon Mistakes
Measuring the angle of incidence from the surface rather than from the normal.
All angles in Snell's law and the law of reflection are measured from the normal (perpendicular) to the surface, not from the surface itself.
Thinking total internal reflection can occur when light goes from air into glass.
TIR only occurs when light travels from a denser medium (higher n) to a less-dense medium (lower n). Air → glass cannot produce TIR.
Math Tips
For Snell's law problems: always draw the normal first, measure all angles from it, and check whether n increases or decreases to predict which way the ray bends. For TIR: use sin θ_c = n₂/n₁ only when n₁ > n₂ (going from denser to less-dense medium).