1bRefraction and Snell's Law
Apply Snell's Law to calculate how light bends at boundaries, and discover total internal reflection and its role in fiber optics.
Refraction explains why lenses focus light, why diamonds sparkle, and how fiber-optic cables carry internet data around the world at the speed of light.
Lesson Overview
When light passes from one medium into another it changes speed and bends — this is refraction. In this lesson you will apply Snell's Law to calculate bending angles, understand the index of refraction, and explore total internal reflection and its applications in fiber optics.
Key Concepts
Snell's Law
n₁ sin θ₁ = n₂ sin θ₂ — relates the angles of incidence and refraction to the indices of refraction of each medium.
Index of Refraction
n = c/v; the ratio of the speed of light in vacuum (c) to its speed in the medium (v). Always ≥ 1.
Bending Toward Normal
When light enters a denser medium (n₂ > n₁) it slows and bends toward the normal (θ₂ < θ₁).
Bending Away from Normal
When light enters a less dense medium (n₂ < n₁) it speeds up and bends away from the normal (θ₂ > θ₁).
Total Internal Reflection
When θ₁ exceeds the critical angle, all light is reflected back into the denser medium — none is transmitted.
Critical Angle
sin θc = n₂/n₁ (for n₁ > n₂); the minimum angle of incidence at which total internal reflection occurs.
Light travels from air (n = 1.00) into glass (n = 1.50) at an angle of incidence of 30°. Find the angle of refraction.
Light travels from glass (n = 1.50) into air (n = 1.00) at θ₁ = 25°. Find θ₂.
What is the speed of light in water if n_water = 1.33? (c = 3.00 × 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 fiber-optic cable uses glass with n = 1.60. What is the critical angle for the glass–air interface?
Light goes from water (n = 1.33) into air (n = 1.00) at θ₁ = 40°. Find θ₂ using Snell's Law.
Hint: Set up n₁ sin θ₁ = n₂ sin θ₂ and solve for sin θ₂. Is θ₂ larger or smaller than θ₁?
A diamond has n = 2.42. Find the critical angle for a diamond–air interface.
Hint: Use sin θc = n_air/n_diamond = 1.00/2.42.
Light in air hits a glass surface (n = 1.60) at 45°. Does total internal reflection occur? Explain.
Hint: Total internal reflection only occurs when light is going from a denser medium to a less dense medium. Check which direction the light is traveling.
If the speed of light in a medium is 2.00 × 10⁸ m/s, what is the index of refraction?
Hint: Use n = c/v with c = 3.00 × 10⁸ m/s.
Why does a straw in a glass of water appear bent at the water surface?
Hint: Think about how light changes direction when it crosses from water into air on its way to your eye.
Key Vocabulary
Refraction
The bending of light as it passes from one medium to another due to a change in speed.
Example: A pencil in a glass of water appears bent because light refracts at the water–air interface.
Index of Refraction (n)
A dimensionless number equal to c/v that describes how much a medium slows light; always ≥ 1.
Example: Glass with n = 1.5 slows light to two-thirds of its vacuum speed.
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 transmit data using total internal reflection to keep light inside the glass fiber.
Critical Angle (θc)
The angle of incidence (in the denser medium) above which total internal reflection occurs; sin θc = n₂/n₁.
Example: For glass (n=1.5) to air, θc ≈ 41.8°; rays hitting the surface at 42° or more are totally reflected.
Interactive Practice — 5 Questions
Light goes from air (n=1.00) into water (n=1.33) at 45°. Which way does it bend?
A medium has n = 2.00. What is the speed of light in this medium? (c = 3×10⁸ m/s)
Snell's Law is written as:
Total internal reflection occurs when:
For glass (n=1.5) to air, sin θc = 1.00/1.50 = 0.667. What is θc?
Independent Practice
Light in air hits a glass surface (n = 1.60) at 35°. Find the angle of refraction inside the glass.
Light travels from glass (n = 1.50) into water (n = 1.33) at 20°. Find the refracted angle.
Calculate the critical angle for a water–air interface (n_water = 1.33).
Explain in your own words why fiber-optic cables can transmit light signals over long distances without the light escaping.
★ A ray of light in glass (n = 1.65) strikes the glass–air boundary at 37°. Determine whether total internal reflection occurs. If not, find the refracted angle. If yes, confirm by computing the critical angle.
ChallengeCommon Mistakes
Applying the critical angle formula when light is going from air into glass.
Total internal reflection only occurs when light travels from a denser medium (higher n) to a less dense medium (lower n).
Using degrees directly in Snell's Law without taking the sine.
Always compute sin θ₁ and sin θ₂; never substitute the angle in degrees directly into the equation.
Thinking n can be less than 1 for ordinary materials.
For all ordinary transparent materials n ≥ 1, because light cannot travel faster than c in a vacuum.
Math Tips
When solving Snell's Law, isolate sin θ₂ = (n₁/n₂) sin θ₁, then take the inverse sine. Keep at least 4 significant figures in intermediate steps to avoid rounding errors.
Quick check: if n₂ > n₁, then θ₂ < θ₁ (bends toward normal). If n₂ < n₁, then θ₂ > θ₁ (bends away). Use this to verify your answer makes sense.