Unit 13 · Topic 13.3

13.3 Refraction

Refraction is the bending of a light ray as it crosses from one material into another, and it happens because light changes speed. The index of refraction $n = \frac{c}{v}$ measures how much a material slows light, and Snell's law, n₁ sin θ₁ = n₂ sin θ₂, gives the new angle. Light going into a lower index beyond the critical angle reflects completely, which is called total internal reflection.

Key terms

refraction
index of refraction
Snell's law
critical angle
total internal reflection

Why light bends

Light travels at c = 3.00 × 10⁸ m/s in a vacuum and more slowly in any material. When a beam crosses a boundary at an angle, the side of the beam that enters the new material first changes speed first, so the wavefront pivots and the beam changes direction. Think of a lawn mower rolling at an angle from pavement onto thick grass: the wheel that hits the grass first slows first, and the mower turns.

The index of refraction is the ratio $n = \frac{c}{v}$ , where v is the speed of light in the material. Because v is never faster than c, n is always at least 1. Some typical values: air ≈ 1.00, water 1.33, glass about 1.5, diamond 2.42. A bigger n means slower light.

The frequency of light doesn't change when it enters a new material, because each crest that arrives at the boundary makes one crest that leaves it. Since v = fλ, slower light has a shorter wavelength: $\lambda_{\text{material}} = \frac{\lambda_{\text{vacuum}}}{n}$ .

Snell's law

Snell's law connects the angles on both sides of a boundary: n₁ sin θ₁ = n₂ sin θ₂. Both angles are measured from the normal.

The rule of thumb is \"slow means toward\". Going into a higher index (slower), the ray bends toward the normal. Going into a lower index (faster), it bends away from the normal.

A ray that hits the boundary straight on, along the normal, doesn't bend at all, because sin 0° = 0 on both sides. It still changes speed.

At any real boundary some light also reflects, following the law of reflection. A window shows a faint reflection for that reason.

Total internal reflection

When light goes from a higher index to a lower one (glass to air, water to air), it bends away from the normal. As you increase the angle of incidence, the refracted ray swings closer to the surface.

At the critical angle $\theta_c$ , the refracted ray travels right along the surface (θ₂ = 90°). Setting sin 90° = 1 in Snell's law gives $\sin\theta_c = \frac{n_2}{n_1}$ .

Beyond the critical angle, no light is transmitted at all. Every bit of it reflects back into the first material. This is total internal reflection, and it's how optical fibers carry light around bends with almost no loss, and why diamonds sparkle.

Total internal reflection can never happen when light goes into a higher index, because then $\frac{n_2}{n_1} > 1$ and no angle has a sine bigger than 1.

Measuring n in the lab

Aim a laser or ray box into a block of plastic at several angles of incidence. Trace each ray on paper, draw the normal, and measure θ₁ in air and θ₂ inside the block with a protractor.

Snell's law with n₁ = 1.00 says sin θ₁ = n sin θ₂. So a graph of sin θ₁ (vertical) against sin θ₂ (horizontal) should be a straight line through the origin, and its slope is n. Using the slope of a best-fit line averages out errors in individual angle readings.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Light entering water
Light in air hits a pool's surface at an angle of incidence of 40°. The water has n = 1.33. Find the angle of refraction and the speed of light in the water.
Show the solutionHide the solution
1. Step 1: Snell's law: (1.00) sin 40° = 1.33 sin θ₂, so sin θ₂ = $\dfrac{0.643}{1.33}$ = 0.483 and θ₂ = 28.9°.
2. Step 2: The ray bends toward the normal (40° down to 28.9°), as it should going into a higher index.
3. Step 3: Speed: $v = \dfrac{c}{n} = \dfrac{3.00 \times 10^8}{1.33} = 2.26 \times 10^8$ m/s.
Answer: θ₂ ≈ 29°; v ≈ 2.26 × 10⁸ m/s.
Example 2Calculator allowed
Critical angle and a changing surrounding
A ray inside a glass block (n = 1.50) hits the glass's flat bottom face at 45° from the normal. Does any light leave the glass if the block is in air (n = 1.00)? What if the block sits in water (n = 1.33)?
Show the solutionHide the solution
1. Step 1: In air: $\sin\theta_c = \dfrac{1.00}{1.50} = 0.667$ , so θc = 41.8°. The ray arrives at 45°, which is beyond 41.8°, so it undergoes total internal reflection and no light leaves.
2. Step 2: In water: $\sin\theta_c = \dfrac{1.33}{1.50} = 0.887$ , so θc = 62.5°. Now 45° is less than the critical angle, so some light passes into the water.
3. Step 3: Find where it goes: 1.50 sin 45° = 1.33 sin θ₂, so sin θ₂ = 0.797 and θ₂ = 52.9°, bent away from the normal.
4. Step 4: The lesson: the critical angle depends on both materials. The closer the two indices are, the bigger the critical angle.
Answer: In air, total internal reflection (θc = 41.8°). In water, light escapes at 52.9° (θc = 62.5°).
Example 3Calculator allowed
Finding n from a graph
A student shines a laser into a plastic block and records angles of incidence of 20°, 30°, 40° and 50°, with matching angles of refraction of 13.3°, 19.6°, 25.6° and 30.9°. Find the index of refraction of the plastic.
Show the solutionHide the solution
1. Step 1: Compute the sines. sin θ₁: 0.342, 0.500, 0.643, 0.766. sin θ₂: 0.230, 0.335, 0.432, 0.514.
2. Step 2: Graph sin θ₁ on the vertical axis against sin θ₂ on the horizontal axis. Snell's law with air as medium 1 predicts a straight line through the origin with slope n.
3. Step 3: The best-fit line has slope $\dfrac{0.766 - 0.342}{0.514 - 0.230} = \dfrac{0.424}{0.284} \approx 1.49$ , and its intercept is very close to zero.
4. Step 4: So n ≈ 1.49, typical of acrylic plastic.
Answer: n ≈ 1.49

Common mistakes

Measuring angles from the boundary instead of the normal. Snell's law only works with angles from the normal.
Thinking the frequency or color changes inside glass. The speed and wavelength change; the frequency stays the same.
Trying to get total internal reflection going from air into glass. It only happens going toward a lower index.
Mixing up the direction of bending. Slower (higher n) bends toward the normal; faster bends away.

On the exam

Expect ranking and comparison questions: given rays bending in several materials, rank the indices or the speeds. A smaller angle from the normal means a bigger n and a slower speed.
Lab questions often ask you to design a way to measure n. Describe varying the angle of incidence, measuring both angles from the normal, and graphing sin θ₁ against sin θ₂ so the slope is n.

Connected topics

Videos

Refraction & Snell’s Law Explained | AP Physics 2 - Unit 13 - Lesson 4
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Refraction of light | Physics | Khan Academy
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Refraction of Light
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Total Internal Reflection of Light and Critical Angle of Refraction Physics
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Total internal reflection | Geometric optics | Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 13.3 Refraction. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

Light travels from air into glass with an index of refraction of 1.50. What is the speed of light in the glass?

Question 2 of 4Calculator allowed

A ray of light in air strikes a calm water surface (n = 1.33) with an angle of incidence of 40°. What is the angle of refraction?

Question 3 of 4Calculator allowed

What is the critical angle for light inside glass (n = 1.50) striking a boundary with air?

Question 4 of 4Calculator allowed

A glass block (n = 1.50) is placed in water (n = 1.33). How does the critical angle for light inside the glass hitting the glass–water boundary compare with the critical angle at a glass–air boundary?

0 of 4 answered