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3.4.2 Refraction

By the end of this lesson you can:

  • Explain refraction and refractive index
  • Use Snell's law to calculate angles
  • Explain total internal reflection and the critical angle
Find your course: NSW Year 11: Wave behavioursNSW Year 12QLD Unit 2: WavesVIC Unit 1: Light and heatAP Physics 2: Unit 13IB: C.3 Wave phenomena
🎥 Watch📝 Practise🕹️ Explore🎯 Exam ready🍎 Teachers
🎥Watch
Learn the idea: a short explanation, then a check.

Refraction Explained

Image description



Have you ever noticed a spoon in a glass of water 'appears' distorted. It's refraction.

Have you ever noticed that that with a glass bottle of soda, the drink appears at the very edge, even though the glass is quite thick

Watch the short

Refraction is the altering of the speed of light (or any wave) as it enters a different medium. When the incident ray strikes the boundary at an angle from the normal, this also result in a change in direction.

Interactive
Before we look at the theory of refraction, let's play with this animation.
From pHET, this  is a great way to explore refraction. Play with it, but better still, use the settings to determine variables that determine the amount of refraction.
Don't worry too much about measuring at this stage, just appreciate what factors determine the amount of bending.

Hopefully you have come to realise that the amount of bending is dependent on the incident angle and the variation between the different media.
You should have also noted that if the wave enters a denser medium, it bends toward the normal. If it enters a less dense medium, it bends away from the normal.

Lets now look at the video
Ensure you take good notes to retain information

Check Your Understanding: Refraction

1. A ray of light travels from air into glass, meeting the surface at an angle. What happens to the ray as it enters the glass?
2. What happens to the speed and wavelength of a light wave as it enters an optically denser medium (higher refractive index)?
3. A swimming pool always looks shallower than it really is. What causes this?
📝Practise
Worked examples first, then problems to try, then worksheets.

We are now ready to try a sample problem
Below is a sample problem with a video that explain how to solve it. It is suggested you try the problem beforehand, as this actually aids understanding, even if you are unsure if you are correct.

Picture

Practice Problems: Refraction

1. Suppose that light rays travelling through air reach quartz at an angle of 31°. The index of refraction of quartz is 1.54. At what angle do the light rays travel within the quartz?

Given: n1 (air) = 1.00, θ1 = 31°, n2 (quartz) = 1.54

Formula: n1 · sin(θ1) = n2 · sin(θ2)

Calculation:

1.00 · sin(31°) = 1.54 · sin(θ2)

0.5150 = 1.54 · sin(θ2)

sin(θ2) = 0.3344 → θ2 = sin−1(0.3344) ≈ 19.53°

Answer: 19.5°
2. A light ray enters a substance from the air at an angle of 35°. The light is refracted inside the substance and travels at an angle of 26°. What is the index of refraction of the substance?

Given: n1 (air) = 1.00, θ1 = 35°, θ2 = 26°

Formula: n2 = (n1 · sin(θ1)) / sin(θ2)

Calculation:

n2 = (1.00 · sin(35°)) / sin(26°)

n2 = 0.5736 / 0.4384 ≈ 1.308

Answer: 1.308
3. Diamond has an index of refraction of 2.42. If it is immersed in water, which has an index of 1.33, and light rays in the water enter the diamond at a 34° angle, what is the angle of refraction inside the diamond?

Given: n1 (water) = 1.33, θ1 = 34°, n2 (diamond) = 2.42

Formula: n1 · sin(θ1) = n2 · sin(θ2)

Calculation:

1.33 · sin(34°) = 2.42 · sin(θ2)

0.7437 = 2.42 · sin(θ2)

sin(θ2) = 0.3073 → θ2 = sin−1(0.3073) ≈ 17.89°

Answer: 17.9°
4. The index of refraction of a substance is 1.83. What is the speed of light in that material?

Given: n = 1.83, c = 3.00 × 108 m/s

Formula: v = c / n

Calculation:

v = (3.00 × 108) / 1.83 ≈ 1.639 × 108 m/s

Answer: 1.64 × 108 m/s
5. A ray of light strikes a mirror at an angle of incidence of 24°. What is the angle of reflection?

Law of Reflection: The angle of incidence equals the angle of reflection (θi = θr).

Calculation: θr = 24°

Answer: 24°
6. A ray of light passes from an unknown substance into the air. If the angle in the unknown substance is 39.0° and the angle in the air is 53.0°, what is the index of refraction of the unknown substance?

Given: θ1 = 39.0°, n2 (air) = 1.00, θ2 = 53.0°

Formula: n1 = (n2 · sin(θ2)) / sin(θ1)

Calculation:

n1 = (1.00 · sin(53.0°)) / sin(39.0°)

n1 = 0.7986 / 0.6293 ≈ 1.269

Answer: 1.27
7. A ray of light has an angle of incidence of 25.0° upon the surface of a piece of quartz (n = 1.54). What is the angle of refraction?

Given: n1 (air) = 1.00, θ1 = 25.0°, n2 (quartz) = 1.54

Formula: sin(θ2) = (n1 · sin(θ1)) / n2

Calculation:

sin(θ2) = (1.00 · sin(25.0°)) / 1.54 = 0.4226 / 1.54 = 0.2744

θ2 = sin−1(0.2744) ≈ 15.93°

Answer: 15.9°
8. A beam of light passes from water (n = 1.33) into polyethylene (n = 1.50). If the angle in the water is 36.0°, what is the angle in the polyethylene?

Given: n1 (water) = 1.33, θ1 = 36.0°, n2 (polyethylene) = 1.50

Formula: sin(θ2) = (n1 · sin(θ1)) / n2

Calculation:

sin(θ2) = (1.33 · sin(36.0°)) / 1.50 = (1.33 · 0.5878) / 1.50 = 0.5212

θ2 = sin−1(0.5212) ≈ 31.41°

Answer: 31.4°
9. Mi-ling makes some hydrogen sulfide, index of refraction = 1.000 644. If Mi-ling measures an angle of 85.000 000° in the hydrogen sulfide, what angle will Mi-ling measure in air if the index of refraction of air is 1.000 292 6?

Given: n1 = 1.000644, θ1 = 85.000000°, n2 = 1.0002926

Formula: sin(θ2) = (n1 · sin(θ1)) / n2

Calculation:

sin(θ2) = (1.000644 · 0.9961947) / 1.0002926 = 0.9965447

θ2 = sin−1(0.9965447) ≈ 85.236°

Answer: 85.236°
10. Light travelling inside a crown glass block (n = 1.52) strikes the boundary with surrounding water (n = 1.33). What is the critical angle for total internal reflection to occur at this glass-water boundary?

Given: n1 (glass) = 1.52, n2 (water) = 1.33

Formula: sin(θc) = n2 / n1

Calculation:

sin(θc) = 1.33 / 1.52 ≈ 0.875

θc = sin−1(0.875) ≈ 61.04°

Answer: 61.0°
11. A laser ray in air strikes a flat glass slab (n = 1.50) at an incident angle of 40°. It passes through the glass slab and exits back into air through a parallel rear face. At what angle does the ray exit the slab relative to the normal?

Entry surface (air → glass):

1.00 · sin(40°) = 1.50 · sin(θglass) → θglass ≈ 25.37°

Exit surface (glass → air):

Because the slab faces are parallel, the angle of incidence at the rear boundary is also 25.37°.

1.50 · sin(25.37°) = 1.00 · sin(θexit) → sin(θexit) = sin(40°)

Answer: 40° (the emergent ray travels parallel to the original incident ray)
🍎For teachers
Ideas for using this lesson in class, and where to find the answers.

The pHET animation above is a great way to demonstrate students how to use the light boxes in order to use Snell's law. It also has an option where students have to find the unknown refractive index of an unknown substance.