E. Physics: Light and Geometric Optics · Light and Geometric OpticsInvestigating Refraction: Snell's Law
E2.4 E3.7 E2.6 A1.2 A1.5 A1.6 A1.8 A1.13 A2.2About 75 minutes
Learning goals- I can analyse refraction data to find a trend and an index of refraction.
- I can state Snell's law and use it to calculate angles of refraction and indices of refraction.
- I can graph sin θi against sin θR and interpret the slope.
- I can explain how measurement uncertainty affects my results.
Words to know
Snell's law
n₁ sin θ₁ = n₂ sin θ₂: the relationship between angles and indices of refraction at a boundary.
sine (sin)
For an angle in a right triangle, the ratio of the opposite side to the hypotenuse; calculators give it directly.
inverse sine (sin⁻¹)
The calculator function that finds an angle from its sine.
line of best fit
A straight line drawn as close as possible to all the data points.
uncertainty
The range within which the true value of a measurement probably lies.
percentage error
|measured − accepted| ÷ accepted × 100%.
How much does light bend?
In the last lesson you learned which way light bends. Today you will find out how much. You will analyse real-style data from a classic investigation, discover the pattern, and use it to predict angles of refraction.
The investigation
Question: How is the angle of refraction related to the angle of incidence when light passes from air into acrylic?
Hypothesis: If the angle of incidence increases, then the angle of refraction will also increase, but by less, because light slows down in acrylic and bends towards the normal.
Equipment: a ray box (or a flashlight with a slit), a semicircular acrylic block, a polar (circular protractor) sheet.
Method: Place the flat face of the block along the 90°–270° line of the protractor sheet, with the centre of the flat face at the centre of the protractor. Aim the ray at the centre of the flat face at angles of incidence from 10° to 70°. For each angle, read the angle of refraction where the ray leaves the curved side.
Planning the investigation (A1.2)
This question is best answered with a controlled experiment, because we can change one thing and measure its effect.
- Independent variable (what you change): the angle of incidence, θi.
- Dependent variable (what you measure): the angle of refraction, θR.
- Controlled variables (kept the same): the material of the block, the colour of the light, the position of the block on the protractor sheet.
Taking many trials across a wide range of angles makes a pattern easier to see, and repeating each trial lets you check how reliable your readings are.
| θi (°) | θR (°) | sin θi | sin θR | sin θi ÷ sin θR |
|---|
| 10 | 7 | 0.174 | 0.122 | 1.42 |
| 20 | 13 | 0.342 | 0.225 | 1.52 |
| 30 | 20 | 0.500 | 0.342 | 1.46 |
| 40 | 26 | 0.643 | 0.438 | 1.47 |
| 50 | 31 | 0.766 | 0.515 | 1.49 |
| 60 | 35 | 0.866 | 0.574 | 1.51 |
| 70 | 39 | 0.940 | 0.629 | 1.49 |
Looking for the pattern
The ratio θi ÷ θR is not constant (10 ÷ 7 = 1.43, but 70 ÷ 39 = 1.79). But look at the last column: sin θi ÷ sin θR stays close to 1.5 every time. The spread (1.42 to 1.52) comes mostly from measurement uncertainty: a 1° reading error matters much more for a 7° angle than for a 39° angle, which is why the 10° trial is the furthest off.
Graphing sin θi against sin θR gives a straight line through the origin. Its slope is about 1.49, the index of refraction of acrylic.
Snell's law
The pattern holds for any two media. It is called Snell's law:
n₁ sin θ₁ = n₂ sin θ₂
- n₁ = index of refraction of the first medium (where the light comes from), θ₁ = angle of incidence
- n₂ = index of refraction of the second medium, θ₂ = angle of refraction
When the first medium is air (n₁ = 1.00), this simplifies to n₂ = sin θ₁ ÷ sin θ₂, exactly the ratio in the table.
Who discovered it? The Persian mathematician Ibn Sahl described this law in a book on burning mirrors and lenses written in Baghdad around 984. The Dutch scientist Willebrord Snell found it again in 1621, and it is named after him. Science often moves forward this way, with discoveries made, lost and made again in different parts of the world.
Worked exampleExample 1: From air into water
Light in air strikes the surface of a lake at an angle of incidence of 30.0°. Find the angle of refraction in the water (n = 1.33).
- Given: n₁ = 1.00, θ₁ = 30.0°, n₂ = 1.33. Required: θ₂.
- Snell's law: n₁ sin θ₁ = n₂ sin θ₂, so sin θ₂ = n₁ sin θ₁ ÷ n₂.
- sin θ₂ = (1.00)(sin 30.0°) ÷ 1.33 = 0.500 ÷ 1.33 = 0.376.
- θ₂ = sin⁻¹(0.376) = 22.1°. (On a calculator: 2nd or SHIFT, then sin.)
θ₂ = 22.1°: smaller than 30.0°, so the light bent towards the normal, as expected for slowing down.
Worked exampleExample 2: From glass into air
Light inside a glass block (n = 1.52) reaches the surface with an angle of incidence of 25.0°. Find the angle of refraction in air.
- Given: n₁ = 1.52, θ₁ = 25.0°, n₂ = 1.00.
- sin θ₂ = n₁ sin θ₁ ÷ n₂ = (1.52)(sin 25.0°) ÷ 1.00 = (1.52)(0.4226) = 0.642.
- θ₂ = sin⁻¹(0.642) = 40.0°.
θ₂ = 40.0°: larger than 25.0°, so the light bent away from the normal as it sped up.
Example 2 drawn to scale: light leaving glass at 25.0° bends away from the normal to 40.0° in air.Worked exampleExample 3: Identify a material
A ray in air hits a clear block at 45.0° and refracts at 28.0°. Find n and suggest what the block is made of.
- n₂ = n₁ sin θ₁ ÷ sin θ₂ = (1.00)(sin 45.0°) ÷ sin 28.0°
- = 0.7071 ÷ 0.4695
- = 1.51
n = 1.51, very close to crown glass (1.52); the block is probably glass.
Worked exampleExample 4: Between two materials
Light passes from water (n = 1.33) into glass (n = 1.52) with θ₁ = 40.0°. Find θ₂.
- sin θ₂ = n₁ sin θ₁ ÷ n₂ = (1.33)(sin 40.0°) ÷ 1.52 = (1.33)(0.6428) ÷ 1.52 = 0.5625.
- θ₂ = sin⁻¹(0.5625) = 34.2°.
θ₂ = 34.2°. The light bends towards the normal, but only a little, because 1.33 and 1.52 are fairly close.
Calculator and significant-digit tips
- Make sure your calculator is in degree (DEG) mode, not radians.
- Keep extra digits in the middle of a calculation; round only the final answer.
- Angles measured with a protractor to the nearest degree support 2 to 3 significant digits in n.
- Check: entering a slower medium (larger n) must give a smaller angle.
Just for funSnell's law is the only law in physics that tells light it's okay to bend the rules... as long as it does it at the boundary.
Hands-on activityMeasure the index of refraction of water
You need: a clear, straight-sided rectangular plastic food container (or glass dish), water with 2 or 3 drops of milk, a flashlight and a comb or slit card (a ray box), a protractor, a ruler, a marker, a dark room.
- Fill the container with milky water. Put it on a sheet of paper and trace its outline.
- Shine a thin beam through the side wall of the container at an angle. Mark the point where the beam enters and two points along the beam inside the water (look from above). Mark the point where the incident beam starts on the paper.
- Remove the container. Draw the normal at the entry point. Measure θi (in air) and θR (in water).
- Repeat for at least five angles between 15° and 65°. Record a table with sin θi, sin θR and their ratio.
- Plot sin θi (y-axis) against sin θR (x-axis), draw a line of best fit and find its slope. Compare your value with 1.33 and calculate the percentage error.
- Explain at least two sources of error. (The thin plastic wall has its own index of refraction, but because its two faces are parallel it only shifts the ray sideways slightly; it does not change the final angle in the water.)
What to look for: a slope between about 1.25 and 1.40; small angles give the least reliable ratios.
Safety (grown-up note): Keep electrical devices away from the water and wipe up spills. If you use a laser pointer instead of a flashlight, supervise: low power only, never towards anyone's eyes.
Your turn
1Write Snell's law and explain each symbol.
n₁ sin θ₁ = n₂ sin θ₂: n₁ and n₂ are the indices of refraction of the first and second media; θ₁ is the angle of incidence and θ₂ the angle of refraction, both measured from the normal.
2Light passes from air into water at 50.0°. Find the angle of refraction.
sin θ₂ = sin 50.0° ÷ 1.33 = 0.766 ÷ 1.33 = 0.576 → θ₂ = 35.2°.
3Light passes from air into glass (n = 1.52) at 35.0°. Find θR.
sin θ₂ = 0.5736 ÷ 1.52 = 0.3774 → θ₂ = 22.2°.
4Light passes from water into air with θi = 20.0°. Find θR.
sin θ₂ = 1.33 × sin 20.0° = 1.33 × 0.342 = 0.455 → θ₂ = 27.1° (away from the normal).
5A ray in air at 60.0° refracts to 35.3° in an unknown liquid. Find n.
n = sin 60.0° ÷ sin 35.3° = 0.866 ÷ 0.578 = 1.50.
6Why is the ratio θi ÷ θR not used to find n?
It is not constant: Snell's law involves the sines of the angles, not the angles themselves.
More practice: warm-up
1Who first described the law of refraction, as far as we know, and around when?
Ibn Sahl, in Baghdad, around 984.
θ₂ = θ₁: no bending (the light does not change speed).
3sin θ₂ = 0.500. What is θ₂?
4In the graph of sin θi against sin θR for air into a material, what does the slope equal?
The index of refraction of the material.
5Light goes from air into diamond. Will θR be larger or smaller than θi?
Smaller (diamond slows light a lot).
More practice: on your own
1Light passes from air into diamond (n = 2.42) at 10.0°. Find θR. Then explain why such a small angle of incidence still gives a noticeable change.
sin θ₂ = 0.1736 ÷ 2.42 = 0.0718 → θ₂ = 4.1°. The angle is less than half of θi because diamond's index is so large.
2Light passes from water (n = 1.33) into ethanol (n = 1.36) at 30.0°. Find θR.
sin θ₂ = 1.33 × 0.500 ÷ 1.36 = 0.489 → θ₂ = 29.3°: only a tiny bend, because the indices are close.
3A student records θi = 40° and θR = 25° for light entering a block. Find n and name a likely material.
n = sin 40° ÷ sin 25° = 0.643 ÷ 0.423 = 1.52: probably crown glass.
4In a sample data table, the trial at θi = 10° gives a ratio of 1.42 while the others give about 1.49. Explain why the small-angle trial is the least reliable.
A reading error of about ±1° is a large percentage of a small angle like 7°, but a small percentage of 35°. So the ratio for small angles has the largest uncertainty.
5Light in air hits acrylic (n = 1.49) at 30.0°. Find θR and then the speed of light in the acrylic.
sin θ₂ = 0.500 ÷ 1.49 = 0.336 → θ₂ = 19.6°. v = 3.00 × 10⁸ ÷ 1.49 = 2.01 × 10⁸ m/s.
Challenge
1A ray enters a rectangular glass block (n = 1.52) from air at 40.0° and leaves through the parallel opposite face. Find the angle inside the block and the angle at which it leaves. What do you notice?
Inside: sin θ₂ = 0.643 ÷ 1.52 = 0.423 → 25.0°. At the second face θ₁ = 25.0° (alternate angles), so sin θ = 1.52 × 0.423 = 0.643 → 40.0°. The ray leaves parallel to its original direction, shifted sideways.
2Show that Snell's law can be written as sin θ₁ ÷ sin θ₂ = v₁ ÷ v₂.
n = c/v, so n₁ sin θ₁ = n₂ sin θ₂ becomes (c/v₁) sin θ₁ = (c/v₂) sin θ₂. Divide both sides by c and rearrange: sin θ₁ ÷ sin θ₂ = v₁ ÷ v₂. The angle is larger where light is faster.
3Light travels from glass (n = 1.52) into air at 45.0°. Try to calculate θR. What happens, and what do you think it means?
sin θ₂ = 1.52 × sin 45.0° = 1.52 × 0.707 = 1.07. The sine of an angle cannot be greater than 1, so there is no refracted ray: the light cannot leave the glass. It is all reflected back inside: total internal reflection (next lesson).
Quiz
2. Light in air hits water (n = 1.33) at 40.0°. The angle of refraction is about:
3. A graph of sin θi (air) against sin θR (material) is a straight line with slope 1.36. The material is most likely:
4. Light passes from glass into air. Compared with θi, θR is:
5. Light hits a block at 50.0° and refracts at 30.0°. The index of refraction of the block is:
6. In the investigation, why does the ray not bend when it leaves through the curved side of the semicircular block?
Big idea: Expectations E2.4 and E3.7 ask students to investigate refraction, analyse data for trends and describe the factors that affect refraction quantitatively. The trend is Snell's law, n₁ sin θ₁ = n₂ sin θ₂. Students also practise graphing, finding a slope, and judging uncertainty (A1.6, A1.8, A1.13).
Common difficulties:
- Calculator mode. Answers that look strange are often from radian mode. Check for "DEG" on the screen.
- Inverse sine. To find an angle from its sine, use sin⁻¹ (often SHIFT or 2nd, then sin).
- Order of media. n₁ always belongs to the medium the light is coming from.
How to help: Help set up the water-container investigation; it takes about 30 minutes and gives real data to graph. Ask your teen to explain why small angles give less reliable ratios.