Knowing that light bends is one thing; predicting exactly how much it bends is what lets us design lenses, fibre optics and prisms. Snell's law turns the qualitative idea of refraction into a single equation that connects the refractive indices of two media to the angles a ray makes on each side of a boundary. Learn to use it both ways and you can solve almost any refraction problem.
Light passes from air ($n = 1.00$) into water ($n = 1.33$) at an angle of incidence of $40°$.
Before reading: will the angle of refraction in the water be larger than, smaller than, or equal to $40°$? Write your prediction and one sentence explaining why.
The quantity $n\sin\theta$ is the same on both sides of a boundary
Snell's law says that when light crosses a boundary, the product of the refractive index and the sine of the angle to the normal stays the same, so if the index goes up, the angle must come down.
The law of refraction connects the two angles (measured from the normal) to the two refractive indices:
$$n_1\sin\theta_1 = n_2\sin\theta_2$$
Each index pairs with the angle in its own medium: $n_1$ with $\theta_1$, $n_2$ with $\theta_2$. Because the product is fixed, raising the index forces the angle (and its sine) to fall.
Law of refraction (Snell's law): $n_1\sin\theta_1 = n_2\sin\theta_2$, with all angles measured from the normal. Each refractive index pairs with the angle in its own medium; a higher index gives a smaller angle from the normal.
Pause, copy the highlighted law into your book before moving on.
Did you get this? True or false: in $n_1\sin\theta_1 = n_2\sin\theta_2$, the angle $\theta_1$ is the angle in the medium the ray ends up in.
Rearranging Snell's law to find the angle of refraction
We just saw that $n_1\sin\theta_1 = n_2\sin\theta_2$. That raises a question: given three of the four quantities, how do we find the fourth? This card answers it → rearrange for $\sin\theta_2$, then take the inverse sine.
To find the angle of refraction, isolate $\sin\theta_2$ on one side, work out its value, then use $\sin^{-1}$ to get the angle itself.
Rearranging Snell's law for the angle in medium 2:
$$\sin\theta_2 = \frac{n_1\sin\theta_1}{n_2} \qquad\Rightarrow\qquad \theta_2 = \sin^{-1}\!\left(\frac{n_1\sin\theta_1}{n_2}\right)$$
Light passes from air ($n_1 = 1.00$) into water ($n_2 = 1.33$) at an angle of incidence of $40°$. Find the angle of refraction in the water.
To find the refraction angle, rearrange to $\sin\theta_2 = \dfrac{n_1\sin\theta_1}{n_2}$ then take $\theta_2 = \sin^{-1}(\dots)$. When the second medium is denser ($n_2 > n_1$), the result satisfies $\theta_2 < \theta_1$.
Add the highlighted rearrangement to your notes before the check below.
Quick check: Light goes from glass ($n_1 = 1.50$) into air ($n_2 = 1.00$). Compared with the angle of incidence, the angle of refraction is:
Pattern A, Calculate then construct
Light passes from air ($n_1 = 1.00$) into a glass block ($n_2 = 1.52$) at an angle of incidence of $55°$. In your book: (1) calculate the angle of refraction using Snell's law; (2) draw a labelled ray diagram showing the boundary, normal, incident ray and refracted ray; (3) state in one sentence which way the ray bends and why.
Odd one out: Three of these are correct statements of, or rearrangements of, the law of refraction. Which one is the odd one out?
Dot point: Solve problems by applying the law of refraction of light using n₁sinθ₁ = n₂sinθ₂
Turning the numbers into an accurate, labelled diagram
We just learned to calculate the angle of refraction. That raises a question: how do we represent the whole event clearly on paper, including any partial reflection? This card answers it → a step-by-step ray diagram convention.
A good refraction ray diagram always starts with the boundary and the normal, then shows the incident ray, the refracted ray bending the correct way, and the angles measured from the normal.
To construct a ray diagram for refraction at a boundary:
The same conventions cover reflection from a plane mirror: incident ray, normal, and reflected ray with equal angles. Keeping angles measured from the normal makes both diagrams consistent.
A refraction ray diagram shows: the boundary, a dashed normal, the incident ray with $\theta_1$, and the refracted ray with $\theta_2$ bending towards the normal into a denser medium (away into a faster one). All angles are measured from the normal, the same convention used for plane-mirror reflection.
Pause, copy the highlighted diagram steps into your book before the check below.
Last check: Decide if the statement is true or false.
When constructing a refraction ray diagram, the angles should be measured from the boundary surface.
Match them up: Match each step in constructing a refraction ray diagram to its purpose.
Dot point: Construct ray diagrams to show reflection from plane mirrors and refraction at a boundary
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 5(4 marks) 1. A ray of light travels from water ($n_1 = 1.33$) into air ($n_2 = 1.00$), striking the boundary at an angle of incidence of $30°$. Calculate the angle of refraction in the air, and explain whether the ray bends towards or away from the normal.
1 mark: applies $n_1\sin\theta_1 = n_2\sin\theta_2$ · 1 mark: $\sin\theta_2 = 0.665$ · 1 mark: $\theta_2 = 41.7°$ · 1 mark: bends away from the normal because it enters a less dense (faster) medium
ApplyBand 5(3 marks) 2. Light travels from air ($n = 1.00$) into water ($n = 1.33$), striking the surface at an angle of incidence of $40^\circ$. Calculate the angle of refraction, showing your working.
1 mark: correct substitution into $n_1\sin\theta_1 = n_2\sin\theta_2$ · 1 mark: $\sin\theta_2 = 0.483$ · 1 mark: $\theta_2 = 29^\circ$ (2 s.f.)
Dot point: Solve problems by applying the law of refraction of light using n₁sinθ₁ = n₂sinθ₂
AnalyseBand 5(3 marks) 3. Describe, step by step, how to construct an accurate ray diagram for light travelling from air into glass at a boundary, and explain how the completed diagram shows that the light has slowed down.
1 mark: draw the boundary and construct the normal perpendicular to it at the point of incidence · 1 mark: draw the incident ray to that point and mark the angle of incidence from the normal, with an arrowhead showing the direction of travel · 1 mark: draw the refracted ray closer to the normal, and explain that bending towards the normal is what shows the light has slowed
Dot point: Construct ray diagrams to show reflection from plane mirrors and refraction at a boundary
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): Applying Snell's law, $n_1\sin\theta_1 = n_2\sin\theta_2$, so $\sin\theta_2 = \dfrac{n_1\sin\theta_1}{n_2} = \dfrac{1.33 \times \sin 30°}{1.00} = \dfrac{1.33 \times 0.500}{1.00} = 0.665$. Therefore $\theta_2 = \sin^{-1}(0.665) = 41.7°$. Because the ray passes from water into air, a less dense, faster medium, it bends away from the normal, so the angle of refraction ($41.7°$) is larger than the angle of incidence ($30°$).
Q2 (3 marks): Applying the law of refraction, $n_1\sin\theta_1 = n_2\sin\theta_2$. Substituting the known values: $1.00 \times \sin 40^\circ = 1.33 \times \sin\theta_2$. Since $\sin 40^\circ = 0.643$, this gives $\sin\theta_2 = 0.643/1.33 = 0.483$. Taking the inverse sine, $\theta_2 = 28.9^\circ$, which is $29^\circ$ to two significant figures. The angle of refraction is smaller than the angle of incidence, which is the expected result: the light has passed into a medium of higher refractive index, so it has slowed down and bent towards the normal.
Q3 (3 marks): First draw the boundary as a straight line separating the two media, and label which side is air and which is glass. At the point where the light will strike it, construct the normal: a line perpendicular to the boundary, drawn dashed so it is not mistaken for a ray. Draw the incident ray in air arriving at that point, put an arrowhead on it to show the direction of energy travel, and mark the angle of incidence $\theta_1$ between the incident ray and the normal, never between the ray and the boundary. Then draw the refracted ray continuing into the glass from the same point, and mark the angle of refraction $\theta_2$ from the normal on that side. The refracted ray must be drawn closer to the normal than the incident ray, so that $\theta_2 < \theta_1$, which is what Snell's law gives when $n_2 > n_1$. That bending towards the normal is precisely what the diagram shows about the physics: light bends towards the normal when it enters a medium in which it travels more slowly. A diagram in which the refracted ray bent away from the normal would be asserting that the light had sped up on entering the glass.
You can now apply $n_1\sin\theta_1 = n_2\sin\theta_2$ to find any angle and construct a clear refraction ray diagram. Next lesson uses the same law to find the special angle where the refracted ray vanishes entirely, total internal reflection.
↩ Back to focus-area hubAt the start you predicted the angle of refraction in water for a $40°$ incident ray. The answer is smaller than $40°$ (about $29°$): water has a higher refractive index, so the ray bends towards the normal. If you predicted a larger angle, revisit Card 1, a higher index means a smaller angle from the normal.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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