IB Physics HLTopic 3 — Oscillations & WavesPaper 1 & 2Snell’s law & TIR~15 min read
Refraction
Stand a straw in a glass of water and it looks snapped in half. Last time we said light bends because it changes speed — true, but hand-wavy. Now we put numbers on it. One quantity, the refractive index, tells you how much a material slows light down, and one equation, Snell’s law, turns that into an angle. Push the angle far enough and something dramatic happens: the light stops escaping altogether.
📘 What you need to know
Refractive indexn = c/v — how much slower light travels in a medium. It has no units
Higher n → more optically dense → slower light. For air, n = 1.00
Snell’s law:n1 sin θ1 = n2 sin θ2
The ratio form links speeds too: n1/n2 = sin θ2/sin θ1 = v2/v1
The critical angle satisfies sin θc = n2/n1 (with n1 > n2)
Total internal reflection needs both: n2 < n1andθ1 > θc
Refractive index
The refractive index of a material, n, tells you how optically dense it is — which really means how much it slows light down.
Refractive indexn = c / v
Where c = 3.00 × 108 m s−1 is the speed of light in a vacuum, and v is its speed in the medium. Because it’s a ratio of two speeds, n is dimensionless — it has no units at all. And since nothing beats light in a vacuum, n is always greater than 1 for real materials.
Air is so thin that we take n = 1.00
A biggern means a more optically dense medium, and slower light inside it
Medium
Refractive index n
Speed of light v = c/n
Critical angle to air
Air
1.00
3.00 × 108 m s−1
—
Water
1.33
2.26 × 108 m s−1
48.8°
Glass
1.50
2.00 × 108 m s−1
41.8°
Diamond
2.42
1.24 × 108 m s−1
24.4°
Bigger n
means
Slower light v = c/n
so ray bends
Towards the normal
Snell’s law
Snell’s law is the equation that connects the two angles at a boundary to the two refractive indices. The most useful form is the tidiest one:
Snell’s lawn1 sin θ1 = n2 sin θ2
Written as a set of ratios it also carries the speeds, which is handy when a question gives you v rather than n:
The ratio formn1 / n2 = sin θ2 / sin θ1 = v2 / v1
Where θ1 is the angle of incidence in medium 1 and θ2 is the angle of refraction in medium 2 — both measured from the normal.
Here n2 > n1, so the light slows and bends towards the normal: θ2 < θ1. The angles in this diagram are drawn to scale for n2/n1 = 1.50.
Look at the shape of the equation: n and sinθ sit on opposite sides of the see-saw. Big n forces small sinθ. So the ray always makes the smaller angle in the denser medium. If your answer comes out with a bigger angle inside the glass, you’ve flipped the equation — check before you write it down.
Through a glass block: two boundaries
Most exam questions involve two boundaries. Watch what happens to a ray crossing a rectangular block: it bends towards the normal going in, and away from the normal coming out. Because the two faces are parallel, the two bends cancel exactly — the emergent ray runs parallel to the original, just shifted sideways.
The refracted ray at the first face becomes the incident ray at the second. The emergent ray leaves at the same angle it arrived, so it is parallel to the original — only displaced sideways.
The critical angle
Now send light the other way: from inside the dense medium out towards the less dense one. It bends away from the normal, so θ2 > θ1. Increase the angle of incidence and the refracted ray swings further and further round… until it lies flat along the boundary, at θ2 = 90°.
The angle of incidence that does this is the critical angle, θc. Push past it and the light cannot get out at all: it is totally internally reflected back into the dense medium.
Light travelling inside the denser medium (shaded). As the angle of incidence grows past the critical angle c, refraction stops and every bit of the light reflects back inside.
Put θ2 = 90° into Snell’s law, so sin θ2 = 1, and you get the critical-angle equation straight away:
The critical anglesin θc = n2 / n1(light starts in medium 1, and n1 > n2)
The larger the refractive index, the smaller the critical angle. Diamond’s huge n = 2.42 gives θc = 24.4°, so light rattles around inside instead of leaking out.
That’s the sparkle: because a diamond’s critical angle is only 24.4°, almost every ray that gets in hits a facet beyond the critical angle and is totally internally reflected. Light bounces around inside and finally bursts out of the top. Glass, at 41.8°, lets far more light dribble out of the sides.
Total internal reflection
TIR is a special case of refraction — the case where refraction fails. It needs two conditions at once:
The light must be going into a less optically dense medium: n2 < n1
The angle of incidence must exceed the critical angle: θ1 > θc
When it happens, the light obeys the ordinary law of reflection: angle of incidence = angle of reflection. Nothing is transmitted, which is why it’s called total.
🧮 Solving a Snell’s law problem
Label the media. Medium 1 is where the light starts. If air isn’t given, take nair = 1.00.
Check the angles are from the normal. If a question gives the angle to the surface, use 90 − θ first.
Write n1 sin θ1 = n2 sin θ2 and substitute before rearranging.
Calculator in degrees. Then sanity-check: is n > 1, and is the angle smaller in the denser medium?
WE 1
The refractive index of water is 1.33. Calculate the speed of light in water. Take c = 3.00 × 108 m s⁻¹.
Step 1 — write down the definition
n = c / v
Step 2 — rearrange for v
v = c / n
Step 3 — substitutev = (3.00 × 10⁸) / 1.33v = 2.26 × 10⁸ m s⁻¹About three-quarters of its vacuum speed. The answer must always come out smaller than c — if it doesn’t, you’ve divided the wrong way.
WE 2
A ray of light travels from air into water at an angle of incidence of 50°. Calculate the angle of refraction. (nair = 1.00, nwater = 1.33)
Step 1 — medium 1 is air, medium 2 is water
n₁ = 1.00, θ₁ = 50°, n₂ = 1.33
Step 2 — write Snell’s law
n₁ sin θ₁ = n₂ sin θ₂
Step 3 — substitute and rearrangesin θ₂ = (1.00 × sin 50°) / 1.33 = 0.576Step 4 — take the inverse sineθ₂ = sin⁻¹(0.576)θ₂ = 35°35° < 50° — the smaller angle is in the denser medium, exactly as it should be.
WE 3
A diver shines a torch upwards from under water (n = 1.33) towards the surface. Calculate the critical angle, and state what happens to a ray striking the surface at 55°.
Step 1 — the light starts in water, so medium 1 is water
n₁ = 1.33 (water), n₂ = 1.00 (air)
Step 2 — use the critical angle equationsin θᶜ = 1.00 / 1.33 = 0.752Step 3 — inverse sineθᶜ = sin⁻¹(0.752) = 48.8°Step 4 — compare 55° with the critical angle
55° > 48.8°, and the light is going into a less dense medium.
θᶜ = 48.8°; the ray is totally internally reflected at 55°Both TIR conditions are met. The reflected ray leaves at 55° too — TIR obeys the law of reflection. This is why a diver looking up sees a mirrored ceiling away from the bright circle overhead.
💡 Top tips
Always check n > 1 at the end of a calculation. If it isn’t, something went wrong.
The critical angle equation is not in the data booklet — learn it. Snell’s law and n = c/v are.
Calculator in degrees. Radians will silently ruin every answer on this page.
Medium 1 is always where the light starts. Get that right and the algebra follows.
At two boundaries, the refracted ray of the first becomes the incident ray of the second. Label the diagram.
Assume nair = 1.00 if a question doesn’t give it.
⚠ Common mistakes
Putting n1 and n2 the wrong way round — giving a bigger angle inside the denser medium
Measuring angles from the surface rather than the normal
Thinking TIR happens going into a denser medium — it never can
Forgetting the second TIR condition; a big angle alone is not enough
Giving nunits — it is a ratio of two speeds, so it has none
Believing the frequency or colour changes on refraction — only v and λ do
Quick recap:n = c/v measures how much a medium slows light; it is dimensionless and always > 1. Snell’s law, n1 sin θ1 = n2 sin θ2, fixes the angles, with the smaller angle always in the denser medium. Going from dense to less dense, the angle of refraction reaches 90° at the critical angle, sin θc = n2/n1. Beyond it you get total internal reflection, obeying i = r.
We’ve now taken a single wave and thrown everything at it — boundaries, gaps, obstacles. What we haven’t done is let two waves meet. When they overlap, their displacements simply add: crest on crest gives a giant, crest on trough gives nothing at all. That’s the principle of superposition, the next page, and it unlocks interference, Young’s slits and diffraction gratings.
Snell’s law and TIR still slippery?
Book a free meeting and we’ll drill refractive index, critical angles and past-paper ray diagrams together.
