IB Physics SLTopic C.3 — How Waves BehavePaper 1 & 2n₁ sin θ₁ = n₂ sin θ₂~9 min read
Refraction
Now we put numbers to the bending. Every transparent material has a refractive index that fixes how much it slows light down — and once you know that, Snell’s law tells you exactly how the ray bends. Push the angle far enough and the light stops escaping altogether: total internal reflection.
📘 What you need to know
The refractive indexn measures how optically dense a material is: n = c/v
Air has n ≈ 1; a higher n means denser and slower light; n has no units
Snell’s law: n1 sin θ1 = n2 sin θ2, with all angles from the normal
The critical angleθc is the angle of incidence that gives a 90° angle of refraction: sin θc = n2/n1
Total internal reflection occurs when light in the denser medium hits the boundary beyond the critical angle
The two conditions for TIR: n1 > n2 and θi > θc
Refractive Index
The refractive index of a material is the factor by which it slows light compared with a vacuum. It’s just a ratio of two speeds:
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 material. Since v is always less than c in matter, n is always greater than 1 (air is so close to a vacuum that we take n = 1). The bigger the refractive index, the more optically dense the material and the slower light travels through it — and being a ratio of speeds, n has no units.
Snell’s Law
Snell’s law ties together the two refractive indices and the two angles (both measured from the normal) at a boundary:
Snell’s lawn1 sin θ1 = n2 sin θ2
An incident ray at θ1 in material 1 refracts to θ2 in the denser material 2. Snell’s law links the two angles to the refractive indices.
The law can also be written as n1/n2 = sin θ2 / sin θ1 = v2/v1, which conveniently connects the angles, the refractive indices and the wave speeds all in one line.
Critical Angle & Total Internal Reflection
Send light from a denser medium towards a less dense one and increase the angle of incidence. The refracted ray bends further and further from the normal until, at one special angle, it skims right along the boundary — a 90° angle of refraction. That angle of incidence is the critical angleθc:
Critical angle
sin θc = n2 / n1
Push past that angle and the light can no longer escape — it all reflects back into the denser medium, obeying the law of reflection. This is total internal reflection (TIR).
From the denser medium: below θc the ray refracts out (with a weak reflection), at θc it runs along the boundary, and above θc it is totally internally reflected.
For TIR to happen, both conditions must hold: the light must be going into a less dense medium (n1 > n2), and the angle of incidence must be greater than the critical angle (θi > θc). A larger refractive index gives a smaller critical angle, making TIR easier to achieve — the principle behind optical fibres.
🧭 Working a refraction problem
Find any speed with n = c/v (and take nair = 1 if it isn’t given)
Apply Snell’s lawn1 sin θ1 = n2 sin θ2, measuring angles from the normal
For a critical angle, set the refraction angle to 90° and use sin θc = n2/n1
Sanity-check: a refractive index should come out greater than 1
Keep your calculator in degrees for these sines
Quick recap:n = c/v sets how much a material slows light; Snell’s law n1 sin θ1 = n2 sin θ2 gives the bend; and beyond sin θc = n2/n1 the light is totally internally reflected.
WE 1
Light travels from air into glass of refractive index 1.50.
Calculate the speed of light in the glass.
Rearrange n = c/v
v = c/n, with c = 3.00 × 10⁸ m s⁻¹
v = (3.00 × 10⁸) ÷ 1.50v = 2.0 × 10⁸ m s⁻¹Light is two-thirds as fast in glass as in a vacuum — that slowing is what bends the ray.
WE 2
A ray of light travels from air into a glass block. The angle of incidence is 39° and the angle of refraction is 25°.
Show that the refractive index of the glass is about 1.5.
Apply Snell’s law
n₁ sin θ₁ = n₂ sin θ₂, with n₁ = 1 (air)
n₂ = sin 39° ÷ sin 25°n₂ = 1.49 ≈ 1.5In a “show that” question, quote an extra significant figure (1.49) to prove it rounds to 1.5.
WE 3
Light travels from a material of refractive index 1.2 into air.
Determine the critical angle of the material.
Use the critical-angle equation
sin θc = n₂/n₁, with n₂ = 1.0 (air), n₁ = 1.2
sin θc = 1.0 ÷ 1.2 = 0.833θc = sin⁻¹(0.833)θc ≈ 56°Beyond 56°, light in this material hitting the boundary is totally internally reflected.
💡 Top tips
Always measure angles from the normal; if given from the surface, use 90° − θ
A refractive index must be greater than 1 — if it isn’t, check your working
Take nair = 1 whenever air isn’t assigned a value
The critical-angle formula sin θc = n2/n1 is not in the data booklet — memorise it
Calculator in degrees for every sine and inverse-sine here
⚠ Common mistakes
Measuring angles from the boundary instead of the normal
Swapping n1 and n2 in Snell’s law or the critical-angle formula
Forgetting TIR needs both conditions: denser-to-less-dense and incidence above the critical angle
Leaving the calculator in radians when finding sin θ or sin⁻¹
Reporting a refractive index below 1 and not noticing the error
Up next: what happens when two waves land in the same place at once — Superposition of Waves, the idea behind interference.
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