Optical fibres carry the entire internet across oceans by trapping light inside thin glass threads. The trick is total internal reflection: beyond a special angle, light hitting the inside of a boundary cannot escape and is reflected perfectly back. Find that critical angle from the two refractive indices and you understand fibre optics, prisms and the sparkle of a diamond.
Light inside glass approaches the glass–air boundary and you slowly increase the angle of incidence.
Before reading: can total internal reflection happen when light goes from glass into air, or only from air into glass? Write your prediction and one sentence explaining why.
The angle of incidence at which the refracted ray bends to 90°
As light leaving a denser medium tilts flatter, the refracted ray bends further from the normal, until at the critical angle it lies right along the boundary, and beyond it cannot escape at all.
Going from a denser medium ($n_1$) to a less dense one ($n_2$), the ray bends away from the normal. As the angle of incidence grows, the angle of refraction grows faster, reaching $90°$ at the critical angle $\theta_c$. Setting $\theta_2 = 90°$ in Snell's law ($n_1\sin\theta_c = n_2\sin 90°$) gives:
$$\sin\theta_c = \frac{n_2}{n_1} \qquad (n_1 > n_2)$$
The critical angle is where the refracted ray reaches $90°$. From Snell's law with $\theta_2 = 90°$: $\sin\theta_c = \dfrac{n_2}{n_1}$, valid only when $n_1 > n_2$ (denser to less dense).
Pause, copy the highlighted critical-angle relation into your book before moving on.
Did you get this? True or false: at exactly the critical angle, the refracted ray travels along the boundary at $90°$ to the normal.
Using the two refractive indices to find θc and predict TIR
We just saw that $\sin\theta_c = n_2/n_1$. That raises a question: for a real glass–air boundary, what is the actual critical angle, and how do we use it to decide whether a ray escapes? This card answers it → calculate $\theta_c$, then compare it with the angle of incidence.
Once you know the critical angle, the rule is simple: if the angle of incidence is greater than $\theta_c$, the ray is totally internally reflected; if it is smaller, the ray refracts and escapes.
Find the critical angle for light passing from glass ($n_1 = 1.50$) into air ($n_2 = 1.00$).
Critical angle: $\theta_c = \sin^{-1}\!\left(\dfrac{n_2}{n_1}\right)$. If the angle of incidence exceeds $\theta_c$, the ray is totally internally reflected; if it is less, the ray refracts out. For a glass–air boundary ($n=1.50$), $\theta_c \approx 42°$.
Add the highlighted method and the rule for TIR to your notes before the check below.
Quick check: A glass–air boundary has a critical angle of $42°$. A ray inside the glass hits the boundary at $50°$. What happens?
Total internal reflection needs two things: a slower medium to start in, and an angle past θc
Use the simulation. With glass ($n = 1.50$) above air ($n = 1.00$), beyond which angle of incidence is there no refracted ray?
Use the simulation. True or false: light travelling from air into water undergoes total internal reflection if the angle of incidence is large enough.
Pattern A, Calculate then reason
An optical fibre core has refractive index $n_1 = 1.48$ surrounded by cladding of $n_2 = 1.46$. In your book: (1) calculate the critical angle at the core–cladding boundary; (2) state whether a ray hitting that boundary at $84°$ undergoes total internal reflection; (3) explain in one sentence how this keeps light travelling along the fibre.
Two truths, one lie: Two truths and a lie about total internal reflection. Which statement is the lie?
Dot point: Solve total internal reflection and critical angle problems using sinθc = n₂/n₁
Why the difference in indices is really a difference in wave speed
We just learned to predict TIR from the indices. That raises a question: what physically differs between two media that have different refractive indices? This card answers it → the speed of light is different in each, and the bigger the speed change, the bigger the bend.
A refractive index is just a coded form of wave speed, a larger difference in refractive indices at an interface means a larger change in the speed of light, and therefore a larger change in direction.
Since $n = c/v$, a higher refractive index means a lower speed $v$. When light crosses an interface, the ratio of speeds equals the inverse ratio of indices:
$$\frac{v_1}{v_2} = \frac{n_2}{n_1}$$
So the change in velocity across an interface is set entirely by the difference between the two refractive indices: a small index difference (like fibre core and cladding) means only a small speed change and a gentle bend, which is why the critical angle is large and TIR happens readily. A large index difference means a big speed change, a sharper bend, and a smaller critical angle.
Because $n = c/v$, the change in a light wave's velocity at an interface is governed by the difference in refractive indices: $\dfrac{v_1}{v_2} = \dfrac{n_2}{n_1}$. A larger index difference means a larger speed change and a sharper change of direction.
Pause, copy the highlighted velocity–index link into your book before the check below.
Last check: Decide if the statement is true or false.
Two media with very different refractive indices have almost the same speed of light inside them.
Match them up: Match each pair of media to what happens to the light's speed and direction at the boundary.
Dot point: Relate the change in velocity of a light wave crossing an interface to differences between refractive indices
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 diamond has a refractive index of $2.42$ and is surrounded by air ($n = 1.00$). Calculate the critical angle for the diamond–air boundary, and explain why total internal reflection causes a cut diamond to sparkle brilliantly.
1 mark: uses $\sin\theta_c = n_2/n_1$ · 1 mark: $\sin\theta_c = 0.413$ · 1 mark: $\theta_c = 24.4°$ · 1 mark: small critical angle means most rays exceed it, so light is repeatedly internally reflected before exiting
ApplyBand 5(3 marks) 2. An optical fibre core has a refractive index of $1.52$ and is surrounded by cladding of refractive index $1.44$. Calculate the critical angle at the core–cladding boundary, and state what happens to a ray striking that boundary at $75^\circ$.
1 mark: correct substitution into $\sin\theta_c = n_2/n_1$ · 1 mark: $\theta_c = 71^\circ$ (2 s.f.) · 1 mark: $75^\circ$ exceeds the critical angle, so the ray is totally internally reflected and stays in the core
Dot point: Solve total internal reflection and critical angle problems using sinθc = n₂/n₁
AnalyseBand 6(3 marks) 3. Explain why total internal reflection can occur when light travels from glass into air, but never when it travels from air into glass, relating your answer to the change in the light's velocity at the interface.
1 mark: going from glass to air the light speeds up, so it bends away from the normal · 1 mark: the refracted angle can therefore reach $90^\circ$ at a finite angle of incidence, defining a critical angle · 1 mark: going from air to glass the light slows and bends towards the normal, so the refracted angle is always smaller and can never reach $90^\circ$
Dot point: Relate the change in velocity of a light wave crossing an interface to differences between refractive indices
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): The critical angle is found from $\sin\theta_c = \dfrac{n_2}{n_1} = \dfrac{1.00}{2.42} = 0.413$, so $\theta_c = \sin^{-1}(0.413) = 24.4°$. This is a very small critical angle, which means almost any ray of light that travels inside the diamond will hit a facet at an angle greater than $24.4°$ and undergo total internal reflection rather than escaping. The light therefore bounces around inside the stone many times before finally exiting, producing the intense sparkle and "fire" of a cut diamond.
Q2 (3 marks): The critical angle is given by $\sin\theta_c = n_2/n_1$, where $n_1$ is the denser medium the light is travelling in. Substituting, $\sin\theta_c = 1.44/1.52 = 0.9474$, so $\theta_c = 71.4^\circ$, which is $71^\circ$ to two significant figures. A ray meeting the boundary at $75^\circ$ has an angle of incidence greater than the critical angle. No light is refracted out into the cladding; instead the ray undergoes total internal reflection and remains within the core. This is precisely how an optical fibre guides light along its length, with the ray reflecting repeatedly from the core–cladding boundary and losing very little energy.
Q3 (3 marks): When light passes from glass into air it enters a medium of lower refractive index, so its velocity increases and the ray bends away from the normal. The angle of refraction is therefore always larger than the angle of incidence. As the angle of incidence is increased, the angle of refraction grows faster and reaches $90^\circ$ while the angle of incidence is still less than $90^\circ$. That particular angle of incidence is the critical angle; beyond it there is no refracted ray at all and all the light is reflected back into the glass. Going the other way, from air into glass, the light enters a medium of higher refractive index, so its velocity decreases and the ray bends towards the normal. The angle of refraction is then always smaller than the angle of incidence, and even at a grazing angle of incidence approaching $90^\circ$ the refracted angle only reaches the critical angle's value, never $90^\circ$. There is no angle at which the refracted ray disappears, so total internal reflection is impossible in that direction.
You can now calculate a critical angle, decide whether a ray refracts or totally internally reflects, and link the velocity change at an interface to the refractive indices. Next lesson is a practical: measuring the refractive index of a material and assessing the accuracy of the result.
↩ Back to focus-area hubAt the start you predicted whether TIR happens from glass into air or air into glass. It happens only going from the denser to the less dense medium (glass → air), and only when the angle of incidence exceeds the critical angle. If you said air → glass, revisit Card 1, light must be trying to leave the slower, denser medium.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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