Year 11 PhysicsModule 3⏱ ~40 min5 MC · 3 Short AnswerLesson 15 of 18
Refraction, Snell's Law and Total Internal Reflection
In 1992, Corning launched its SMF-28 single-mode optical fibre, the glass used in Australia's NBN backbone and the Sydney–Los Angeles undersea cable (12,000 km). The core has $n_{core} = 1.4677$ and the cladding $n_{clad} = 1.4640$, giving a critical angle of $\theta_c = \sin^{-1}(1.4640/1.4677) = 86.9°$. Light pulses travel through the 12,000 km cable at $0.67c$ (200,000 km/s), every bounce off the core-cladding interface undergoes total internal reflection because the angle always exceeds 86.9°. Without TIR, the cable would be useless.
Today's hook: Corning's 1992 SMF-28 optical fibre carries internet traffic across the Sydney–Los Angeles undersea cable (12,000 km) at 0.67c. The core ($n = 1.4677$) and cladding ($n = 1.4640$) differ by only 0.0037 in refractive index, yet that tiny difference creates a critical angle of 86.9°, every light pulse bounces along the fibre by total internal reflection, losing less than 0.2 dB per kilometre. Snell's law ($n_1\sin\theta_1 = n_2\sin\theta_2$) and the critical angle formula are the two tools that make this engineering possible.
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Warm up first
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Before you read, predict
Light travels from glass ($n = 1.5$) into air. At what angle of incidence do you think all the light will be reflected back into the glass rather than passing through? Predict and explain.
Warm-up, the refractive index of a medium is defined as:
TIR occurs when $\theta_i > \theta_c$ going from denser to less-dense medium
Understand
Why higher refractive index → more bending at a boundary
The two conditions required for TIR
How optical fibres use TIR
Can Do
Apply Snell's law to calculate refraction angles
Calculate critical angle from refractive indices
Determine whether TIR will occur for a given situation
Cross-lesson links: Refraction was introduced qualitatively in L05. This lesson adds Snell's law and the critical angle formula. The Corning SMF-28 fibre (1992, $n_{core}=1.4677$, $n_{clad}=1.4640$, $\theta_c=86.9°$) connects to the NBN context from L05. TIR also appears in L14 (why mirrors outperform lenses for large telescopes) and L16 (prism binoculars use TIR instead of mirrors).
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Snell's Law
Core Content
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Snell's Law
+5 XP
Push a pencil into a glass of water at an angle and watch it: the pencil appears to bend sharply at the water surface, the submerged portion shifted toward the normal. Pull the pencil out slowly and the bend disappears the instant it leaves the water. This visible kink is refraction, the pencil has not bent, but the light travelling from the submerged part changes direction as it crosses the water-air boundary. The faster the light moves in the second medium, the more it bends away from the normal at that interface.
Snell's law
$n_1 \sin\theta_1 = n_2 \sin\theta_2$
$n$ = refractive index · $\theta$ = angle from normal · $n_{air} \approx 1.00$ · $n_{glass} \approx 1.5$ · $n_{water} \approx 1.33$
Worked example
Light in air hits glass ($n = 1.5$) at 40°. Find the refracted angle.
$1.00 \times \sin 40° = 1.5 \times \sin\theta_2$
$\sin\theta_2 = \sin 40°/1.5 = 0.643/1.5 = 0.429$
$\theta_2 = \sin^{-1}(0.429) = 25.4°$
Snell's law: $n_1\sin\theta_1 = n_2\sin\theta_2$, where $n = c/v$ (dimensionless). Refractive index measures how much slower light travels in a medium compared to vacuum; higher $n$ means greater bending toward the normal when entering.
Pause, copy the highlighted law and definition into your book before moving on.
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Total Internal Reflection and Critical Angle
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Total Internal Reflection and Critical Angle
+5 XP
We just saw that Snell's law predicts how much light bends at a boundary. That raises a question: is there a situation where refraction is impossible, where the light cannot cross into the lower-index medium? This card answers it → yes, when the angle of incidence exceeds the critical angle, all light is reflected back into the higher-index medium (total internal reflection).
TIR occurs when light travels from a higher refractive index to a lower refractive index AND the angle of incidence exceeds the critical angle. At $\theta_c$, the refracted ray grazes the surface ($\theta_2 = 90°$):
For glass ($n = 1.5$) to air: $\sin\theta_c = 1/1.5 = 0.667 \Rightarrow \theta_c = 41.8°$. Any ray hitting the glass-air boundary at more than 41.8° undergoes total internal reflection.
A transparent optical fibre has a higher-index solid core surrounded by lower-index cladding. An exact ray reflects repeatedly inside the core. Three interface diagrams show partial reflection and refraction below the critical angle, a refracted ray along the boundary at the critical angle, and reflection only above the critical angle.
Total internal reflection occurs only when light travels from higher to lower refractive index and the incidence angle exceeds the critical angle. Every angle is measured from the normal.
Read the rays: Cover the panel headings. Identify each state from its ray behaviour, then explain why the fibre core must have the higher refractive index.
TIR conditions: (1) light must travel from a higher-index to a lower-index medium, and (2) angle of incidence must exceed $\theta_c$, where $\sin\theta_c = n_2/n_1$. Applications: optical fibres, endoscopes, diamonds ($\theta_c = 24.4°$), prism binoculars.
Add the highlighted TIR conditions and critical angle formula to your notes before the check below.
Light travels from water ($n = 1.33$) to air. The critical angle is approximately:
TIR can occur when light travels from air into glass.
A higher refractive index means light travels slower in that medium.
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Activity 3, Snell's Law Calculations
Activities
Activity 3, Snell's Law Calculations
ApplyBand 4
Show all working:
Light in air hits glass ($n = 1.5$) at 30°. Find the refracted angle.
Light in glass ($n = 1.5$) hits air at 25°. Find the refracted angle.
Light in air hits diamond ($n = 2.42$) at 45°. Find the refracted angle.
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Activity 4, Critical Angle and TIR
Activity 4, Critical Angle and TIR
ApplyBand 4
For each situation, calculate the critical angle and state whether TIR occurs:
Glass ($n = 1.5$) to air; angle of incidence = 40°
Glass ($n = 1.5$) to air; angle of incidence = 50°
Diamond ($n = 2.42$) to air; angle of incidence = 30°
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Activity 2, Optical Fibre Explanation
Activity 2, Optical Fibre Explanation
ExplainBand 4
Explain how an optical fibre uses total internal reflection to transmit a light pulse from one end to the other without the light escaping through the side of the fibre.
TIR cannot occur in which situation?
A medium has $n = 2.0$. Its critical angle to air is:
Light in air strikes glass ($n = 1.5$) at 60° to the normal. Applying Snell's law, the refracted angle is approximately:
✓
Show what you have learned
Multiple Choice, Snell's law and TIR
+5 XP
Short Answer, 10 marks
+5 XP
UnderstandBand 3(3 marks) 1. State the two conditions required for total internal reflection to occur. Explain what happens at the critical angle.
ApplyBand 4(3 marks) 2. Light travels from glass ($n = 1.5$) to water ($n = 1.33$). Calculate the critical angle for this interface.
AnalyseBand 5(4 marks) 3. Explain why optical fibres can transmit light signals over thousands of kilometres with minimal loss. Your answer should include Snell's law, the critical angle, and the structure of the fibre.
Q1 (3 marks): Condition 1: Light must travel from a medium with higher refractive index to one with lower refractive index. Condition 2: The angle of incidence must be greater than the critical angle. At the critical angle itself, the refracted ray travels along the boundary ($\theta_r = 90°$); TIR begins only beyond $\theta_c$, when all light is reflected back into the higher-index medium.
Q3 (4 marks): An optical fibre has a high-$n$ silica core surrounded by lower-$n$ cladding. Light enters the core end at a shallow angle (nearly parallel to the axis). At every core-cladding boundary, the angle of incidence exceeds the critical angle (by Snell's law: $\sin\theta_c = n_{clad}/n_{core} < 1$). TIR occurs at each reflection, no light escapes through the side. The only significant loss mechanisms are absorption in the glass and scattering at impurities, not refraction. Modern fibres achieve losses as low as 0.2 dB/km.
✓
Retrieve, reflect and finish
Check what actually stuck
How did your thinking change?
The Corning SMF-28 fibre launched in 1992 makes the numbers real: core $n = 1.4677$, cladding $n = 1.4640$, critical angle $\theta_c = \sin^{-1}(1.4640/1.4677) = 86.9°$. Because the core-cladding refractive index difference is very small, the critical angle is very large, nearly grazing the surface, so light must travel almost parallel to the fibre axis to bounce by TIR, which is exactly how a single-mode fibre works. The Sydney–Los Angeles cable (12,000 km) carries your internet traffic at 0.67c entirely by this mechanism. You can now calculate the critical angle for any pair of media using $\sin\theta_c = n_2/n_1$.
