3.5.5Stoichiometry
Total Internal Reflection — Cambridge IGCSE Chemistry 0620 Revision Notes & Topical Questions
5 min read 5 parts Lesson 11 of 28 Unit quiz
Section 1 of 5Critical Angle
The critical angle depends on the refractive index of the denser material.
Critical angle
- The angle of incidence in the denser medium that makes the refracted ray travel along the boundary at exactly 90° to the normal.
Equation
- The critical angle c is linked to the refractive index n by sin c = 1 ÷ n, and n must be greater than 1 for this to work.
Refractive index effect
- A material with a larger refractive index has a smaller critical angle, so light is trapped inside it more easily.
Requirement
- Total internal reflection needs light travelling from a denser to a less dense medium at an angle larger than the critical angle.
Section 2 of 5Behaviour at the Boundary
1
When the angle of incidence is smaller than the critical angle, the light is refracted out of the denser material.
2
When the angle of incidence equals the critical angle, the refracted ray grazes along the boundary between the two materials.
3
When the angle of incidence is larger than the critical angle, no light leaves the denser material, so all of it is reflected back inside.
4
The totally reflected ray obeys the law of reflection, so its angle of reflection equals its angle of incidence.
Section 3 of 5Uses of Total Internal Reflection
Uses of Total Internal Reflection
Endoscopes
- Bundles of optical fibres carry light into the body and return images, so doctors can see inside a patient without surgery.
Communications
- Electrical signals are converted into light pulses that travel along optical fibres carrying internet and telephone data.
Fibre advantages
- Fibres use less energy, need fewer boosters and suffer no interference from nearby cables, unlike copper cables.
Fibre structure
- Each fibre has a glass core, which is the denser medium, surrounded by a less dense material that keeps light inside it.
Section 4 of 5Worked Example
Question
- Light travels in water (n = 1.33) and meets the water–air surface.
- Calculate the critical angle and state whether a ray at 55° to the normal is totally internally reflected.
- Step 1: Write sin c = 1 ÷ n = 1 ÷ 1.33.
- Step 2: sin c = 0.752, so c = sin⁻¹(0.752) = 48.8°, which is 49° to 2 significant figures.
- Step 3: Compare the ray: 55° is larger than 49°, so the ray is totally internally reflected.
Answer The critical angle is 49°, and the ray at 55° is totally internally reflected.
Section 5 of 5Exam Checklist & Tips
3.5.5 – Key Points for the Exam
- Critical angle The angle of incidence in the denser medium that gives a refraction angle of 90°.
- Total internal reflection Complete reflection of light back into a denser medium when the angle of incidence exceeds the critical angle.
- Denser medium A material with a higher refractive index, in which light travels more slowly.
- Optical fibre A thin glass fibre that guides light along its length by total internal reflection.
- Law of reflection The rule that the angle of incidence equals the angle of reflection at a surface.
Common Mistakes
- Saying total internal reflection happens at any angle;
- it happens only when the angle of incidence exceeds the critical angle.
- Using sin c = n instead of sin c = 1 ÷ n, which would give an impossible value for a material with n greater than 1.
- Describing the critical angle as an angle of refraction of 0°;
- at the critical angle the refraction angle is 90°.
- Thinking total internal reflection happens in the less dense medium; it occurs inside the denser material.
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