Total internal reflection and optical fibres
Light going from a denser to a rarer medium is reflected completely when its angle of incidence exceeds the critical angle. Optical fibres carry light signals this way.
From refraction to total reflection
Light going from a denser medium (like glass) into a rarer medium (like air) bends away from the normal, so r > i. As i increases, r reaches 90° first. The angle of incidence at which this happens is the critical angle, ic. The refracted ray then just grazes along the boundary. For i > ic, no light is refracted at all. All of it is reflected back into the denser medium. This is total internal reflection (TIR). It is called total because no light is lost to refraction. At an ordinary glass surface, most light passes through and only a little reflects.
Derivation: sin ic = 1/n
- 1Let light go from a denser medium (refractive index n₂) into a rarer medium (n₁). Snell's law: n₂ sin i = n₁ sin r.
- 2At the critical angle, i = ic and r = 90°, so sin r = 1.
- 3Then n₂ sin ic = n₁, so sin ic = n₁/n₂.
- 4★For a medium of refractive index n surrounded by air (n₁ = 1): sin ic = 1/n, or n = 1/sin ic.
★ key step
Critical angles of some media (with air)
| Medium | Refractive index n | Critical angle ic |
|---|---|---|
| Water | 1.33 | about 48.8° |
| Glass | 1.50 | about 41.8° |
| Crown glass | 1.52 | about 41.1° |
| Diamond | 2.42 | about 24.4° |
Where total internal reflection is used
- Totally reflecting prisms: a right-angled isosceles glass prism has angles of 90°, 45° and 45°. It turns light through 90° or 180°, or inverts an image without changing its size. Inside, the light meets a face at 45°, more than the critical angle of glass (about 42°). Used in periscopes and binoculars.
- Optical fibres: thin, flexible strands of very pure glass or quartz. Each has a core surrounded by a cladding of lower refractive index.
- Light entering one end meets the core-cladding boundary at more than the critical angle. It is totally reflected again and again, and leaves the far end with very little loss, even when the fibre bends.
- Fibres carry telephone, TV and internet signals as pulses of light. Bundles of fibres (light pipes) let doctors look inside the body in endoscopy.
- Extra reading, not in your syllabus: the sparkle of a cut diamond (critical angle about 24.4°) and the mirage on a hot road are also linked to total internal reflection.
How to draw an optical fibre and a totally reflecting prism
- Fibre: draw two long parallel lines for the core, and two outer parallel lines for the cladding. Label both and write 'n(core) > n(cladding)'.
- Draw a ray entering the left end of the core at a small angle to the axis.
- Draw the ray zigzagging along the core, reflecting at the core-cladding boundary each time. Draw a dotted normal at one reflection and mark i > ic.
- Show the ray leaving the far end, and put arrows on every part of the ray.
- Prism: draw a right-angled isosceles triangle. Draw a ray entering one short face at 90°, meeting the long face at 45°, and leaving the other short face at 90°. Mark '45° > ic' at the long face.
Worked example: critical angle
(a) Find the critical angle for a medium of refractive index √2. (b) Light travels at 2 × 10⁸ m s⁻¹ in a glass. Find n and sin ic for this glass in air (c = 3 × 10⁸ m s⁻¹). (c) An optical fibre has a core of n = 1.50 and a cladding of n = 1.44. Find sin ic at the core-cladding boundary.
- (a) sin ic = 1/n = 1/√2, so ic = 45°.
- (b) n = c/v = (3 × 10⁸) ÷ (2 × 10⁸) = 1.5.
- sin ic = 1/n = 1/1.5 = 2/3 ≈ 0.67, so ic ≈ 41.8°.
- (c) sin ic = n(cladding)/n(core) = 1.44 ÷ 1.50 = 0.96, so ic ≈ 74°.
- Any ray that meets the boundary at more than about 74° stays inside the core.
(a) 45°. (b) n = 1.5, sin ic = 2/3 (ic ≈ 41.8°). (c) sin ic = 0.96 (ic ≈ 74°)
Optical fibre cables carry India's internet and phone calls, including the lines that link village panchayats. Doctors pass a thin bundle of fibres, an endoscope, down the throat to look inside the stomach without surgery.
2 marks: “State two conditions for total internal reflection. Write the relation between the critical angle and the refractive index.” Light must travel from an optically denser to a rarer medium, and the angle of incidence must be greater than the critical angle. sin ic = 1/n for a medium of refractive index n in air.
3 marks: “With a labelled diagram, explain how an optical fibre carries light. Why is the refractive index of the cladding kept lower than that of the core?” Core and cladding with n(core) > n(cladding); light meets the boundary at i > ic and is totally reflected again and again, with little loss. A lower cladding index makes light go from denser to rarer at the boundary, the first condition for TIR. Uses: communication, endoscopy.
2 marks: “The critical angle for a medium is 30°. Find its refractive index and the speed of light in it (c = 3 × 10⁸ m s⁻¹).” n = 1/sin 30° = 2. v = c/n = 3 × 10⁸ ÷ 2 = 1.5 × 10⁸ m s⁻¹.
Denser to rarer, beyond critical: both must be true for total internal reflection.