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Total internal reflection as light conduit

Short answer

An optical fibre works as a light conduit by total internal reflection (TIR): when a ray in the core (n1) strikes the cladding (n2) at an angle φ > critical angle φc = sin⁻¹(n2/n1), it is completely reflected with no loss. Light entering the end face within the acceptance angle θa is refracted so that φ>φc and propagates by repeated TIR.

Describe with the aid of neat diagram the basic principle of total internal reflection that enables the fiber to work as a "light conduit".

An optical fibre works as a light conduit (light pipe) because light launched into its core is trapped by total internal reflection (TIR) at the core–cladding boundary and travels along the fibre, even when the fibre bends gently.

Refraction and the critical angle

When light passes from a denser medium n1 to a rarer medium n2 (n1 > n2), Snell's law gives n1 sin φ1 = n2 sin φ2, and the refracted ray bends away from the normal.

   (a) φ < φc         (b) φ = φc         (c) φ > φc
       n2  /              n2                 n2
   -------/-------   ----------->      ---------------
     φ   /             φc /              φ \    / φ
   n1   /             n1 /             n1   \  /
       /                /                    \/
   partly refracted  grazes boundary   totally reflected
  • (a) For a small angle of incidence, most light refracts into the rarer medium (some is reflected).
  • (b) At the critical angle φc, the refracted ray travels along the boundary (φ2 = 90°):
n1 sin φc = n2 sin 90°   →   sin φc = n2 / n1
  • (c) For φ > φc, no refraction is possible and all light is reflected back into the denser medium: total internal reflection. No energy is lost at the reflection.

TIR in a fibre

 cladding n2 _________________________________
              \        /\        /\        /
 core n1   θa  \  φ>φc/  \      /  \      /
   ----*------->\    /    \    /    \    /--->
                 \  /      \  /      \  /
 cladding n2 _____\/________\/________\/______
  • The core (n1) is made slightly denser than the cladding (n2), e.g. n1 = 1.48, n2 = 1.46, giving φc = sin⁻¹(1.46/1.48) = 80.6°.
  • A ray entering the end face within the acceptance angle θa is refracted so that it meets the core–cladding boundary at φ > φc. It is totally reflected, crosses the core, meets the opposite boundary at the same angle and is reflected again.
  • Thousands of such reflections per metre guide the light along the whole length with very little loss, so the fibre acts as a "light conduit".
  • The acceptance angle follows from Snell's law: NA = sin θa = √(n1² − n2²) (here 0.24, θa ≈ 14°).
  • If the fibre is bent too sharply, φ falls below φc at the outer wall and light escapes (bending loss). The cladding also keeps the reflecting surface clean, since dust or scratches on a bare core would spoil TIR.

✓ Answer checked by Claude Opus 5.5 (high effort)

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