--- title: "Total internal reflection as light conduit" subject: "Propagation and Antenna" chapter: 6 key: "semester_6/propagation_and_antenna:total-internal-reflection-light-conduit" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/total-internal-reflection-light-conduit site: BCT Engineering --- # Total internal reflection as light conduit Propagation and Antenna · Chapter 6: Optical Fibres - Asked in: 2075 Bhadra, 6 marks - Video: [Total Internal Reflection of Light and Critical Angle of Refraction Physics](https://www.youtube.com/watch?v=5bkiQob8ikc) — The Organic Chemistry Tutor - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/total-internal-reflection-light-conduit - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-6 ## Question Describe with the aid of neat diagram the basic principle of total internal reflection that enables the fiber to work as a "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. ## Answer 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. ```text (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°): ```text 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 ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What happens at the critical angle when light goes from n1 to n2 (n1 > n2)? A. The ray is totally reflected B. The ray goes straight through C. The ray bends towards the normal D. The refracted ray travels along the boundary (90 degrees to the normal) Answer: D. At phi_c, n1 sin(phi_c) = n2 sin 90, so sin(phi_c) = n2/n1. (Key point: Critical angle) 2. What happens for an angle of incidence greater than the critical angle? A. All light is reflected back into the denser medium with no energy lost B. Part of the light refracts out C. Light is absorbed D. Light travels along the boundary Answer: A. No refraction is possible, so reflection is total. (Key point: TIR above critical) 3. For n1 = 1.48 and n2 = 1.46, what is phi_c? A. 14 degrees B. 80.6 degrees C. 9.4 degrees D. 45 degrees Answer: B. phi_c = sin^-1(1.46/1.48) = 80.6 degrees. (Key point: Critical angle value) 4. Why does light escape if the fibre is bent too sharply? A. The core index changes B. The cladding becomes denser C. The angle at the outer wall falls below the critical angle D. The light speeds up Answer: C. Below phi_c, TIR fails and light refracts into the cladding. 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