Total Internal Reflection for A Level Physics - interactive
Total Internal Reflection (TIR) is one of the most practical wave phenomena in A-Level Physics (AQA, OCR, Edexcel). It governs everything from high-speed fibre optic telecommunications to medical endoscopes. Let's explore the conditions required for TIR and test them using our interactive visual simulation below.
⚡ Interactive Optics Lab: Total Internal Reflection
Adjust the angle of incidence (θ1) and medium density to see refraction transition into total internal reflection.
1. What is Total Internal Reflection?
When light travels from an optically denser medium with a higher refractive index (n1) into an optically less dense medium with a lower refractive index (n2), the light ray speeds up and refracts away from the normal line.
As you increase the angle of incidence (θ1), the angle of refraction (θ2) also increases until it reaches exactly 90°. The angle of incidence that causes refraction along the boundary is called the Critical Angle (θc).
The Two Strict Conditions for TIR to Occur:
- The light wave must be traveling from a medium of higher refractive index (n1) into a medium of lower refractive index (n2) (e.g., glass to air).
- The angle of incidence (θ1) must be strictly greater than the critical angle (θc).
2. Deriving the Critical Angle Formula
Snell's Law of Refraction states:
n1 sin(θ1) = n2 sin(θ2)
At the critical point where θ1 = θc, the refracted angle θ2 = 90°. Since sin(90°) = 1:
n1 sin(θc) = n2 (1) ⇒ sin(θc) = n2 / n1
When Medium 2 is air (n2 ≈ 1.00), the formula simplifies to:
sin(θc) = 1 / n1
💡 A-Level Exam Trick: Optical Fibres & Modal Dispersion
In A-Level exam questions on Step-Index Optical Fibres, you are often asked why the cladding is needed. The core has a high n1, while the cladding has a slightly lower n2. The cladding:
- Protects the core from scratches (which allow light to escape).
- Ensures TIR occurs inside the core.
- Reduces signal degradation across long distances.
3. Challenging A-Level Exam Questions
Question 1 (Prism Geometry & Critical Proof)
The two non-right angles of a right-angled glass prism are x° and y° respectively. A light ray enters the prism perpendicularly through its hypotenuse (the longest side), undergoes Total Internal Reflection at the side adjacent to angle x°, and then refracts into air at the side adjacent to angle y°.
Show that x° > y°.
Click to view full mathematical proof
Proof:
- Since the prism is right-angled, x + y = 90°.
- The ray enters the hypotenuse perpendicularly (at 90°), so it passes through unbent and strikes the side adjacent to angle x°.
- By interior geometry, the angle of incidence at the side adjacent to x° is i1 = x°.
- Since the ray undergoes Total Internal Reflection at this face, i1 > θc ⇒ x° > θc.
- The reflected ray then strikes the side adjacent to angle y°. By geometry, its angle of incidence at this second face is i2 = y°.
- Because the ray refracts out into air at this second face rather than reflecting, its angle of incidence must be less than the critical angle: i2 < θc ⇒ y° < θc.
- Combining these two inequalities gives: y° < θc < x° ⇒ x° > y°. Q.E.D.
Question 2 (Optical Fibre Modal Dispersion)
A step-index optical fibre has a core refractive index n1 = 1.48 and a cladding refractive index n2 = 1.44. The core length is 2.5 km.
- Calculate the maximum angle of incidence θmax at the air-core boundary for which light is guided along the core.
- Calculate the time delay Δt between the fastest (axial) ray and the slowest (maximally reflected) ray traveling down the fibre.
Question 3 (Temperature-Dependent Refraction)
A container holds a liquid whose refractive index decreases linearly with increasing temperature according to n(T) = 1.45 - 0.002T (where T is in °C). A glass rod of refractive index ng = 1.35 is immersed in the liquid.
Determine the minimum temperature Tmin of the liquid required for Total Internal Reflection to occur when light attempts to pass from the liquid into the glass rod.
Question 4 (Composite Prisms & Critical Boundaries)
An equilateral triangular glass prism (nA = 1.60) is glued along one face to a second triangular glass prism (nB = 1.38). A monochromatic ray enters Prism A perpendicular to its outer face.
Determine whether Total Internal Reflection occurs at the internal interface between Prism A and Prism B, showing all your working.
Question 5 (Endoscope Acceptance Semi-Angle)
A medical endoscope bundle consists of unclad glass fibres (n = 1.52) surrounded by air.
Explain mathematically why any light ray entering the flat entrance face of an unclad fibre at an angle of incidence in air θ0 < 38.5° will automatically undergo Total Internal Reflection at the core boundary inside the fibre.