Total Internal Reflection
Total internal reflection requires n1 > n2 and i > C. At i = C the transmitted ray grazes the interface; real interfaces can still have evanescent fields or losses.
Why this shows up in the exam
Right-angle prisms · Light pipes · Gemstone cutting
Learn the idea
Light is totally reflected when it travels from higher to lower refractive index and incidence exceeds the critical angle. Beyond the escape cone, no propagating refracted ray can satisfy Snell's law, so the energy returns by reflection in the ideal model.
🧠 Memory hook: TIR needs both: denser to rarer, and i above C.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- i > C — angular condition after confirming n1 > n2
- R = 1 — ideal lossless power reflectance under total internal reflection
How to approach it
- 1Order the refractive indices
- 2Find or infer C
- 3Compare i with C and trace the reflected ray
Common slip-ups that cost marks
- •Checking only the angle
- •Including i = C as total reflection
- •Assuming ordinary reflection has no transmitted energy at all angles
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A real object is placed 30 cm from a converging lens of focal length 10 cm. Find the real image distance.
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