A light ray travels from glass (n = 1.5) toward air (n = 1.0). The angle of incidence is gradually increased from 0°. What happens when the angle exactly equals the critical angle?
AAll light is reflected back into the glass and the refracted ray disappears
BThe refracted ray travels along the glass-air interface at 90° to the normal; TIR begins for any larger angle
CThe reflected and refracted rays have exactly equal intensities
DThe angle of refraction equals the angle of incidence
At the critical angle, sin θ_c = n₂/n₁ = 1.0/1.5, so θ₂ = 90° — the refracted ray just grazes the boundary. This is the threshold: at this angle there is still a (grazing) refracted ray. For any angle greater than θ_c, the math requires sin θ₂ > 1, which is impossible, so no refracted ray exists and all energy reflects back. Option A describes what happens beyond the critical angle, not at it exactly.
Question 2 Multiple Choice
A student shines light from air into water and increases the angle of incidence all the way to 85°. Why does total internal reflection not occur?
AThe angle is not large enough — the critical angle for the air-water interface is greater than 85°
BTIR requires light to travel from the denser medium into the less dense medium; light going from air into water cannot undergo TIR regardless of angle
CTIR only occurs in manufactured materials like optical fiber glass, not in water
DThe light is partially absorbed by the water before it can reflect
TIR is fundamentally asymmetric: it only occurs when light travels from a denser medium (higher n) to a less dense medium (lower n). Going from air (n ≈ 1.0) into water (n ≈ 1.33) is the wrong direction — the refracted ray bends toward the normal, and the refraction angle is always less than the incidence angle. There is no critical angle for this direction because sin θ₂ = (n₁/n₂) sin θ₁ < sin θ₁, never reaching 90°.
Question 3 True / False
At angles greater than the critical angle, the interface between two media acts as a perfect mirror — all incident light is reflected with no energy loss.
TTrue
FFalse
Answer: True
TIR is lossless in principle because there is no refracted ray into which energy can be transmitted. The evanescent field technically penetrates a fraction of a wavelength into the second medium, but no net energy is carried away. This is fundamentally different from ordinary reflection off a metal mirror (which absorbs a few percent) or off a glass surface near normal incidence (which reflects only ~4%). The lossless nature of TIR is what makes it so valuable for optical fibers.
Question 4 True / False
Total internal reflection can occur when light travels from air into glass if the angle of incidence is large enough.
TTrue
FFalse
Answer: False
TIR requires traveling from a higher-index medium into a lower-index medium. Air has n ≈ 1.0; glass has n ≈ 1.5. Light going from air into glass is moving into a denser medium, where refraction bends the ray toward the normal — the refraction angle is always less than the incidence angle and can never reach 90°. The sin factor (n_air/n_glass < 1) ensures no critical angle exists for this direction.
Question 5 Short Answer
Explain why optical fibers can transmit light over kilometers with minimal loss, using the concept of total internal reflection.
Think about your answer, then reveal below.
Model answer: An optical fiber has a glass core with higher refractive index surrounded by cladding with a slightly lower index. Light entering the fiber at a shallow enough angle hits the core-cladding interface at an angle exceeding the critical angle, triggering TIR. The light reflects perfectly back into the core — no energy escapes into the cladding. This process repeats at every bounce along the entire length of the fiber. Since TIR is lossless, the only significant attenuation comes from absorption within the glass itself, not from the reflections.
The key insight is that TIR is not just 'good reflection' — it is theoretically perfect reflection. Conventional mirrors and partial reflections lose a small percentage at each bounce, which compounds over thousands of reflections across kilometers. TIR loses essentially nothing per bounce. The fiber geometry is designed so that any light ray within the acceptance cone strikes the boundary beyond the critical angle, guaranteeing TIR for the entire guided signal.