Critical Angle (Total Internal Reflection)

Equivalent forms

\sin\theta_c = n_2 / n_1

A simple arcsine separates 'most light escapes' from 'every photon stays inside' — the principle behind every fiber-optic cable on Earth.

Symbol	Name	SI unit	Dimension	Range
$theta_c$	Critical angleoutput Smallest incidence angle producing total internal reflection	rad	1	0 – 1.5708
$n_1$	Refractive index (denser medium) Refractive index of the medium light is currently in	dimensionless	1	1 – 3
$n_2$	Refractive index (rarer medium) Refractive index of the medium on the other side	dimensionless	1	1 – 3

Unit systems

Symbol	SI	CGS	Imperial
$theta_c$	rad	rad	deg

Where it holds

Valid for light going from optically denser to rarer medium (n_1 > n_2), monochromatic and paraxial-equivalent geometry. Evanescent wave penetrates

\sim \lambda

into rarer medium even under TIR.

Dimensional analysis

\theta _c \to [1]

(radian)

\arcsin (n_2/n_1) \to \arcsin ([1]/[1]) = [1]

Both sides dimensionless.

\checkmark

Discovery

Johannes Kepler · 1611

Kepler observed and documented total internal reflection in 'Dioptrice'. The formal angle condition was derived from Snell's law shortly after Snellius's discovery.

Try this

How does a thin glass thread carry the entire internet around the planet?

Light travels from glass (n₁ = 1.5) into air (n₂ = 1.0). Find the critical angle at which total internal reflection begins.

Research status: stable

Real-world applications

Optical fibers: cladding n < core n keeps light bouncing inside core (entire internet runs on this).
Diamond sparkle: $n = 2.42$ gives $\theta _c \approx 24^{\circ}$ , trapping most light to escape only from cuts.
Periscopes and binoculars: TIR prisms replace mirrors with no metal coatings (no tarnish).
Endoscopes: bundled fibers carry images out of the human body via TIR.

Common misconceptions

TIR has zero leakage — an evanescent wave extends $\sim \lambda$ into the rarer medium and can couple to another nearby medium (frustrated TIR).
TIR happens at any interface — only when light heads from denser to rarer medium.
Critical angle depends on wavelength — only weakly, through dispersion of n_1 and n_2.

Experimental verification

Demonstrated by Kepler with prism experiments. Modern optical fibers maintain >99.999% reflectivity per bounce, propagating light hundreds of km with minimal loss; verified daily in global telecom infrastructure.

Derivation

Apply Snell's law:

n_1 \sin \theta _1 = n_2 \sin \theta _2

.

As \theta _1

increases,

\theta _2

grows faster (when n_1 > n_2) until

\sin \theta _2 = 1

, i.e.,

\theta _2 = 90^{\circ}

.

The corresponding incidence angle

is \theta _c = \arcsin (n_2/n_1)

.

For \theta _1

>

\theta _c

,

\sin \theta _2

would exceed 1 — impossible for real angles — and Fresnel coefficients show 100% reflection back into medium 1.

Limiting cases

n_1 = n_2

⟶

\theta _c = 90^{\circ}

No critical angle exists — refraction occurs at all incidence angles.

n_2 / n_1 \to 0

⟶

\theta _c \to 0

Even nearly normal incidence reflects internally — used in highly confining waveguides.

n_1 < n_2⟶ undefinedGoing from rare to dense medium — refraction always succeeds, no TIR possible.

What if…

What if another medium is placed near the TIR surface within

\sim \lambda

?

Frustrated TIR: the evanescent wave tunnels into the third medium and some light escapes — used in optical couplers and beam splitters.

What if you exceed

\theta _c

slightly?

Reflection becomes 100%, but the reflected wave acquires a phase shift (Fresnel formulas). Used in quarter-wave plates and Fresnel rhombs.

What if the medium changes (e.g., wet fiber)?

If outer index rises (water $n=1.33\,\mathrm{vs}$ cladding $n=1.46)$ , $\theta _c$ changes — bare fibers leak when wet, hence sealed jacketing.

1

Glass-to-air critical angle

Given ·

n 1:: 1.5
n 2:: 1

Find · theta_c

Steps

Apply $\theta _c = \arcsin (n_2 / n_1)$
Compute $n_2/n_1 = 1.0/1.5 = 0.6667$
$\theta _c = \arcsin (0.6667) \approx 41.8^{\circ} = 0.7297\,\mathrm{rad}$

Result ·

\theta _c = \arcsin (1.0 / 1.5) = \arcsin (0.6667) = 0.7297\,\mathrm{rad} \approx 41.8^{\circ}

2

Critical angle for diamond–air

Given ·

n 1:: 2.42
n 2:: 1

Find · theta_c

Steps

$n_2/n_1 = 1.0/2.42 = 0.4132$
$\theta _c = \arcsin (0.4132) \approx 24.4^{\circ} = 0.4259\,\mathrm{rad}$
Small $\theta _c$ traps most light inside — why diamonds sparkle so brilliantly

Result ·

\theta _c = \arcsin (1.0/2.42) = \arcsin (0.4132) = 0.4259\,\mathrm{rad} \approx 24.4^{\circ}

Snell's Law→refractive indexBrewster's Angle→