Total internal reflection — critical angle derivation: Question bank

Before we start, a one-line reminder of the notation used throughout so nothing is unearned:

$n_{1}$ = index of the medium the light starts in (incident side).
$n_{2}$ = index of the medium the light is trying to enter.
$θ_{1}$ = angle of incidence, $θ_{2}$ = angle of refraction, both measured from the normal (the line perpendicular to the surface).
$θ_{c}$ = critical angle, the special $θ_{1}$ at which $θ_{2} = 90°$ .

A one-picture reminder of why $sin θ_{c} = n_{2} / n_{1}$

Before you tackle the traps, look at the boundary itself — because almost every trap below is really a misreading of this one picture.

Figure — Total internal reflection — critical angle derivation

Read the figure left to right. In the leftmost panel the incidence angle $θ_{1}$ is small: the light crosses into the rarer medium and bends away from the normal, so $θ_{2} > θ_{1}$ . As you slide $θ_{1}$ up (middle panel), the refracted ray tips further and further toward the surface. In the right panel $θ_{1}$ has reached the exact value where the refracted ray lies flat along the boundary — that is $θ_{2} = 90°$ , and that incidence angle is $θ_{c}$ .

Where the formula comes from — a wavefront argument, not a memorised rule. Snell's law $n_{1} sin θ_{1} = n_{2} sin θ_{2}$ is just the statement that wavefronts stay continuous along the boundary (see Snell's Law). Plug in the grazing condition $θ_{2} = 90°$ , for which $sin 90° = 1$ : $n_{1} sin θ_{c} = n_{2} \cdot 1 ⟹ sin θ_{c} = \frac{n _{2}}{n _{1}}$ So the formula is nothing more than "the refracted ray has run out of room to bend." Since we need $n_{1} > n_{2}$ (dense→rare), the right side is always less than 1, which is exactly why a valid angle exists. Keep this panel in mind — the "flat refracted ray" picture is the anchor for every item below.

True or false — justify

TIR can occur when light travels from a rarer medium into a denser medium.

False — with

n_{1} < n_{2}

we get

sin θ_{2} = \frac{n _{1}}{n _{2}} sin θ_{1} < 1

for every

θ_{1}

, so a real refracted ray always exists and refraction never ceases.

At exactly

θ_{1} = θ_{c}

, all the light is already totally internally reflected.

False — at exactly

θ_{c}

the refracted ray still exists but grazes the surface at

θ_{2} = 90°

; TIR (100% reflection) begins only for

θ_{1}

strictly greater than

θ_{c}

A larger refractive index of the denser medium gives a larger critical angle.

False —

sin θ_{c} = n_{2} / n_{1}

, so raising

n_{1}

shrinks the fraction and hence

θ_{c}

; that is why diamond (

n = 2.42

θ_{c} \approx 24°

) traps light far more easily than glass.

The critical angle depends only on the denser medium's index.

False — it depends on the ratio

n_{2} / n_{1}

; the same glass has one

θ_{c}

against air and a completely different one against water.

Below the critical angle, no light reflects at the boundary at all.

False — some light always reflects (partial reflection happens at every boundary); below

θ_{c}

most energy is transmitted, and the reflected fraction merely grows as

θ_{1}

approaches

θ_{c}

If two media have equal refractive indices, there is no critical angle.

True — with

n_{1} = n_{2}

sin θ_{c} = 1

θ_{c} = 90°

; light never bends and can never be trapped, so TIR effectively cannot occur.

TIR only reflects part of the light; some always leaks through.

False — beyond

θ_{c}

100% of the energy is reflected (ideally); no ordinary refracted ray carries away time-averaged energy, which is exactly what "total" means.

Beyond the critical angle the electromagnetic field is exactly zero on the far side of the boundary.

False — a non-propagating evanescent field exists just past the surface, decaying exponentially over a fraction of a wavelength; it carries no net energy across, so reflection is still total (see the edge-case section).

Total internal reflection is the same phenomenon as the reflection off a metal mirror.

False — a metal mirror absorbs a few percent each bounce, while TIR is essentially lossless; this is why fibre optics and prisms use TIR rather than metallic coatings.

Spot the error

sin θ_{c} = n_{1} / n_{2}

, since the incident medium is written first."

The ratio is flipped. Setting

θ_{2} = 90°

n_{1} sin θ_{c} = n_{2} sin 90°

gives

sin θ_{c} = n_{2} / n_{1}

— rarer over denser, which is correctly less than 1.

"Light going from air into glass can undergo TIR if the angle is steep enough."

Backwards direction. Air→glass is rarer→denser, so no critical angle exists on that side; TIR needs the light to be inside the denser medium hitting the boundary.

"The critical angle is where the reflected ray is exactly

90°

from the incident ray."

That describes Brewster's condition — at

θ_{B}

the reflected and refracted rays are perpendicular (

θ_{B} + θ_{2} = 90°

), which polarizes the reflection. The critical angle is different: it is defined by the refracted ray grazing at

θ_{2} = 90°

, a statement about refraction ceasing, not about polarization (see Brewster's Angle).

"Since

sin θ_{c} = n_{2} / n_{1}

, a fibre core with

n = 1.0

against air would trap light perfectly."

With

n_{1} = 1.0 = n_{2}

the fraction is 1 and

θ_{c} = 90°

; nothing is trapped. A working fibre needs a core denser than its cladding (see Optical Fibres).

θ_{c} = sin^{- 1} (1.4/1.2)

for light going from

n_{1} = 1.2

n_{2} = 1.4

The argument

1.4/1.2 > 1

has no valid arcsine — the tell that this direction (rarer→denser) admits no critical angle at all, so the setup is invalid.

"Refraction still happens beyond

θ_{c}

, it just bends more than

90°

There is no such thing as bending "more than

90°

" for a transmitted ray — that would put it back on the incident side. Snell demands

sin θ_{2} > 1

, which is impossible, so refraction genuinely stops (see Refraction of Light).

Why questions

Why does the refracted ray reach

90°

before the incident ray does, as

θ_{1}

grows?

Because dense→rare gives

sin θ_{2} = \frac{n _{1}}{n _{2}} sin θ_{1}

with

n_{1} / n_{2} > 1

, so

θ_{2} > θ_{1}

always;

θ_{2}

races ahead and hits its ceiling of

90°

while

θ_{1}

is still smaller.

Why is

θ_{2} = 90°

chosen as the defining condition for

θ_{c}

Because

90°

is the largest possible refraction angle — the ray lying flat along the surface (rightmost panel of the figure). Any demand for a larger

θ_{2}

is geometrically impossible, so this is the exact tipping point where refraction can no longer exist.

Why do diamonds sparkle more than glass gems of the same cut?

Diamond's high index (

2.42

) gives a small

θ_{c} \approx 24°

, so most internal rays exceed it and bounce many times inside before escaping, concentrating and delaying the light before release.

Why must

sin θ_{c}

always come out less than 1 for a valid critical angle?

Because

sin θ_{c} = n_{2} / n_{1}

and TIR requires

n_{1} > n_{2}

(denser to rarer); a value

\geq 1

signals you either flipped the ratio or chose the wrong direction (see Refractive Index).

Why is TIR preferred over silvered mirrors inside binoculars and periscopes?

TIR reflects essentially 100% of the light with no metallic absorption, so images stay bright; a

45°

prism angle comfortably exceeds glass's

θ_{c} \approx 42°

(see Prisms and Total Internal Reflection).

Why does a road on a hot day look like a wet, reflective surface from far away?

The air near the hot road forms a continuous index gradient (rarest right at the surface); each thin layer refracts the sky-light a little more until a ray is bent back upward. It is not a single sharp interface, so strictly it is gradient refraction, not textbook TIR — though the trapping effect looks similar (see Mirage and Atmospheric Refraction).

Edge cases

What is

θ_{c}

when the two media are identical (

n_{1} = n_{2}

sin θ_{c} = 1

, so

θ_{c} = 90°

; light travels straight through with no bending and can never be trapped — TIR is impossible.

n_{2} \to n_{1}

from below, what happens to the critical angle?

n_{2} / n_{1} \to 1

, so

θ_{c} \to 90°

; the "trapping window"

(θ_{c}, 90°)

shrinks to nothing, meaning it becomes ever harder to achieve TIR as the indices approach each other.

n_{1}

becomes enormous compared to

n_{2}

, what does

θ_{c}

approach?

n_{2} / n_{1} \to 0

, so

θ_{c} \to 0°

; almost any angle of incidence traps the light, which is the ideal limit for lossless light-piping.

What happens exactly at

θ_{1} = θ_{c}

— is the boundary a mirror or a window?

Neither fully — the refracted ray exists but skims along the surface (

θ_{2} = 90°

), a knife-edge case; the true mirror behaviour switches on only just past this angle.

If light hits the boundary at

θ_{1} = 0°

(straight on) in a dense→rare setup, does TIR occur?

No — a normally-incident ray passes straight through with

θ_{2} = 0°

; since

0° < θ_{c}

, refraction happens freely and no trapping occurs.

Can TIR occur if the incident medium is a vacuum?

No — vacuum has the lowest possible index (

n = 1

), so there is no rarer medium to send light into; TIR requires starting in something denser than the destination.

Beyond

θ_{c}

, can any energy cross into the rarer medium?

Not through the flat boundary alone — the field there is evanescent, decaying exponentially with penetration depth

\sim λ / (2 π)

and carrying no net energy across. But if a second dense medium is brought within that decay distance, light can "tunnel" across the gap (frustrated TIR) — the wave equivalent of quantum tunnelling.

How deep does the evanescent field reach past the surface at

θ_{1} > θ_{c}

Only a fraction of a wavelength — its amplitude falls to

1/ e

within roughly

λ / (2 π)

of the boundary, which is why "total" reflection is genuinely total unless another medium is placed right up against the surface.

Connections

Snell's Law — the single law every trap above traces back to.
Refraction of Light — TIR is the failure limit of refraction.
Refractive Index — sets the size of $θ_{c}$ through the ratio.
Optical Fibres — the "identical index" edge case explains why cladding must be rarer.
Brewster's Angle — the orthogonal-rays condition cleared up in Spot-the-error.
Mirage and Atmospheric Refraction — the gradient-refraction edge case.
Prisms and Total Internal Reflection — the lossless-mirror "why" question.

A one-picture reminder of why sinθc​=n2​/n1​

True or false — justify

Spot the error

Why questions

Edge cases

Connections

A one-picture reminder of why $sin θ_{c} = n_{2} / n_{1}$