ARCHETYPE: TIR, Critical-Angle & Optical-Fibre Problems

A diamond blazes, an optical fibre carries a video call under the ocean, a mirage floats a lake of sky onto a hot road, and a periscope bends light around a corner with no silver at all. Every one of these is the same piece of physics — total internal reflection — and on the exam it arrives as a small, reliable family of numericals worth three to five marks. The examiner will hand you a refractive index and ask for a critical angle; hand you a core and a cladding and ask for the light-gathering cone of a fibre; or ask whether a prism will bounce light by TIR at all. Each is mechanical once you have the setup, and each rewards the student who signs the ratio the right way up. This is the problem gym for TIR, the critical angle and the optical fibre. The refraction spine earlier in this chapter proved Snell's law and defined the critical angle; here the setup is the whole point. Total internal reflection happens only when light tries to leave a denser medium for a rarer one at an angle beyond the critical angle $C$ , where $\sin C = n_2/n_1$ . Turn that one condition into a fibre and you get the two quantities every fibre question wants: the numerical aperture $NA = \sqrt{n_1^2 - n_2^2}$ and the acceptance angle whose sine is that $NA$ . We run the drill on the board staples — the critical angles of glass, water and diamond, and the numerical aperture of a real communications fibre — then climb to the two questions that separate top scorers: the minimum index a right-angle prism needs to turn light by TIR, and the apparent-depth cousin that examiners drill alongside. Threading it all is one discipline: decide which medium is denser, put the rarer index on top of the ratio, and never confuse the cone measured in air with the wall reflection inside the glass. By the end you should read a TIR problem and know its shape before you write a line.

Key takeaways

Total internal reflection needs light going from denser to rarer at more than the critical angle $C$ , where $\sin C = n_2/n_1$ (rarer over denser); into air this is $\sin C = 1/n$ .
Critical-angle anchors: glass $n = 1.50 \to C = 41.8°$ ; water $n = 1.33 \to C = 48.75°$ ; diamond $n = 2.42 \to C = 24.41°$ — the tiny diamond angle is why it sparkles.
An optical fibre guides light by repeated TIR at the core–cladding wall; the light-gathering cone has numerical aperture $NA = \sqrt{n_1^2 - n_2^2}$ and half-acceptance angle $\theta_a = \arcsin(NA)$ .
Fibre anchor: core $n_1 = 1.50$ , cladding $n_2 = 1.48 \to NA = 0.2441$ , $\theta_a = 14.1°$ , full cone $2\theta_a = 28.3°$ , fractional index difference $1.33\%$ .
A right-angle TIR prism turns light through $90°$ provided $\sin 45° \ge 1/n$ , i.e. $n \ge \sqrt2 = 1.414$ — so ordinary glass ( $n \approx 1.5$ ) works, which is why binoculars use prisms not mirrors.
The apparent-depth cousin: a coin under $2.0\ \mathrm{m}$ of water ( $n = 1.33$ ) looks $2.0/1.33 = 1.50\ \mathrm{m}$ deep — real depth divided by the index.
Key idea: solve every TIR numerical by finding the denser medium, using $\sin C = n_{\text{rarer}}/n_{\text{denser}}$ , and for a fibre computing $NA = \sqrt{n_1^2 - n_2^2}$ with its acceptance angle in air.

The core insight

There is one total-internal-reflection numerical, solved by one procedure: identify the denser medium (TIR lives only there), compute the critical angle from $\sin C = n_2/n_1$ , and for a fibre find the numerical aperture $NA = \sqrt{n_1^2 - n_2^2}$ and the half-acceptance angle $\theta_a = \arcsin(NA)$ measured in air, checking that any guided ray strikes the wall at more than the critical angle.

Where this lives today

Total internal reflection is one of the quiet workhorses of modern technology. The internet runs on it: hair-thin glass fibres, their cores and claddings chosen for exactly the numerical aperture of Example 2, carry pulses of laser light across oceans by millions of TIR bounces, with losses so low a signal survives a hundred kilometres between amplifiers. A medical endoscope threads a bundle of such fibres into the body to light and image an organ from inside, and a dentist's cure light pipes intense blue light down a fibre guide. Binoculars, periscopes and SLR cameras fold their optics with the TIR prisms of Example 3 — no silvering, no loss. A diamond is Example 1 made jewellery: its tiny $24°$ critical angle traps light through many internal bounces so the cut stone flashes. Even the retro-reflectors on road signs and bicycle pedals, and the edge-lit glow of a phone's display and a swimming pool's underwater lights, are TIR at work. Every one is the same condition — denser to rarer, beyond the critical angle — scaled from a gemstone to a transoceanic cable.

The setup: one condition and its fibre

In an archetype the bookkeeping is the physics, so learn it cold. Total internal reflection is not a new law — it is Snell's law pushed to its limit. When light travels from a denser medium (index $n_1$ ) toward a rarer one (index $n_2 < n_1$ ), the refracted ray bends away from the normal, and at one special angle of incidence — the critical angle $C$ — it grazes along the boundary at $90°$ . Beyond $C$ there is no refracted ray at all; the light is thrown entirely back into the denser medium:

\sin C = \frac{n_2}{n_1} \qquad (1)

where: $n_1$ is the index of the denser medium the light is in, and $n_2$ the index of the rarer medium it fails to enter. The ratio is rarer over denser, so it is always less than one and $C$ is a real angle. When the rarer medium is air, $n_2 = 1$ and this becomes the familiar $\sin C = 1/n$ . Read equation (1) as a threshold: a bigger index $n_1$ (or a smaller $n_2$ ) gives a smaller critical angle, so a wider cone of rays is trapped — which is precisely why a high-index diamond, with $C$ only about $24°$ , holds and re-reflects light so brilliantly. The master recipe for any TIR problem is four moves that never change:

Find the denser medium. TIR can only happen for light already inside the denser medium trying to escape into a rarer one. Name $n_1$ (denser) and $n_2$ (rarer) before anything else.

Compute the critical angle. $\sin C = n_2/n_1$ , so $C = \arcsin(n_2/n_1)$ . Into air, $C = \arcsin(1/n)$ .

For a fibre, find the cone. The numerical aperture is $NA = \sqrt{n_1^2 - n_2^2}$ (core and cladding), and the half-acceptance angle in air is $\theta_a = \arcsin(NA)$ .

Test the angle. For a "will it TIR?" question, demand that the ray meets the surface at more than $C$ : TIR occurs if and only if $\theta \ge C$ .

Figure

An optical fibre in cross-section: rays entering the flat end within the acceptance cone strike the core-cladding wall beyond the critical angle and zig-zag down the fibre by repeated total internal reflection, while a ray entering too steeply refracts into the cladding and is lost

Reading the critical-angle fan

To see where the critical angle comes from, watch a single ray inside the glass as you tilt it. At small angles of incidence the ray splits: most of it refracts out into the air, bent away from the normal, and a little reflects back. As you tilt further, the refracted ray bends closer and closer to the surface and grows brighter while the transmitted beam dims. At exactly the critical angle the refracted ray lies flat along the boundary — it grazes at $90°$ — and beyond that angle it vanishes entirely: all the light reflects back into the glass, obeying the ordinary law of reflection with no loss. That abrupt, loss-free switch is what makes TIR so useful: unlike a metal mirror, which absorbs a few percent at every bounce, a clean glass surface beyond the critical angle reflects essentially all of the light, which is exactly what a kilometres-long fibre needs.

Figure

The critical-angle fan for light inside glass meeting air: below the critical angle the ray refracts out and partly reflects, at the critical angle the refracted ray grazes along the surface at ninety degrees, and above the critical angle all the light is totally internally reflected

The fibre: numerical aperture and acceptance angle

An optical fibre is a thin thread of high-index glass (the core, index $n_1$ ) wrapped in a lower-index sheath (the cladding, index $n_2$ ). Light launched into the flat end and striking the core–cladding wall beyond the critical angle is trapped and bounces its way along, however the fibre curves. But not every ray you shine at the end is captured: a ray entering too steeply refracts into the wall at less than the critical angle and leaks away. The cone of directions that is captured is measured by the numerical aperture,

NA = \sqrt{n_1^2 - n_2^2} \qquad (2)

where: $n_1$ is the core index and $n_2$ the cladding index. The larger the index gap, the wider the cone. The half-angle of that cone, measured in the outside air, is the acceptance angle $\theta_a$ , and it is fixed by the numerical aperture through

\sin\theta_a = \frac{NA}{n_0}, \qquad n_0 = 1 \ (\text{air}) \qquad (3)

where: $n_0$ is the index of the medium outside the fibre end, taken as air, so $\sin\theta_a = NA$ directly. Any ray arriving within $\theta_a$ of the axis is guided; anything steeper is lost. Note the two indices that matter are the core and cladding, not air — a common slip is to compute $NA$ from the core-to-air step, which is the wrong pair.

Derivation AThe critical angle from Snell's law

The goal is to derive the critical-angle relation $\sin C = n_2/n_1$ from Snell's law and use it to land the locked board anchors — glass $C = 41.8°$ , water $C = 48.75°$ and diamond $C = 24.41°$ — showing why total internal reflection is simply Snell's law at the angle where the refracted ray runs out of room. Snell's law is quoted from earlier in this chapter; everything else is one substitution.

Step 1. Write Snell's law for light passing from the denser medium ( $n_1$ ) into the rarer one ( $n_2$ ): $n_1 \sin\theta_1 = n_2 \sin\theta_2$ , where $\theta_1$ is the angle of incidence inside the glass and $\theta_2$ the angle of refraction in the rarer medium.

Step 2. Increase $\theta_1$ . Because $n_1 > n_2$ , the refracted angle $\theta_2$ is always larger than $\theta_1$ , so as $\theta_1$ grows, $\theta_2$ reaches $90°$ first — the refracted ray lies flat along the surface. The angle of incidence at which this happens is, by definition, the critical angle $C$ .

Step 3. Set $\theta_1 = C$ and $\theta_2 = 90°$ , so $\sin\theta_2 = 1$ : Snell's law becomes $n_1 \sin C = n_2 \times 1$ , which rearranges to $\sin C = n_2/n_1$ . For the rarer medium being air ( $n_2 = 1$ ), $\sin C = 1/n_1$ .

Step 4. Substitute the anchors. Glass $n = 1.50$ : $C = \arcsin(1/1.50) = \arcsin(0.6667) = 41.8°$ . Water $n = 1.33$ : $C = \arcsin(1/1.33) = \arcsin(0.7519) = 48.75°$ . Diamond $n = 2.42$ : $C = \arcsin(1/2.42) = \arcsin(0.4132) = 24.41°$ . The trend is the physics: the higher the index, the smaller the critical angle, so diamond traps light in a cone twice as wide as glass — the geometric root of its fire. Verified: $C = 41.8°$ , $48.75°$ and $24.41°$ from $\sin C = 1/n$ are recomputed in verify_hs2-tir-fibre-problems.py.

Go deeper · Derivation B · 1 minNumerical aperture and the acceptance angle

Derivation BNumerical aperture and the acceptance angle

The goal is to derive the fibre's numerical aperture $NA = \sqrt{n_1^2 - n_2^2}$ and its acceptance angle, and to land the locked anchor for a real communications fibre — core $n_1 = 1.50$ , cladding $n_2 = 1.48$ , giving $NA = 0.2441$ and $\theta_a = 14.1°$ — from just two applications of Snell's law: one at the flat end face, one at the core–cladding wall. No calculus, only trigonometry and the critical-angle condition.

Step 1. A ray enters the end face from air at angle $\theta_a$ to the axis and refracts into the core at angle $r$ : Snell at the face gives $\sin\theta_a = n_1 \sin r$ (with $n_0 = 1$ ). This $r$ is the angle the ray makes with the fibre axis inside the core.

Step 2. That same ray strikes the core–cladding wall at angle $(90° - r)$ to the wall's normal. For the ray to be guided, this must be at least the critical angle for the core–cladding boundary: $90° - r \ge C$ , where $\sin C = n_2/n_1$ . The widest accepted ray sits exactly at $90° - r = C$ .

Step 3. At that extreme, $r = 90° - C$ , so $\sin r = \cos C = \sqrt{1 - \sin^2 C} = \sqrt{1 - (n_2/n_1)^2}$ . Feed this into the face equation from Step 1: $\sin\theta_a = n_1 \sin r = n_1\sqrt{1 - (n_2/n_1)^2} = \sqrt{n_1^2 - n_2^2}$ . The right-hand side is, by definition, the numerical aperture, so $NA = \sqrt{n_1^2 - n_2^2}$ and $\sin\theta_a = NA$ .

Step 4. Substitute the anchor: $NA = \sqrt{1.50^2 - 1.48^2} = \sqrt{2.25 - 2.1904} = \sqrt{0.0596} = 0.2441$ . Then $\theta_a = \arcsin(0.2441) = 14.1°$ , and the full acceptance cone is $2\theta_a = 28.3°$ . The fractional index difference $(n_1 - n_2)/n_1 = 0.02/1.50 = 1.33\%$ is deliberately tiny in real fibres, which keeps the cone narrow and the light travelling almost straight — the secret to sending a signal for kilometres. Verified: $NA = 0.2441$ , $\theta_a = 14.1°$ , $2\theta_a = 28.3°$ and the $1.33\%$ index difference are recomputed in verify_hs2-tir-fibre-problems.py.

Variants and where they trap you

variant	how to handle it	the trap
critical angle into air	$\sin C = 1/n$ ; higher $n$ , smaller $C$	inverting the ratio, $\sin C = n$
critical angle between two media	$\sin C = n_{\text{rarer}}/n_{\text{denser}}$	using the wrong index on top
fibre numerical aperture	$NA = \sqrt{n_1^2 - n_2^2}$ (core, cladding)	using core-and-air instead of core-and-cladding
acceptance angle	$\theta_a = \arcsin(NA)$ , measured in air	forgetting it is measured in air, not the core
right-angle TIR prism	demand $\sin 45° \ge 1/n$ , so $n \ge \sqrt2$	thinking it reflects off silver, not by TIR
apparent depth	apparent $=$ real depth $/\,n$	multiplying by $n$ instead of dividing

Common trapTIR only happens denser to rarer, and the ratio is rarer over denser

The single most expensive mistake in this whole topic is getting the critical-angle ratio upside down or applying TIR in the wrong direction. Total internal reflection can only occur for light already inside the denser medium trying to enter a rarer one — never the other way round. Light going from air into glass simply refracts; there is no critical angle for that direction. And the formula is $\sin C = n_{\text{rarer}}/n_{\text{denser}}$ , a ratio that must come out less than one: for glass to air it is $\sin C = 1/1.5 = 0.667$ , giving $C = 41.8°$ , not $\sin C = 1.5$ (which is impossible and a dead giveaway of a flipped ratio). Two more slips ride alongside. First, the numerical aperture uses the core and cladding indices, $NA = \sqrt{n_1^2 - n_2^2}$ — not the core-to-air step. Second, the acceptance angle is measured in the outside air, so $\sin\theta_a = NA$ ; the angle inside the core is different. Name the denser medium first, put the rarer index on top, keep the fibre's two glasses straight, and the family behaves.

Worked examples

1 · Board level — three critical angles (CBSE and ICSE pattern)

Find the critical angle for the glass–air, water–air and diamond–air boundaries, given the refractive indices $n_{\text{glass}} = 1.50$ , $n_{\text{water}} = 1.33$ and $n_{\text{diamond}} = 2.42$ . Comment on why diamond sparkles.

Step 1. Each is a denser medium meeting air, so $\sin C = 1/n$ . Sign check: the ratio is always less than one, so each $C$ is a genuine angle.

Step 2. Glass: $C = \arcsin(1/1.50) = \arcsin(0.6667) = \boxed{41.8°}$ .

Step 3. Water: $C = \arcsin(1/1.33) = \arcsin(0.7519) = \boxed{48.75°}$ . Diamond: $C = \arcsin(1/2.42) = \arcsin(0.4132) = \boxed{24.41°}$ .

Step 4. Comment: diamond's tiny critical angle of about $24°$ means light entering it is trapped whenever it hits a facet at more than $24°$ — a very wide range — so it bounces many times inside before escaping, and the cut concentrates that escape into brilliant flashes. Glass, with $C = 41.8°$ , traps a much narrower cone, so it never sparkles the same way.

2 · Board level — the numerical aperture of a fibre (CBSE and NEET pattern)

An optical fibre has a core of refractive index $1.50$ and a cladding of refractive index $1.48$ . Find its numerical aperture, its half-acceptance angle in air, the full acceptance cone, and the fractional index difference.

Step 1. Identify the pair: core $n_1 = 1.50$ , cladding $n_2 = 1.48$ . Use $NA = \sqrt{n_1^2 - n_2^2}$ .

Step 2. $NA = \sqrt{1.50^2 - 1.48^2} = \sqrt{2.25 - 2.1904} = \sqrt{0.0596} = \boxed{0.2441}$ .

Step 3. Acceptance angle in air: $\theta_a = \arcsin(NA) = \arcsin(0.2441) = \boxed{14.1°}$ , and the full cone is $2\theta_a = \boxed{28.3°}$ .

Step 4. Fractional index difference: $\dfrac{n_1 - n_2}{n_1} = \dfrac{1.50 - 1.48}{1.50} = \dfrac{0.02}{1.50} = \boxed{1.33\%}$ . This small difference is deliberate: it keeps the acceptance cone narrow so rays travel nearly straight and the signal survives kilometres of fibre.

3 · Competitive level — the minimum index for a TIR prism (JEE pattern)

A right-angle glass prism is used to turn a beam through $90°$ by total internal reflection at its hypotenuse, where the light strikes at $45°$ . What is the minimum refractive index the glass must have for this to work, and does ordinary glass ( $n \approx 1.5$ ) qualify?

Step 1. For TIR at the hypotenuse, the $45°$ angle of incidence must be at least the critical angle: $45° \ge C$ . Since a larger $C$ means a smaller index, the borderline case is $C = 45°$ .

Step 2. At the borderline, $\sin 45° = 1/n_{\min}$ , so $n_{\min} = \dfrac{1}{\sin 45°} = \dfrac{1}{1/\sqrt2} = \sqrt2 = \boxed{1.414}$ .

Step 3. Compare with ordinary glass, $n \approx 1.50 > 1.414$ : its critical angle is $C = \arcsin(1/1.5) = 41.8° < 45°$ , so light striking the hypotenuse at $45°$ is safely beyond the critical angle and totally reflects. Ordinary glass qualifies with margin to spare.

Step 4. This is why binoculars, periscopes and single-lens-reflex cameras use glass prisms rather than silvered mirrors to fold their light paths: a TIR prism reflects essentially $100\%$ of the light with no metal coating to tarnish or absorb, whereas even a good mirror loses a few percent at every bounce. Timing note: under a minute. The whole question is the one line $n_{\min} = 1/\sin 45° = \sqrt2$ ; the trap is thinking the prism needs silvering — it does not, the $45°$ incidence already exceeds the critical angle of any common glass.

4 · Competitive level — the apparent-depth cousin (JEE and NEET pattern)

A coin lies at the bottom of a tank under $2.0\ \mathrm{m}$ of water of refractive index $1.33$ . How deep does the coin appear to an observer looking straight down, and how is this related to total internal reflection?

Step 1. For near-normal viewing, refraction at the water surface makes an object look shallower by a factor of the refractive index: apparent depth $=$ real depth $/\,n$ . This is Snell's law applied to the small angles of a vertical line of sight, quoted from earlier in this chapter.

Step 2. Apparent depth $= \dfrac{2.0}{1.33} = \boxed{1.50\ \mathrm{m}}$ . The coin appears about half a metre closer to the surface than it truly is.

Step 3. Connect to TIR: apparent depth and the critical angle are two faces of the same refraction. The very bending that lifts the coin toward you is what, at a steep enough angle, refuses to let light out at all — the critical angle. A fish looking up sees the entire sky squeezed into a bright cone of half-angle $C = 48.75°$ (Snell's window); outside that cone the underside of the surface acts as a mirror by TIR.

Step 4. Sanity check the direction: dividing by $n > 1$ makes the apparent depth smaller than the real depth, which matches everyday experience — a pool always looks shallower than it is. Multiplying by $n$ would deepen it, the wrong way, and is the classic error.

Exam traps

Inverting the critical-angle ratio. It is $\sin C = n_{\text{rarer}}/n_{\text{denser}}$ , always less than one. Writing $\sin C = n$ (bigger than one) is impossible and the surest sign of a flip — into air it is $\sin C = 1/n$ .

Applying TIR the wrong way. TIR needs light going from denser to rarer. Light from air into glass just refracts; there is no critical angle for that direction. Name the denser medium first.

Wrong indices in the numerical aperture. $NA = \sqrt{n_1^2 - n_2^2}$ uses the core and cladding, not the core and air. Using the wrong pair inflates the aperture and the acceptance angle.

Confusing the cone with the wall. The acceptance angle is measured in the outside air; the angle a guided ray makes inside the core is different. Keep the two separate.

Thinking a prism needs silver. A right-angle prism turns light by TIR at $45°$ , needing only $n \ge \sqrt2 = 1.414$ — no coating. That is the whole reason it beats a mirror.

Apparent depth upside down. Apparent depth $=$ real depth $/\,n$ : divide by the index to get a shallower image. Multiplying deepens it, the wrong way.

Board scope. This numerical archetype is set by CBSE, ICSE, JEE-Main, JEE-Advanced, NEET and AP Physics 2, all of which examine the critical angle and TIR-prism problems. IGCSE 0625 owns the critical angle $\sin C = 1/n$ and the uses of optical fibres (endoscopes, communications) but not the numerical-aperture formula — so IGCSE candidates should learn the critical-angle and prism parts here and treat the $NA$ derivation as enrichment. IB SL and HL carry the critical angle only as a wave phenomenon, not this fibre-aperture gym; do the ideas, skip the drill. Wherever a topic splits by board, that split is noted so you never revise something off your syllabus.

Origins · 1 minCharles Kao and how this was discovered

Charles Kao

1933–2018 · Shanghai · Hong Kong · Harlow

Charles Kao is the father of fibre-optic communications. Working at Standard Telecommunication Laboratories in Harlow, England, he and George Hockham published in 1966 the paper that changed how the world moves information: they showed that the crippling losses in glass fibres of the day were caused not by any fundamental limit but by impurities, and that glass pure enough would let light carry signals over kilometres by total internal reflection. That single insight — that the fibre's problem was solvable chemistry, not physics — launched the entire optical-communications industry, and the transoceanic cables now carrying essentially all the internet's traffic are its direct descendants. Kao received the 2009 Nobel Prize in Physics for the achievement. Honest scope: the critical-angle law that traps light in a fibre is Snell's, centuries older; Kao's contribution was showing that ultra-pure glass could make that trapping practical at continental scale, which is why the fibre carries the world's data.

History

1621

— Willebrord Snellius states the law of refraction from which the critical angle follows.

1842

— Daniel Colladon demonstrates light guided along a curving jet of water, the ancestor of the optical fibre.

1854

— John Tyndall popularises the light-guiding water-jet experiment in his lectures at the Royal Institution.

1954

— Abraham van Heel and Harold Hopkins independently develop clad fibre bundles for image transmission.

1966

— Charles Kao and George Hockham show that pure enough glass fibre can carry light over kilometres, founding fibre-optic communications.

Going further

Texts. NCERT Physics Class 12, Ch. 9 ("Ray Optics and Optical Instruments"), total internal reflection, the critical angle and optical fibres · H. C. Verma, Concepts of Physics Vol. 1, "Geometrical Optics" — TIR, prisms and apparent depth · D. C. Pandey / Cengage optics problem sets for the JEE and NEET fibre and prism variants · AP Physics 2 (College Board CED, Unit 13) TIR and critical-angle problems.

Provenance. Standard geometric optics. Recompute: critical angles $\arcsin(1/1.50) = 41.8°$ , $\arcsin(1/1.33) = 48.75°$ , $\arcsin(1/2.42) = 24.41°$ ; fibre $NA = \sqrt{1.50^2 - 1.48^2} = 0.2441$ , $\theta_a = \arcsin(0.2441) = 14.1°$ , $2\theta_a = 28.3°$ , index difference $0.02/1.50 = 1.33\%$ ; TIR prism $n_{\min} = 1/\sin 45° = \sqrt2 = 1.414$ ; apparent depth $2.0/1.33 = 1.50\ \mathrm{m}$ ; Snell's-window half-angle $= 48.75°$ .

Prerequisites. The Mirror Formula and Sign Convention — the ray-optics reasoning and sign discipline this page shares. Leads to. Lens–Mirror Systems and the Silvered Lens, where refraction and reflection act together; and the university seam below.

Go deeper (UG). The full electromagnetic account of total internal reflection — the evanescent wave that leaks a fraction of a wavelength into the rarer medium, frustrated TIR, and the phase shift on reflection — is developed in Total Internal Reflection in the university optics track. It is enrichment only; every result here follows from Snell's law and trigonometry, and no exam on this page asks for the wave treatment.

Why is the critical angle formula sin C = n2/n1 and not the other way up?

Because it comes straight from Snell's law at the moment the refracted ray reaches ninety degrees. Snell's law says $n_1 \sin\theta_1 = n_2 \sin\theta_2$ ; at the critical angle the refracted ray grazes the surface, so $\theta_2 = 90°$ and $\sin\theta_2 = 1$ , leaving $n_1 \sin C = n_2$ , i.e. $\sin C = n_2/n_1$ — the rarer index ( $n_2$ ) over the denser one ( $n_1$ ). Because $n_2 < n_1$ , this ratio is always less than one, so $C$ is a real angle. If you ever write $\sin C = n_1/n_2$ (bigger than one) you have flipped it, and the impossible "sine greater than one" is your warning. Into air, $n_2 = 1$ , so it simplifies to the familiar $\sin C = 1/n$ .

What exactly is the numerical aperture, and why does it use the core and cladding?

The numerical aperture measures how wide a cone of light a fibre can capture and guide. A ray is only guided if it hits the core–cladding wall beyond the critical angle for that boundary, and that critical angle depends on the core index $n_1$ and the cladding index $n_2$ — the two glasses the wall separates. Tracing the widest such ray back out through the end face gives $NA = \sqrt{n_1^2 - n_2^2}$ , built entirely from those two indices. Using the core-and-air step instead is a common error: air is not what the guided ray reflects off inside the fibre. The bigger the gap between core and cladding, the wider the cone the fibre accepts — but real communications fibres keep the gap tiny (a fractional difference near $1\%$ ) so the cone is narrow and the light stays nearly straight over long distances.

Is the acceptance angle measured inside the fibre or outside in the air?

Outside, in the air. The acceptance angle $\theta_a$ is the half-angle of the cone of incoming rays, measured from the fibre axis in the outside medium (air, $n_0 = 1$ ), and $\sin\theta_a = NA$ . A ray arriving within that cone refracts into the core and then strikes the wall beyond the critical angle, so it is guided; a ray arriving steeper than $\theta_a$ leaks into the cladding and is lost. The angle the ray makes inside the core is smaller, because refraction at the end face bends it toward the axis — so do not confuse the internal ray angle with the external acceptance angle. When a problem says "the greatest angle at which light can enter and still be guided", it means $\theta_a$ , measured in air.

Why does a diamond sparkle so much more than glass?

Because its refractive index is so high that its critical angle is tiny. Diamond's index of $2.42$ gives a critical angle of only about $24°$ , against glass's $42°$ . That means light entering a diamond is totally internally reflected whenever it meets a facet at more than $24°$ — an enormous range of angles — so a ray bounces around many times inside the stone before it finally escapes. A skilled cut arranges the facets so that this trapped light exits through the top in concentrated flashes, and because the index also varies with colour (dispersion), the flashes break into spectral fire. Glass, with its wider $42°$ critical angle, lets light escape after far fewer bounces, so a glass "gem" looks dull by comparison even when cut identically.

Does an optical fibre reflect light off a metal coating like a mirror?

No — and that is the whole point. A fibre guides light by total internal reflection at the boundary between its glass core and its glass cladding, with no metal involved. Beyond the critical angle, a clean glass–glass boundary reflects essentially one hundred percent of the light, losing nothing to absorption; a metal mirror, however good, absorbs a few percent at every bounce. Over the millions of reflections in a long fibre, those tiny metal losses would extinguish the signal completely, whereas TIR loses almost nothing per bounce. That loss-free reflection, combined with the ultra-pure glass Charles Kao showed was possible, is what lets a fibre carry a signal for tens of kilometres between amplifiers.

What happens to light that hits the surface at exactly the critical angle?

At exactly the critical angle the refracted ray runs along the boundary itself, grazing at ninety degrees to the normal, and in the ideal limit it carries vanishingly little energy — essentially all the light is reflected back. In practice the critical angle is the sharp threshold between two regimes: at even a fraction of a degree less, a refracted ray escapes and the reflection is only partial; at even a fraction more, no light escapes and the reflection is total. Because the exactly-critical case is a knife-edge of zero width, problems treat it as the boundary condition — "TIR occurs if and only if the angle is at least $C$ " — rather than a physical situation to compute. It is the dividing line, not a working point.

How is apparent depth related to total internal reflection?

They are two consequences of the same refraction at a water surface. Apparent depth comes from light bending away from the normal as it leaves the water, which makes objects look shallower by the factor $1/n$ — a coin under $2\ \mathrm{m}$ of water ( $n = 1.33$ ) appears at $1.5\ \mathrm{m}$ . Push the viewing angle steeper and steeper and that same outward bending eventually reaches the critical angle, beyond which no light escapes at all: an underwater observer then sees the whole sky compressed into a bright circle (Snell's window) of half-angle $C = 48.75°$ , ringed by the totally reflected, mirror-like underside of the surface. So the gentle lifting of apparent depth and the abrupt cut-off of total internal reflection are the same law of refraction read at small and at large angles.