Reflection vs Total Internal Reflection Explained

Ordinary reflection happens at every boundary, at every angle, and sends back only part of the light while the rest goes through. Total internal reflection sends back all of it, and only when the light heads into a lower index medium at an angle beyond the critical angle.

AP Physics: Unit 13 (topics 13.1 Reflection, 13.3 Refraction). Ordinary reflection is AP Physics 2 Topic 13.1: essential knowledge 13.1.B.1 says light that is incident on a surface can be reflected, and 13.1.B.2 gives the law of reflection, that the angle between the incident ray and the normal, the line perpendicular to the surface, equals the angle between the reflected ray and the normal, with theta_i = theta_r as the relevant equation. Its two named kinds are diffuse reflection from a rough surface (13.1.B.3) and specular reflection from a smooth surface (13.1.B.4), each explained by whether the surface normal varies over the illuminated area. Total internal reflection is the last part of Topic 13.3, Refraction: 13.3.A.5 says it may occur when light passes from one medium into another medium with a lower index of refraction; 13.3.A.5.i says it occurs beyond a critical angle of incidence and gives theta_critical = arcsin(n2/n1) as a DERIVED equation, not a relevant one; 13.3.A.5.ii says that for incident rays at the critical angle the ray refracts at 90 degrees and travels along the surface of the material; and 13.3.A.5.iii says that for incident rays beyond the critical angle all light is reflected and no light is transmitted into the other medium. Supporting statements: 13.3.A.4.i and 13.3.A.4.ii fix the direction of the bend, 13.3.A.4.iii says a ray incident along the normal is not refracted, 13.3.A.3 gives n = c/v, and 13.3.A.4 gives Snell's law. Essential knowledge 14.3.A.1 and 14.3.A.1.i establish that an ordinary boundary produces both reflected and transmitted waves. Neither the law of reflection nor the critical angle relation is printed on the AP Physics 2 equation sheet: the waves, sound, and optics block prints fifteen lines and neither appears among them, checked line by line against the rendered appendix page. Neither Topic 13.1 nor Topic 13.3 carries a boundary statement; in Unit 13 only Topic 13.2 has one, limiting the study of mirrors to plane, convex spherical and concave spherical. Unit 13, Geometric Optics, is weighted at 12 to 15 percent of the multiple-choice section over a suggested 8 to 12 class periods. Topic 13.1 lists suggested skills 1.A, 2.B, 2.C and 3.B; Topic 13.3 lists 1.B, 2.B, 2.D, 3.A and 3.B.

The distinction, stated once

Both are the same reflection. Only the share of the light differs.

AP Physics 2 essential knowledge 13.1.B.1 says light that is incident on a surface can be reflected, and 13.1.B.2 gives the law of reflection: the angle between the incident ray and the normal, the line perpendicular to the surface, is equal to the angle between the reflected ray and the normal, so $\theta_i = \theta_r$ . That law has no conditions attached. No index appears in it, no direction of travel, no threshold angle.

Essential knowledge 13.3.A.5 then says total internal reflection may occur when light passes from one medium into another medium with a lower index of refraction, and 13.3.A.5.i says it occurs beyond a critical angle of incidence. Essential knowledge 13.3.A.5.iii states the outcome: for incident rays beyond the critical angle, all light is reflected, and no light is transmitted into the other medium.

Read those side by side and the difference is a single word. Total.

At an ordinary boundary between transparent media, both a reflected ray and a transmitted ray leave the surface. Essential knowledge 14.3.A.1 says a wave that travels from one medium to another can be transmitted or reflected depending on the properties of the boundary, and 14.3.A.1.i says such a wave will result in reflected and transmitted waves. Plural, and at the same time. A window shows you the street by transmission and your own face by reflection at the same instant.

Total internal reflection is what is left when the transmitted ray disappears. The reflected ray was always there, obeying $\theta_i = \theta_r$ at every angle from zero upward. Past the critical angle it stops sharing.

So the right mental model is not that reflection switches on. It is that refraction switches off, and reflection inherits all the light because there is nowhere else for it to go.

Side by side

	Ordinary reflection	Total internal reflection
CED statement	13.1.B.1 and 13.1.B.2	13.3.A.5 with 13.3.A.5.i to 13.3.A.5.iii
Which topic	13.1 Reflection	13.3 Refraction, not the reflection topic
Angle rule	$\theta_i = \theta_r$	$\theta_i = \theta_r$ , the same rule
Fraction of light returned	Part of it	All of it, 13.3.A.5.iii
Is there a transmitted ray	Yes, 14.3.A.1.i	No, none at all
Conditions on the indices	None	$n_2 < n_1$ , from the higher index into the lower, 13.3.A.5
Conditions on the angle	None. Works at every angle	Must be beyond $\theta_{\text{critical}}$
Works both ways across a boundary	Yes	No. One direction only
Happens at a metal mirror	Yes	Not the mechanism. A mirror is a reflecting surface, not a lower index medium
Relation you would quote	$\theta_i = \theta_r$ , the relevant equation for 13.1.B.2	$\theta_{\text{critical}} = \sin^{-1}(n_2/n_1)$ , labelled a derived equation at 13.3.A.5.i
Printed on the AP Physics 2 sheet	No	No
Kinds the CED names	Diffuse, 13.1.B.3, and specular, 13.1.B.4	None. There is one case

The row that surprises people is which topic. Total internal reflection is taught inside the refraction topic, not the reflection topic, and that placement is a hint about what is actually going on. The examinable question is never "is the light reflected", because it always is. The question is whether anything got through, and that is a refraction question answered with Snell's law.

The row about metal mirrors closes a common wrong association. A bathroom mirror reflects almost all the light that reaches it, at every angle, from either side. That is not total internal reflection, because there is no second transparent medium with a lower index and no critical angle involved. Returning nearly all the light and returning all of it for the reason 13.3.A.5 gives are different facts.

The three conditions, and the one people drop

Ordinary reflection has no conditions. Total internal reflection has conditions, and they are the content of the topic.

Condition one: the light must be going from higher index to lower. Essential knowledge 13.3.A.5 says total internal reflection may occur when light passes from one medium into another medium with a lower index of refraction. Glass to air, yes. Air to glass, never. The relation contains its own test: an inverse sine needs an argument of at most one, so $\sin^{-1}(n_2/n_1)$ exists only when $n_2 < n_1$ . Try it the other way and the calculator refuses before the physics does.

There is a physical reason as well as an algebraic one. Essential knowledge 13.3.A.4.ii says a ray entering a higher index medium refracts toward the normal, so the refracted angle is always smaller than the incident angle and can never reach $90^\circ$ . There is no angle at which the transmitted ray could be squeezed out of existence.

Condition two: the angle of incidence must be beyond the critical angle. Essential knowledge 13.3.A.5.i. Below it, the boundary behaves entirely normally: a refracted ray leaves at the Snell's law angle and a reflected ray leaves at $\theta_i$ .

Condition three, the one that gets dropped: at exactly the critical angle it has not happened yet. Essential knowledge 13.3.A.5.ii says that for incident rays at the critical angle, the ray refracts at 90 degrees and travels along the surface of the material. So a refracted ray still exists there, grazing along the boundary. Essential knowledge 13.3.A.5.iii puts the total case strictly beyond the critical angle. An answer that says total internal reflection begins at the critical angle contradicts a statement the CED wrote separately for that exact case.

One more piece of the CED's own wording is worth noticing. Essential knowledge 13.3.A.5 says total internal reflection may occur, not will occur. The two conditions above are necessary; the framework is careful not to promise more than that.

The case that separates them: one glass block, three angles

Take a block of glass with index $n_1 = 1.52$ sitting in air, $n_2 = 1.00$ , and send light out through the flat top surface from inside. Only the angle changes.

The critical angle first, from the derived equation at 13.3.A.5.i:

\theta_{\text{critical}} = \sin^{-1}\!\left(\frac{n_2}{n_1}\right) = \sin^{-1}\!\left(\frac{1.00}{1.52}\right) = \sin^{-1}(0.658) = 41.1^\circ

Angle of incidence inside the glass	Reflected ray	Transmitted ray	Which process
$30.0^\circ$	Yes, at $30.0^\circ$	Yes, at $49.5^\circ$ , bent away from the normal	Ordinary reflection and refraction together
$41.1^\circ$ , the critical angle	Yes, at $41.1^\circ$	Yes, grazing at $90^\circ$ along the surface	Still not total, 13.3.A.5.ii
$50.0^\circ$	Yes, at $50.0^\circ$ , carrying all the light	None	Total internal reflection, 13.3.A.5.iii

Read down the reflected column and notice that it never changes character. At every angle the reflected ray leaves at $\theta_i$ , exactly as 13.1.B.2 requires. Nothing about the reflection is different at $50^\circ$ from what it was at $30^\circ$ except how much light is in it.

Read down the transmitted column and the whole story is there: a normal refracted ray, then a grazing one, then nothing.

Now send the same light the other way, from the air into the glass at $50.0^\circ$ . Snell's law gives $\sin\theta_2 = \sin 50.0^\circ / 1.52 = 0.504$ , so $\theta_2 = 30.3^\circ$ , a perfectly ordinary refraction toward the normal. The same angle, the same pair of materials, and no total internal reflection, because the direction of travel is the other condition and it fails.

The ratio is what matters, not either index

The critical angle depends on $n_2/n_1$ and on nothing else. Not on the colour of the light in this course, not on the brightness, not on the shape of the boundary. That single fact explains both applications the topic is famous for.

Push the indices apart and the critical angle collapses. Glass of index $1.52$ against air gives $41.1^\circ$ . Anything steeper than that is trapped. A $45^\circ$ prism therefore turns a beam through a right angle with no silvering at all, because $45^\circ$ is beyond $41.1^\circ$ , and it does it by total internal reflection rather than by a coating.

Bring the indices together and the critical angle climbs. Glass of index $1.50$ against a cladding of index $1.45$ gives $\sin^{-1}(1.45/1.50) = 75.2^\circ$ . Only rays that hit the wall of the fibre at a very glancing angle stay inside, which is exactly the design: light launched nearly along the axis strikes the wall at a large angle to the normal and is trapped for kilometres.

Change what is on the other side and the answer changes. Put that same $45^\circ$ glass prism, index $1.50$ , under water instead of in air. Now $n_2 = 1.33$ and the critical angle is $\sin^{-1}(1.33/1.50) = 62.5^\circ$ . The $45^\circ$ face is no longer beyond it, so the light refracts out into the water and the prism stops working. Nothing about the glass changed.

This is the strongest argument against the mental shortcut "dense materials trap light". Nothing traps light on its own. A boundary traps light, and only relative to what is on the other side.

The word denser is worth a caution of its own. An optically denser medium is one with a higher index of refraction, which need not have anything to do with mass density. See index of refraction for that distinction.

What the equation sheet prints, and what it does not

Neither relation on this page is printed on the AP Physics 2 equation sheet, and knowing that in advance is worth a minute in the exam room.

The waves, sound, and optics block of the Table of Information prints fifteen lines. In order: $\lambda = v/f$ ; $n = c/v$ ; $n_1\sin\theta_1 = n_2\sin\theta_2$ ; the mirror and lens equation $1/s_i + 1/s_o = 1/f$ ; the magnitude of the magnification; $\Delta D = m\lambda$ ; $\Delta D = a\sin\theta$ ; $a(y_{\text{min}}/L) \approx m\lambda$ ; $\Delta D = d\sin\theta$ ; $d(y_{\text{max}}/L) \approx m\lambda$ ; $v_{\text{string}}$ ; $T = 1/f$ ; $x(t)$ ; $y(x)$ ; and the beat frequency. That list was read off the rendered appendix page and checked line by line.

The law of reflection is not on it. Essential knowledge 13.1.B.2 gives $\theta_i = \theta_r$ as a relevant equation inside the framework, but the exam booklet does not reprint it. It is short enough to know.

The critical angle relation is not on it either, and the CED signals this itself. It labels $\theta_{\text{critical}} = \sin^{-1}(n_2/n_1)$ a derived equation at 13.3.A.5.i, where every other optics relation in the unit is labelled a relevant equation. Derived is the framework's way of saying you are expected to produce it.

Producing it takes one line. Snell's law is printed, and the critical angle is by definition the incidence angle whose refracted angle is $90^\circ$ :

n_1\sin\theta_c = n_2\sin 90^\circ = n_2 \quad \Rightarrow \quad \sin\theta_c = \frac{n_2}{n_1}

That derivation also explains 13.3.A.5.ii for free. The critical angle is defined as the case where the refracted ray reaches $90^\circ$ , so of course a refracted ray still exists there. It is the last angle at which one does.

For the same reason, the failure mode past $\theta_c$ is visible in the algebra. At $50.0^\circ$ inside glass of index $1.52$ , Snell's law gives $\sin\theta_2 = 1.52\sin 50.0^\circ = 1.16$ , and no angle has a sine above one. The equation returning nothing is 13.3.A.5.iii telling you there is no transmitted ray.

When it costs a mark

Saying total internal reflection begins at the critical angle. It begins beyond it. Essential knowledge 13.3.A.5.ii puts a refracted ray grazing along the surface at exactly the critical angle, and 13.3.A.5.iii puts the total case beyond.

Computing a critical angle for light entering a denser medium. There is not one. $\sin^{-1}(n_2/n_1)$ with the larger index on top has no value, and 13.3.A.4.ii explains why: the ray bends toward the normal, so the refracted angle can never reach $90^\circ$ .

Saying reflection only happens beyond the critical angle. Reflection happens at every angle, and 14.3.A.1.i says a wave crossing a boundary produces reflected and transmitted waves. What is exclusive to the beyond-critical case is that nothing is transmitted.

Measuring angles from the surface. Essential knowledge 13.1.B.2 defines the normal as the line perpendicular to the surface, and every angle in this topic is measured from it. A ray $10^\circ$ from a surface is at $80^\circ$ incidence, which flips the verdict on whether it is trapped.

Using a different angle rule for the total case. The reflected ray obeys $\theta_i = \theta_r$ whether or not the reflection is total. There is no separate law.

Calling a mirror an example of total internal reflection. A mirror reflects strongly at a surface, with no lower index medium and no critical angle in the story. Diffuse and specular reflection, 13.1.B.3 and 13.1.B.4, are the CED's two named kinds of ordinary reflection, and neither of them is the total case.

Assuming a material has one critical angle. It has one per boundary. The same glass gives $41.1^\circ$ against air and $62.5^\circ$ against water, because only the ratio matters.

Quoting the critical angle equation as though it were on the sheet. It is a derived equation in the CED and it is not in the Table of Information. Derive it from Snell's law with $\theta_2 = 90^\circ$ and the mark is safe either way.

When they coincide, and why that lulls you

Three overlaps keep the two looking like one phenomenon, and each hides a different part of the distinction.

They share an angle rule exactly. The reflected ray leaves at $\theta_i$ in both cases, so a ray diagram drawn for the total case looks identical to one drawn for an ordinary reflection at the same angle. Every geometric prediction you would make is the same. Only the transmitted ray, or its absence, distinguishes them on the page, and that is the ray students leave off the diagram.

Both return light toward where it came from. In everyday terms both are "the light bounces", which is a description that fits both and separates neither.

Ordinary reflection can return almost all the light. Approach a water surface from below at a shallow angle just under the critical angle and very little escapes; a polished mirror returns nearly everything at any angle. So "nearly all the light came back" is not evidence of total internal reflection. Only the two CED conditions are.

There is also a real continuity worth naming, because it is the thing that makes the topic feel arbitrary. As the incidence angle rises toward $\theta_c$ , the transmitted ray swings further from the normal and grows fainter, and at $\theta_c$ it lies flat along the surface. The transition is not a switch, it is the end of a trend. The CED marks the endpoint precisely at 13.3.A.5.ii and only calls the reflection total after it.

The three questions that settle any case:

Which way is the light going? Higher index into lower, or the reverse. If the reverse, the answer is ordinary reflection with refraction, and there is nothing more to check.
What is the critical angle for this boundary? From the ratio of the two indices, not from either one alone.
Is the angle of incidence beyond it, or at it, or below it? Three answers, and the middle one still has a refracted ray.

Where this sits on the AP exam

Both belong to Unit 13, Geometric Optics in AP Physics 2, weighted at 12 to 15 percent of the multiple-choice section over a suggested 8 to 12 class periods. They sit in different topics: ordinary reflection is Topic 13.1, Reflection, and total internal reflection is the last objective of Topic 13.3, Refraction. Neither topic carries a boundary statement; across all four topics in Unit 13, only Topic 13.2 has one, and it limits the study of mirrors to plane, convex spherical, and concave spherical.

The suggested skills differ between the two topics in a way that predicts the question type. Topic 13.1 lists 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.B, calculate or estimate an unknown quantity with units from known quantities; 2.C, compare physical quantities between two or more scenarios; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim. Topic 13.3 lists 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.B; 2.D, predict new values or factors of change using functional dependence between variables; 3.A, create experimental procedures appropriate for a given scientific question; and 3.B.

Skill 2.D on the refraction topic is the tell for this pair. "Predict new values or factors of change using functional dependence" is the prism-in-water question above: change one index and say what happens to the critical angle. Skill 3.A is why the CED's own sample activities for Unit 13 have students measuring an index of refraction experimentally, and the third free-response question on the AP Physics 2 exam is the Experimental Design and Analysis question.

For the pair one level up, where the transmitted ray is still present, see reflection vs refraction. For the definitions alone, see total internal reflection, critical angle and reflection. For the relation the critical angle is derived from, see Snell's law.

One glass block, three angles, and the reverse trip

Light travels inside glass of index $n_1 = 1.52$ and reaches a flat boundary with air, $n_2 = 1.00$ . (a) Find the critical angle. (b) Describe every ray leaving the boundary for an incidence angle of $30.0^\circ$ . (c) Do the same for $50.0^\circ$ . (d) Send light the other way, from the air into the glass at $50.0^\circ$ , and say whether total internal reflection can occur.

(a) Use the derived equation at 13.3.A.5.i: $\theta_{\text{critical}} = \sin^{-1}(n_2/n_1) = \sin^{-1}(1.00/1.52) = \sin^{-1}(0.6579)$ , so $\theta_{\text{critical}} = 41.1^\circ$ . If you prefer not to memorise it, set $\theta_2 = 90^\circ$ in Snell's law: $1.52\sin\theta_c = 1.00\sin 90^\circ$ , giving $\sin\theta_c = 1.00/1.52$ , the same thing.
(b) At $30.0^\circ$ , which is below the critical angle, both rays exist. The reflected ray leaves at $\theta_r = \theta_i = 30.0^\circ$ by 13.1.B.2. For the transmitted ray, Snell's law gives $\sin\theta_2 = (1.52)(\sin 30.0^\circ)/1.00 = (1.52)(0.5000) = 0.7600$ , so $\theta_2 = 49.5^\circ$ .
The refracted ray bent away from the normal, from $30.0^\circ$ to $49.5^\circ$ , which is what 13.3.A.4.i requires for light entering a lower index medium.
(c) At $50.0^\circ$ , which is beyond $41.1^\circ$ , try Snell's law anyway: $\sin\theta_2 = (1.52)(\sin 50.0^\circ)/1.00 = (1.52)(0.7660) = 1.164$ . No angle has a sine greater than one, so there is no transmitted ray. By 13.3.A.5.iii all the light is reflected, and it leaves at $\theta_r = 50.0^\circ$ , the same law of reflection as in part (b).
(d) Going from air into glass, $n_1 = 1.00$ and $n_2 = 1.52$ . Snell's law gives $\sin\theta_2 = (1.00)(\sin 50.0^\circ)/1.52 = 0.7660/1.52 = 0.5040$ , so $\theta_2 = 30.3^\circ$ , an ordinary refraction toward the normal. Total internal reflection is impossible in this direction: 13.3.A.5 requires passage into a medium with a lower index, and $\sin^{-1}(1.52/1.00)$ has no value.

The critical angle is $41.1^\circ$ . At $30.0^\circ$ there is a reflected ray at $30.0^\circ$ and a refracted ray at $49.5^\circ$ . At $50.0^\circ$ there is only a reflected ray at $50.0^\circ$ , carrying all the light. Entering from the air at $50.0^\circ$ simply refracts to $30.3^\circ$ , with no possibility of total internal reflection.

The prism that stops working under water

A right-angle prism made of glass with $n = 1.50$ is used to turn a laser beam through $90^\circ$ . The beam enters one short face along the normal, strikes the hypotenuse at an angle of incidence of $45.0^\circ$ , and leaves through the other short face. (a) Find the critical angle for the glass-to-air boundary and show that the prism works in air. (b) Find the critical angle for the same glass against water, $n = 1.33$ . (c) State what happens to the beam when the prism is submerged. (d) Explain why the glass itself is not what decides the outcome.

(a) At the entry face the beam is along the normal, so by 13.3.A.4.iii it is not refracted and reaches the hypotenuse at $45.0^\circ$ . For glass against air, $\theta_{\text{critical}} = \sin^{-1}(1.00/1.50) = \sin^{-1}(0.6667) = 41.8^\circ$ .
Since $45.0^\circ > 41.8^\circ$ , the incidence is beyond the critical angle, so 13.3.A.5.iii applies: all the light is reflected and none is transmitted. The reflected ray leaves at $45.0^\circ$ on the other side of the normal, which turns the beam through $90.0^\circ$ . No silvering is needed.
(b) For the same glass against water, $\theta_{\text{critical}} = \sin^{-1}(1.33/1.50) = \sin^{-1}(0.8867) = 62.5^\circ$ .
(c) Now $45.0^\circ$ is below $62.5^\circ$ , so the boundary behaves ordinarily. Snell's law gives $\sin\theta_2 = (1.50)(\sin 45.0^\circ)/1.33 = (1.50)(0.7071)/1.33 = 0.7975$ , so a refracted ray leaves into the water at $52.9^\circ$ . Some light is still reflected at $45.0^\circ$ , by 14.3.A.1.i, but the reflection is no longer total and the device leaks.
(d) Because the critical angle depends on the ratio $n_2/n_1$ . The glass was identical in both parts, at index $1.50$ , and the answer changed because the second medium changed from $1.00$ to $1.33$ . There is no such thing as the critical angle of a material, only the critical angle of a boundary.

In air the critical angle is $41.8^\circ$ , below the $45.0^\circ$ incidence, so the beam is totally internally reflected and turns through $90.0^\circ$ . Under water the critical angle rises to $62.5^\circ$ , above the $45.0^\circ$ incidence, so light refracts out at $52.9^\circ$ and the prism no longer works.

Designing a fibre: how close the two indices have to be

An optical fibre has a core of index $1.50$ and a cladding of index $1.45$ . Light travelling inside the core strikes the core-cladding wall. (a) Find the critical angle at that wall. (b) A ray travels at $8.0^\circ$ to the axis of the fibre. Find its angle of incidence at the wall and say whether it stays inside. (c) Compare with the critical angle the same core would have against air. (d) Say what the comparison implies about which rays a fibre can carry.

(a) $\theta_{\text{critical}} = \sin^{-1}(n_2/n_1) = \sin^{-1}(1.45/1.50) = \sin^{-1}(0.9667) = 75.2^\circ$ . The two indices are close, so the ratio is near one and the critical angle is large.
(b) The wall runs along the axis, so its normal is perpendicular to the axis. A ray at $8.0^\circ$ to the axis therefore makes an angle of $90.0^\circ - 8.0^\circ = 82.0^\circ$ with that normal. Since $82.0^\circ > 75.2^\circ$ , the ray is beyond the critical angle and 13.3.A.5.iii applies: all of it is reflected back into the core, and it stays inside.
(c) Against air the same core would have $\theta_{\text{critical}} = \sin^{-1}(1.00/1.50) = 41.8^\circ$ , far smaller. A bare core in air would trap a much wider range of rays.
(d) With a cladding, only rays within about $14.8^\circ$ of the axis reach the wall beyond $75.2^\circ$ and stay in, because $90.0^\circ - 75.2^\circ = 14.8^\circ$ . That is a narrow acceptance, and it is the design choice: restricting which rays travel keeps them arriving together. The point for the exam is the dependence, not the engineering. Bringing the two indices closer raises the critical angle and traps less; pushing them apart lowers it and traps more.

The critical angle at the core-cladding wall is $75.2^\circ$ . A ray $8.0^\circ$ from the axis meets the wall at $82.0^\circ$ , beyond the critical angle, so it is totally internally reflected and stays inside. Against air the same core would have a critical angle of only $41.8^\circ$ , so the cladding narrows the range of rays the fibre carries to about $14.8^\circ$ from the axis.

Frequently asked questions

What is the difference between reflection and total internal reflection?

Ordinary reflection happens at every boundary, at every angle, in either direction, and returns only part of the light while the rest is transmitted. Total internal reflection returns all of it and nothing is transmitted. AP Physics 2 essential knowledge 13.1.B.2 gives the law of reflection with no conditions attached, while 13.3.A.5 attaches two conditions to the total case: the light must pass into a medium with a lower index of refraction, and 13.3.A.5.i requires the angle of incidence to be beyond the critical angle. The angle rule is the same in both cases.

Does total internal reflection follow the law of reflection?

Yes, exactly. The reflected ray leaves at an angle to the normal equal to the angle of incidence, by essential knowledge 13.1.B.2, whether or not the reflection is total. That is why the two cases produce identical ray diagrams for the reflected ray. What distinguishes them is the transmitted ray: below the critical angle one exists and takes some of the light, and beyond it, by 13.3.A.5.iii, no light is transmitted into the other medium at all.

When does total internal reflection happen?

Only when both conditions hold. First, the light must be passing from a medium into another medium with a lower index of refraction, per AP Physics 2 essential knowledge 13.3.A.5, so glass to air works and air to glass never does. Second, per 13.3.A.5.i, the angle of incidence must be beyond the critical angle, which is the inverse sine of the ratio of the second index to the first. At exactly the critical angle it has not happened yet: 13.3.A.5.ii says the ray refracts at 90 degrees and travels along the surface of the material.

Is the critical angle equation on the AP Physics 2 equation sheet?

No. The CED gives it at essential knowledge 13.3.A.5.i and labels it a derived equation rather than a relevant equation, and the waves, sound, and optics block of the Table of Information does not print it. Neither does it print the law of reflection. Derive the critical angle from Snell's law, which is printed, by setting the refracted angle to 90 degrees: n1 sin theta c equals n2 sin 90 equals n2, so the sine of the critical angle is n2 divided by n1.

Why is total internal reflection taught in the refraction topic and not the reflection topic?

Because the thing that changes is the refraction, not the reflection. The reflected ray obeys the same law at every angle and is always present. What ends beyond the critical angle is the transmitted ray, and deciding whether one exists is a Snell's law question. That is why the statements live in AP Physics 2 Topic 13.3, Refraction, at 13.3.A.5, and why the algebra that signals the effect is Snell's law returning a sine greater than one.

Is a mirror an example of total internal reflection?

No. A mirror is a reflecting surface and returns most of the light at every angle from either side, with no second transparent medium and no critical angle involved. AP Physics 2 names two kinds of ordinary reflection in essential knowledge 13.1.B.3 and 13.1.B.4, diffuse from a rough surface and specular from a smooth one, and neither is the total case. Returning nearly all the light is not the same as the condition 13.3.A.5 sets.

Can the critical angle be different for the same material?

Yes, because it belongs to a boundary rather than to a material. It is the inverse sine of the ratio of the two indices, so the same glass of index 1.50 has a critical angle of 41.8 degrees against air and 62.5 degrees against water. That is why a 45 degree glass prism turns a beam through a right angle in air by total internal reflection and leaks light when the same prism is submerged: the 45 degree incidence sits above one critical angle and below the other.