How to Decide if a Ray Refracts or Reflects at a Second Boundary

A ray enters a glass block from air and travels towards the opposite face. When it reaches that second boundary, its angle of incidence is \(50^\circ\). Does it leave the glass, or does it bounce back inside?

A student might predict that the ray simply refracts into the air. But that prediction is wrong: if the glass has a refractive index of 1.50 and the next medium is air, the ray undergoes total internal reflection.

Change the next medium to water, however, and the outcome is different. The ray refracts instead.

The decision depends on two things: the refractive indices of the two media at the second boundary, and the angle of incidence measured from the normal.

Here is the rule you need:

• If light travels from a higher refractive index to a lower one, calculate the critical angle.
• If the angle of incidence is smaller than the critical angle, the ray refracts.
• If the angle of incidence equals the critical angle, the refracted ray travels along the boundary.
• If the angle of incidence exceeds the critical angle, total internal reflection occurs.
• If light travels from a lower refractive index to a higher one, total internal reflection cannot occur.

The important detail is that you must make this decision again at every new boundary. The angle and the two media at the first boundary do not automatically determine what happens at the second.

01Start by identifying the second boundary

Consider a ray travelling through a material before reaching another material.

The material the ray is currently travelling through is the incident medium. The material it is approaching is the transmitted medium.

At the second boundary, label their refractive indices:

• \(n_1\): refractive index of the incident medium.
• \(n_2\): refractive index of the transmitted medium.

These labels are specific to the boundary you’re examining. If the ray has already crossed one boundary, the incident medium at the next boundary is the material it has just entered.

Draw the normal before measuring the angle

The normal is an imaginary line drawn perpendicular to the boundary at the point where the ray strikes it.

The angle of incidence, \(i\), is measured between the incoming ray and the normal.

The angle of refraction, \(r\), is measured between the transmitted ray and the normal.

Both angles are measured in degrees from the normal, not from the surface.

This distinction matters more than it might seem.

If a question gives an angle of \(35^\circ\) to the surface, the angle of incidence is:

\[ i=90^\circ-35^\circ \]

\[ i=55^\circ \]

Using \(35^\circ\) directly in Snell’s law would give you the wrong result.

Also, when a ray reaches a second face that isn’t parallel to the first, you must draw a new normal. The ray’s direction might not have changed, but its angle relative to the new normal has.

Which medium allows total internal reflection?

Total internal reflection is only possible when light travels from a material with a higher refractive index into one with a lower refractive index.

In other words:

\[ n_1>n_2 \]

Refractive index describes how much light slows down in a material. It is given by:

\[ n=\frac{c_0}{v} \]

where:

• \(n\) is the refractive index (dimensionless).
• \(c_0\) is the speed of light in a vacuum, measured in \(\text{m s}^{-1}\).
• \(v\) is the speed of light in the material, measured in \(\text{m s}^{-1}\).

A higher refractive index means light travels more slowly through that material.

Don’t confuse optical density with ordinary mass density. A material can have a greater refractive index without having a greater mass per unit volume.

If \(n_1\leq n_2\), there is no critical angle for light travelling in that direction, and total internal reflection cannot occur.

For example, light travelling from water into glass cannot undergo total internal reflection, regardless of how large its angle of incidence becomes.

02Calculate the critical angle

Once you’ve established that \(n_1>n_2\), you can calculate the critical angle.

The critical angle, \(\theta_c\), is the angle of incidence in the higher-index medium that produces a refracted ray travelling at \(90^\circ\) to the normal.

In other words, the refracted ray travels along the boundary between the two materials.

We can derive the critical angle directly from Snell’s law.

Snell’s law states:

\[ n_1\sin i=n_2\sin r \]

At the critical angle:

\[ i=\theta_c \]

and

\[ r=90^\circ \]

Substituting these values:

\[ n_1\sin\theta_c=n_2\sin90^\circ \]

Since \(\sin90^\circ=1\):

\[ n_1\sin\theta_c=n_2 \]

Rearranging gives the critical angle formula:

\[ \boxed{\theta_c= \sin^{-1}\left(\frac{n_2}{n_1}\right)} \]

Remember that this formula is only valid when \(n_1>n_2\).

The decision table

Once you’ve calculated the critical angle, the outcome is straightforward.

Why does the ray stop refracting once the angle exceeds the critical angle?

Consider what Snell’s law would predict:

\[ \sin r=\frac{n_1}{n_2}\sin i \]

When \(i>\theta_c\), the right-hand side becomes greater than 1.

But the sine of a real angle cannot exceed 1.

There is no possible refracted ray described by Snell’s law. Instead, the light is reflected back into the original medium. This is total internal reflection.

The angle of reflection equals the angle of incidence:

\[ \theta_{\text{reflection}}=i \]

03How the experiment demonstrates total internal reflection

How can you actually observe the transition from refraction to total internal reflection?

A useful investigation uses a semicircular glass block, a narrow laser beam, and a protractor.

The experimental challenge is to change the angle at the second boundary without accidentally introducing another refraction that makes the measurements confusing.

Setting up the apparatus

Place a semicircular glass block on a sheet of paper. Position its flat face along a straight line and mark the midpoint of that face.

Now direct a narrow laser beam through the curved surface so that it travels along a radius towards the midpoint of the flat face.

This arrangement is important.

The normal to a curved surface at any point lies along the radius passing through that point. Therefore, when the beam enters along a radius, it strikes the curved surface at normal incidence.

There is no change in its direction at this first boundary.

The beam then travels through the glass and reaches the flat face, which becomes the second boundary.

You can change the incidence angle at the flat face by directing the laser through different points on the curved edge while keeping its path aligned through the centre.

This allows you to investigate what happens as the internal angle of incidence increases.

What should the experiment show?

Before changing the laser position, make a prediction.

If Snell’s law describes the refraction correctly, what should happen as the angle of incidence increases?

The refracted angle should also increase. The ray should bend progressively further away from the normal as it leaves the glass.

Eventually, the refracted ray should reach \(90^\circ\) to the normal.

Beyond that angle, there should be no propagating refracted ray entering the air. Instead, the ray should reflect back into the glass.

This is the predicted transition to total internal reflection.

Illustrative experimental predictions

Assume the glass has a refractive index of 1.50 and air has a refractive index of 1.00.

The critical angle is:

\[ \theta_c=\sin^{-1}\left(\frac{1.00}{1.50}\right) \]

\[ \theta_c=41.8^\circ \]

The following table gives illustrative predictions from Snell’s law. These are calculated values, not claimed experimental measurements.

Consider the \(40^\circ\) reading.

Using Snell’s law:

\[ 1.50\sin40^\circ=1.00\sin r \]

\[ \sin r=0.9642 \]

\[ r=74.6^\circ \]

The ray still refracts, but it is travelling very close to the boundary.

Now increase the incidence angle to \(45^\circ\):

\[ \sin r=1.50\sin45^\circ \]

\[ \sin r=1.061 \]

This is impossible for a real refracted angle.

The model therefore predicts total internal reflection.

What if the original apparatus is poorly designed?

Suppose you initially use a rectangular glass block and direct the laser into its first face at an angle.

The beam refracts as it enters the glass. By the time it reaches the opposite face, its angle is no longer the angle you originally set in air.

If you compare that original angle with the critical angle, your conclusion may be wrong.

The semicircular block solves this problem by allowing the beam to enter normally at the curved surface. The first refraction no longer changes the beam’s direction, so the second boundary can be investigated independently.

There are still experimental limitations. Imperfect alignment, surface scattering, the finite width of the beam, and protractor uncertainty can make the transition near the critical angle difficult to identify precisely.

The important experimental conclusion is that increasing the internal incidence angle eventually removes the transmitted ray and produces total internal reflection, provided the ray is travelling from higher to lower refractive index.

04The same angle can produce different outcomes

One of the most tempting mistakes is to assume that a large angle of incidence automatically causes total internal reflection.

It doesn’t.

The refractive indices matter just as much as the angle.

Consider these different combinations:

Notice how the critical angle changes when you change the second medium.

For glass-to-air, the critical angle is approximately \(41.8^\circ\).

For glass-to-water, it increases to approximately \(62.5^\circ\).

This means a ray incident at \(50^\circ\) undergoes total internal reflection at a glass-air boundary but refracts at a glass-water boundary.

A contrasting case: large angle, but no total internal reflection

Imagine a ray travelling from water into glass at an incidence angle of \(70^\circ\).

You might see that \(70^\circ\) is much larger than the critical angle for glass-to-air and immediately predict total internal reflection.

But the ray is travelling from water into glass, not glass into water.

The incident medium has a lower refractive index than the transmitted medium:

\[ n_1=1.33 \]

\[ n_2=1.50 \]

Therefore, total internal reflection is impossible.

Instead, Snell’s law gives:

\[ 1.33\sin70^\circ=1.50\sin r \]

\[ \sin r= \frac{1.33}{1.50}\sin70^\circ \]

\[ r=56.4^\circ \]

The ray refracts towards the normal.

The key lesson is that you cannot decide whether total internal reflection occurs by looking at the angle alone.

You must identify the direction of travel and the two refractive indices first.

Another important distinction: reflection does not always mean total internal reflection

At an ordinary boundary below the critical angle, light can produce both a reflected ray and a refracted ray.

This is partial reflection.

Total internal reflection is the special case where no propagating refracted ray enters the second medium.

So if a question shows a reflected ray, don’t automatically label it total internal reflection. Check whether a transmitted ray also exists.

05Worked examples

06A reliable method for HSC exam questions

When a question asks whether a ray refracts or undergoes total internal reflection at a second boundary, follow this sequence.

1. Stop at the second boundary. Ignore the original direction of the ray for a moment.
2. Identify the medium pair. Label the current medium \(n_1\) and the next medium \(n_2\).
3. Draw the new normal. Make sure it is perpendicular to the second surface.
4. Determine the incidence angle. Measure it from the new normal, not from the surface or the original normal.
5. Check the refractive indices. If \(n_1\leq n_2\), total internal reflection is impossible.
6. Calculate the critical angle. If \(n_1>n_2\), use \(\theta_c=\sin^{-1}(n_2/n_1)\).
7. Compare the angles. If \(i>\theta_c\), total internal reflection occurs. If \(i<\theta_c\), refraction occurs. At equality, the refracted ray travels along the boundary.

If refraction occurs, use Snell’s law to calculate the refracted angle.

If total internal reflection occurs, use the law of reflection to calculate the reflected angle.

The most important habit is to repeat this process at every new boundary. Never assume that the incidence angle at the second surface is the same as the angle at the first.

07Questions and solutions

The following questions move from straightforward critical-angle calculations to situations where the ray changes direction before reaching the second boundary.

Question 1

A ray travels through glass with refractive index 1.50 and approaches a glass-air boundary. Its angle to the surface is \(52.0^\circ\).

The refractive index of air is 1.00.

Calculate:

a) The angle of incidence.

b) The critical angle.

c) The angle of refraction.

d) Whether total internal reflection occurs.

Solution 1

The ray refracts into the air; total internal reflection does not occur. The key is to convert the given angle to the correct incidence angle before using Snell’s law.

Part a: Find the angle of incidence.

The given angle is measured from the surface.

Angles of incidence are measured from the normal, so:

\[ i=90^\circ-52.0^\circ \]

\[ i=38.0^\circ \]

Part b: Calculate the critical angle.

The ray travels from glass into air:

\[ n_1=1.50,\qquad n_2=1.00 \]

Therefore:

\[ \theta_c= \sin^{-1}\left(\frac{1.00}{1.50}\right) \]

\[ \theta_c=41.8^\circ \]

Part c: Calculate the refracted angle.

Using Snell’s law:

\[ n_1\sin i=n_2\sin r \]

\[ \sin r= \frac{1.50}{1.00}\sin38.0^\circ \]

\[ \sin r=0.9235 \]

\[ r=67.5^\circ \]

Part d: Determine the outcome.

Compare:

\[ 38.0^\circ<41.8^\circ \]

The incidence angle is below the critical angle, so the ray refracts into the air.

The refracted angle is \(67.5^\circ\) to the normal.

A common mistake here is using the supplied \(52^\circ\) directly in Snell’s law. That angle is measured from the surface, not the normal.

Question 2

A ray travels from air into a parallel-sided acrylic slab with refractive index 1.49.

At the first boundary, its angle of incidence is \(60.0^\circ\). The slab is surrounded by air on both sides.

Determine:

a) The angle of the ray inside the acrylic.

b) Whether total internal reflection occurs at the second boundary.

c) The angle at which the ray emerges into the air.

Assume the two faces of the slab are perfectly parallel and the refractive index of air is 1.00.

Solution 2

The ray emerges from the second face of the acrylic without undergoing total internal reflection. Its internal angle is smaller than the critical angle because it bends towards the normal when entering the acrylic.

Part a: Find the internal angle.

At the first boundary:

\[ n_1=1.00,\qquad n_2=1.49 \]

Using Snell’s law:

\[ 1.00\sin60^\circ=1.49\sin r_1 \]

Rearranging:

\[ \sin r_1=\frac{\sin60^\circ}{1.49} \]

\[ \sin r_1=0.5812 \]

\[ r_1=35.6^\circ \]

The ray therefore travels through the acrylic at \(35.6^\circ\) to the first normal.

Part b: Determine whether total internal reflection occurs at the second boundary.

The second boundary is acrylic-to-air.

Therefore:

\[ n_1=1.49,\qquad n_2=1.00 \]

Calculate the critical angle:

\[ \theta_c= \sin^{-1}\left(\frac{1.00}{1.49}\right) \]

\[ \theta_c=42.2^\circ \]

Because the two faces are parallel, their normals are also parallel.

This means the internal ray meets the second face at the same angle it makes with the first normal:

\[ i_2=35.6^\circ \]

Compare:

\[ 35.6^\circ<42.2^\circ \]

Therefore, the ray refracts out of the acrylic.

Part c: Find the emerging angle.

Apply Snell’s law at the second boundary:

\[ 1.49\sin35.6^\circ=1.00\sin r_2 \]

\[ \sin r_2=0.866 \]

\[ r_2=60.0^\circ \]

The ray emerges at \(60.0^\circ\) to the normal.

This is the same angle as the original incidence angle. The emerging ray is parallel to the incoming ray because the slab has parallel faces and the same medium, air, on both sides.

The tempting mistake is to compare the original \(60^\circ\) angle directly with the critical angle of \(42.2^\circ\). That would incorrectly predict total internal reflection.

Always use the angle the ray actually makes with the normal at the second boundary.

Question 3

A ray travels from air into a triangular glass prism with refractive index 1.50.

At the first face, the angle of incidence is \(50.0^\circ\). The normal to the first face is horizontal.

The ray then reaches a second face, which also separates glass from air. The second-face normal line is rotated \(14.3^\circ\) clockwise from the first-face normal line.

Determine:

a) The angle of the ray inside the glass relative to the first normal.

b) The angle of incidence at the second boundary.

c) Whether total internal reflection occurs at the second boundary.

d) The angle of the reflected ray, if applicable.

Assume the refractive index of air is 1.00.

Solution 3

The ray undergoes total internal reflection at the second face, with an angle of reflection of \(45.0^\circ\). The important detail is that the second normal is tilted relative to the first normal, so the two incidence angles are different.

Part a: Calculate the internal angle at the first boundary.

The ray travels from air into glass:

\[ n_1=1.00,\qquad n_2=1.50 \]

Using Snell’s law:

\[ 1.00\sin50^\circ=1.50\sin r_1 \]

\[ \sin r_1=\frac{\sin50^\circ}{1.50} \]

\[ \sin r_1=0.5107 \]

\[ r_1=30.7^\circ \]

The ray travels through the glass at \(30.7^\circ\) anticlockwise from the first normal.

Part b: Calculate the second incidence angle.

The second normal is tilted \(14.3^\circ\) clockwise from the first normal.

The ray is therefore \(30.7^\circ+14.3^\circ\) from the second normal line.

\[ i_2=30.7^\circ+14.3^\circ \]

\[ i_2=45.0^\circ \]

This is why you must draw a new normal at every boundary.

Part c: Determine whether total internal reflection occurs.

The second boundary separates glass from air:

\[ n_1=1.50,\qquad n_2=1.00 \]

Calculate the critical angle:

\[ \theta_c= \sin^{-1}\left(\frac{1.00}{1.50}\right) \]

\[ \theta_c=41.8^\circ \]

Compare:

\[ 45.0^\circ>41.8^\circ \]

Therefore, total internal reflection occurs.

Part d: Find the reflection angle.

The law of reflection states:

\[ \theta_{\text{reflection}}=i_2 \]

Therefore:

\[ \theta_{\text{reflection}}=45.0^\circ \]

The ray reflects back into the glass at \(45.0^\circ\) to the second normal.

The trap is reusing the first internal angle of \(30.7^\circ\) at the second face. That would ignore the change in surface orientation and produce the wrong result.

Question 4

A ray travels through glass with refractive index 1.50 and reaches a glass-water boundary. Its angle of incidence is initially \(55.0^\circ\).

The refractive index of water is 1.33.

A detector identifies both a reflected ray inside the glass and a transmitted ray in the water.

a) Calculate the critical angle.

b) Calculate the angle of refraction.

c) Explain why the presence of a reflected ray does not mean total internal reflection has occurred.

d) The incidence angle is increased to \(64.0^\circ\). Determine the new outcome and the angle of the reflected ray.

Solution 4

At \(55.0^\circ\), the ray refracts into the water while also producing an ordinary reflected ray. When the incidence angle increases to \(64.0^\circ\), total internal reflection occurs.

Part a: Calculate the critical angle.

The ray travels from glass into water:

\[ n_1=1.50,\qquad n_2=1.33 \]

Therefore:

\[ \theta_c= \sin^{-1}\left(\frac{1.33}{1.50}\right) \]

\[ \theta_c=62.5^\circ \]

Part b: Calculate the refracted angle.

At the original incidence angle:

\[ i=55.0^\circ \]

Using Snell’s law:

\[ 1.50\sin55^\circ=1.33\sin r \]

Rearranging:

\[ \sin r=\frac{1.50}{1.33}\sin55^\circ \]

\[ \sin r=0.9239 \]

\[ r=67.5^\circ \]

The ray refracts into the water at \(67.5^\circ\) to the normal.

Part c: Explain the reflected ray.

The incidence angle is below the critical angle:

\[ 55.0^\circ<62.5^\circ \]

Therefore, refraction is still possible.

The reflected ray is produced by partial reflection at the boundary. It does not indicate total internal reflection because a transmitted ray also exists.

This distinction is important: reflection can occur without total internal reflection.

Part d: Increase the incidence angle to \(64.0^\circ\).

Now compare:

\[ 64.0^\circ>62.5^\circ \]

The ray undergoes total internal reflection.

Its angle of reflection is:

\[ \theta_{\text{reflection}}=64.0^\circ \]

The ray reflects back into the glass, and no propagating refracted ray enters the water.

The change in outcome is caused by crossing the critical angle, not by changing the materials.

Question 5

A glass block with refractive index 1.50 is separated from air by a parallel-sided water layer with refractive index 1.33.

The water layer is \(2.0\text{ cm}\) thick, and the two interfaces are parallel.

A ray inside the glass strikes the glass-water boundary at an angle of \(50.0^\circ\) to the normal.

Consider only the component of the beam transmitted into the water at the first boundary. Assume ideal ray optics, and ignore absorption and interference effects.

A student argues:

“The critical angle for glass-to-water is greater than \(50^\circ\), so the ray enters the water. Therefore, it must also escape into the air.”

Determine whether this conclusion is correct.

Your answer must:

a) Calculate the critical angle at the glass-water boundary.

b) Calculate the angle of the ray inside the water.

c) Determine whether total internal reflection occurs at the water-air boundary.

d) Describe what happens when the reflected ray returns to the glass-water boundary.

e) Determine the ranges of initial incidence angles in the glass that produce each of the following outcomes:

• The ray emerges into air.
• The ray enters water but undergoes total internal reflection at the water-air boundary.
• The ray undergoes total internal reflection at the glass-water boundary.

f) Explain whether changing the thickness of the water layer changes the total internal reflection decision.

Solution 5

The student’s conclusion is incorrect. The ray enters the water but undergoes total internal reflection at the water-air boundary, so it does not escape into the air.

This question is more difficult because the ray must be analysed at two successive boundaries. Passing the first boundary does not guarantee that it will pass the second.

Part a: Calculate the critical angle at the glass-water boundary.

The first boundary separates glass from water:

\[ n_1=1.50,\qquad n_2=1.33 \]

Therefore:

\[ \theta_{c,gw}= \sin^{-1}\left(\frac{1.33}{1.50}\right) \]

\[ \theta_{c,gw}=62.5^\circ \]

The incidence angle is \(50.0^\circ\).

Since:

\[ 50.0^\circ<62.5^\circ \]

the ray refracts into the water.

Part b: Calculate the angle inside the water.

Use Snell’s law at the glass-water boundary:

\[ n_g\sin i_g=n_w\sin\theta_w \]

where:

• \(n_g\) is the refractive index of glass.
• \(n_w\) is the refractive index of water.
• \(i_g\) is the incidence angle in glass.
• \(\theta_w\) is the refracted angle in water.

Substitute:

\[ 1.50\sin50^\circ=1.33\sin\theta_w \]

Rearranging:

\[ \sin\theta_w= \frac{1.50}{1.33}\sin50^\circ \]

\[ \sin\theta_w=0.8639 \]

\[ \theta_w=59.8^\circ \]

The ray now travels through the water at approximately \(59.8^\circ\) to the normal.

Part c: Determine whether total internal reflection occurs at the water-air boundary.

This is a new boundary, so we must start the decision process again.

The incident medium is now water, not glass.

The next medium is air.

Therefore:

\[ n_1=1.33,\qquad n_2=1.00 \]

Calculate the new critical angle:

\[ \theta_{c,wa}= \sin^{-1}\left(\frac{1.00}{1.33}\right) \]

\[ \theta_{c,wa}=48.8^\circ \]

The ray arrives at this second boundary at:

\[ i_2=59.8^\circ \]

Compare:

\[ 59.8^\circ>48.8^\circ \]

Therefore, total internal reflection occurs at the water-air boundary.

The ray enters the water but cannot propagate into the air.

Part d: What happens when the ray returns to the glass-water boundary?

The reflected ray travels back through the water and reaches the glass-water boundary again.

This time, light travels from water into glass.

The refractive indices are:

\[ n_1=1.33,\qquad n_2=1.50 \]

Since the incident medium has the lower refractive index, total internal reflection cannot occur.

The ray refracts back into the glass.

Using Snell’s law:

\[ 1.33\sin59.8^\circ=1.50\sin r \]

\[ \sin r\approx0.7660 \]

\[ r=50.0^\circ \]

The ray re-enters the glass at \(50.0^\circ\) to the normal, on the returning side of the interface.

This is a useful example of why the direction of travel matters. The same two materials can allow total internal reflection in one direction but not the other.

Part e: Determine the ranges of initial incidence angles.

This is the most interesting part of the question.

We have three materials:

\[ n_g=1.50 \]

\[ n_w=1.33 \]

\[ n_a=1.00 \]

There are two important critical angles already established:

Glass to water:

\[ \theta_{c,gw}=62.5^\circ \]

Water to air:

\[ \theta_{c,wa}=48.8^\circ \]

There is also a critical angle for a direct glass-air boundary:

\[ \theta_{c,ga}= \sin^{-1}\left(\frac{1.00}{1.50}\right) \]

\[ \theta_{c,ga}=41.8^\circ \]

Why is this third angle useful?

Because it tells us the original angle in glass at which the ray would be just critical when reaching the air boundary directly.

We can also derive this threshold using Snell’s law.

At the glass-water boundary:

\[ n_g\sin\theta_g=n_w\sin\theta_w \]

At the water-air boundary, the limiting condition for transmission is:

\[ n_w\sin\theta_w=n_a \]

Combining these equations:

\[ n_g\sin\theta_g=n_a \]

Therefore:

\[ \theta_g= \sin^{-1}\left(\frac{n_a}{n_g}\right) \]

This is exactly the direct glass-air critical angle:

\[ \theta_g=41.8^\circ \]

We can now classify every initial incidence angle.

Notice the middle range.

There is a whole interval of incidence angles where the ray successfully enters the water but cannot escape into the air.

This is the feature the student’s original argument misses.

Part f: Does the water layer thickness affect the decision?

No. Changing the thickness does not change the incidence angle at the water-air boundary when the two interfaces remain parallel.

The ray travels through the water in a straight line. Increasing the layer thickness changes the position where it reaches the second boundary, but not its angle relative to the common normal.

Therefore, the total internal reflection decision remains unchanged.

The assumption of parallel interfaces is essential here. If the second boundary were tilted, the normal would change direction, and the incidence angle would need to be recalculated.

The deeper lesson is that a ray passing through several materials must be tracked boundary by boundary. A ray can refract at one interface, undergo total internal reflection at the next, and refract again when it returns to an earlier interface.

08Where this takes you next

Once you can reliably decide whether a ray refracts or undergoes total internal reflection at a second boundary, you can apply the same reasoning to more complicated optical systems.

Prisms introduce multiple faces with different normals, while optical fibres rely on repeated total internal reflection at the boundary between a higher-index core and a lower-index cladding.

The same decision process applies in both cases: identify the two media, draw the local normal, calculate the critical angle when appropriate, and compare it with the actual incidence angle.