Why Total Internal Reflection Cannot Occur from Air into Glass

A ray of light strikes a glass surface at an angle of \(85^\circ\) to the normal. Most of the light might appear to reflect. If you increase the angle even further, could the reflection become total internal reflection?

No. Total internal reflection cannot occur when light travels from air into glass because it is moving from a lower refractive index to a higher refractive index. The ray bends towards the normal as it enters the glass, and a transmitted ray is always possible for any incident angle below \(90^\circ\).

The interesting part is that reversing the direction changes the answer. Light travelling from glass into air can undergo total internal reflection, even though the boundary contains exactly the same two materials.

01An experiment that reveals why direction matters

Imagine testing both directions with a ray box and a glass block.

A rectangular glass block seems like the obvious apparatus. Shine a narrow beam into one face and observe where it emerges.

There’s a problem, though. The beam refracts when it enters the block and again when it leaves. The outgoing direction alone doesn’t isolate what happened at the first boundary.

A semicircular glass block solves this.

First arrangement: Air into glass

Place a semicircular glass block on a sheet of paper with its flat face towards a ray box. Mark the midpoint of the flat face as \(O\), and draw a normal through \(O\), perpendicular to that face.

Connect the ray box to its power supply, select a single narrow beam, and aim the beam from the air towards \(O\).

The beam follows this path:

1. It travels through air towards the flat glass surface.
2. At \(O\), some light reflects into the air, while some refracts into the glass.
3. Inside the block, the transmitted ray travels from \(O\) towards the curved surface along a radius.
4. It meets the curved surface normally, so it leaves without changing direction again.

That last feature is useful. The curved surface doesn’t introduce another change of direction that could confuse the measurement.

Suppose you increase the angle of incidence from \(30^\circ\) to \(60^\circ\), then to \(80^\circ\).

You might predict that eventually the refracted ray disappears because the incident ray is almost parallel to the surface.

But the ray model predicts something different: the refracted ray continues to enter the glass. Its direction approaches a limiting angle inside the glass.

The reflected beam can become much stronger at very large incident angles. That doesn’t establish total internal reflection. Reflection and refraction can occur simultaneously.

Second arrangement: Glass into air

Now move the ray box to the curved side of the block.

Aim the beam along a radius towards \(O\). It enters the curved surface normally, travels straight through the glass, and reaches the flat face.

This time, the incident ray at the flat surface is inside the glass.

As you increase its angle to the normal, the expected behaviour changes:

• At \(30^\circ\), some light refracts into the air.
• At approximately \(41.8^\circ\), the refracted ray reaches the limiting direction along the flat surface.
• At \(50^\circ\), no propagating refracted ray enters the air. The incident light undergoes total internal reflection back into the glass.

These are predicted outcomes from established refraction and reflection laws, not measurements from a particular experiment. In an actual ray-box investigation, the beam has finite width, surfaces scatter some light, and a very faint transmitted beam can be difficult to see.

The important experimental distinction is the direction in which light reaches the flat boundary. The apparatus and materials are otherwise unchanged.

02Why the ray bends towards the normal

First, get the angle convention right.

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

The angle of incidence is measured between the incident ray and that normal, not between the ray and the glass surface.

For a smooth, flat boundary:

• The incident ray approaches the surface.
• The refracted ray travels into the second medium.
• The reflected ray stays in the first medium.
• All angles are measured from the normal, between \(0^\circ\) and \(90^\circ\).

If a ray makes \(15^\circ\) with the surface, its incidence angle is \(75^\circ\).

Why does the refracted ray bend towards the normal?

Light travels more slowly in glass than in air. When a wavefront reaches the boundary at an angle, one part enters the slower medium before the rest. The direction of the wavefront changes, and the ray, which is perpendicular to the wavefront, turns towards the normal.

A useful analogy is a shopping trolley rolling diagonally from smooth flooring onto a rough mat. One wheel slows before the other, changing the trolley’s direction.

The trolley represents the changing direction, not light itself. Light is a wave, and the analogy cannot explain how much light reflects or transmits.

The mathematical limit

Snell’s law describes the change in direction:

\[ n_a\sin\theta_a=n_g\sin\theta_g \]

Here:

• \(n_a\) is the refractive index of air.
• \(n_g\) is the refractive index of glass.
• \(\theta_a\) is the ray’s angle in air, measured from the normal.
• \(\theta_g\) is the ray’s angle in glass, measured from the normal.

Refractive index is dimensionless. Angles in this article are measured in degrees.

For our illustrative calculations, assume ordinary transparent materials with:

\[ n_a=1.00,\qquad n_g=1.50 \]

These are rounded modelling values. Actual refractive indices depend on the material and the wavelength of light.

Rearranging Snell’s law for light entering glass:

\[ \sin\theta_g =\frac{n_a}{n_g}\sin\theta_a \]

Therefore:

\[ \sin\theta_g =\frac{1.00}{1.50}\sin\theta_a \]

The sine of an angle cannot exceed 1. So:

\[ \sin\theta_g\leq\frac{1}{1.50} \]

This means the refracted angle in glass approaches a maximum of:

\[ \theta_g \leq\sin^{-1}\left(\frac{1}{1.50}\right) \]

\[ \theta_g\leq41.8^\circ \]

For a beam actually crossing into the glass with an air incidence angle below \(90^\circ\), the glass angle is strictly less than \(41.8^\circ\).

Notice what the calculation proves. No matter how close the air ray comes to skimming the surface, its refracted direction doesn’t reach \(90^\circ\) inside the glass.

The transmitted direction remains mathematically possible.

What happens as the angle increases?

The following values are predictions from Snell’s law, not measured experimental results.

The refracted angle increases more slowly than the incident angle. It approaches \(41.8^\circ\), rather than continuing towards \(90^\circ\).

At an incidence angle of exactly \(90^\circ\), the ideal incident ray runs along the surface rather than crossing towards it. The \(41.8^\circ\) result is therefore a limiting refracted angle, not evidence that a beam entering from air has reached total internal reflection.

03Reverse the direction: Why glass into air is different

Return to Snell’s law, but let the light approach the boundary from inside the glass:

\[ n_g\sin\theta_g=n_a\sin\theta_a \]

Now the light travels from refractive index \(1.50\) into refractive index \(1.00\).

It bends away from the normal.

As the angle inside the glass increases, the refracted angle in air increases even faster.

Eventually, the refracted angle reaches \(90^\circ\). That’s the critical condition.

For glass into air:

\[ n_g\sin\theta_c=n_a\sin90^\circ \]

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

\[ \sin\theta_c=\frac{n_a}{n_g} \]

\[ \theta_c =\sin^{-1}\left(\frac{1.00}{1.50}\right) =41.8^\circ \]

Here, \(\theta_c\) is the critical angle, measured from the normal inside the glass.

Total internal reflection occurs only when:

\[ \theta_g>\theta_c \]

For example, a ray travelling inside glass at \(50.0^\circ\) to the normal undergoes TIR at a glass-air boundary because:

\[ 50.0^\circ>41.8^\circ \]

Trying to calculate its refracted angle gives:

\[ \sin\theta_a =1.50\sin50.0^\circ \approx1.15 \]

There is no real angle whose sine is \(1.15\). A propagating refracted ray is no longer possible in the air.

That is the mathematical reason for TIR.

The same number has two different meanings

Here’s a subtle point worth remembering.

The angle \(41.8^\circ\) appeared in both directions, but it played a different role.

It is incorrect to call \(41.8^\circ\) a critical incident angle for air entering glass.

The equality comes from reversing the same boundary conditions. It doesn’t mean the two directions allow the same behaviour.

For a full treatment of the threshold itself, see Critical Angle and Total Internal Reflection for HSC Physics.

04What about a ray that entered from air but reflects inside glass later?

This is where the wording of an HSC question can become tricky.

A ray might begin its journey in air, enter a glass prism, and then undergo total internal reflection at a different glass-air surface.

That doesn’t contradict anything we’ve established.

The rule applies to the direction of travel at the particular boundary being tested, not to where the ray originally came from.

Consider two situations.

A parallel-sided glass slab

A ray enters one flat glass surface from air, travels through the glass, and strikes the opposite surface.

The two faces are parallel, and air surrounds the slab.

The normals are also parallel. So the incidence angle at the second face equals the refracted angle inside the glass at the first face.

We already know that angle is less than \(41.8^\circ\).

Therefore, the ray cannot undergo TIR at the second glass-air surface, even though it is now travelling from higher to lower refractive index.

A glass prism with a tilted second face

Now change the geometry.

A ray enters the first glass face from air and bends towards its normal. It then travels to a second face with a different normal.

At this new face, its incidence angle may be much larger than the angle it had inside the glass at the entrance.

If it exceeds the critical angle, TIR can occur.

The key difference is the surface normal at the second boundary.

A ray can have a small angle to the first normal but a large angle to another.

That leads to a more reliable decision process for any multi-boundary problem:

1. Identify which material contains the ray immediately before it reaches the boundary.
2. Identify the material on the other side.
3. Draw the normal at that particular surface.
4. Measure the incidence angle from that normal.
5. If the ray travels from higher to lower refractive index, compare its angle with the critical angle. Otherwise, TIR is impossible.

This prevents two opposite mistakes: claiming TIR at an air-glass entrance because the angle is large, and ruling out TIR later just because the ray originally entered from air.

05Questions and solutions

Assume geometric optics, smooth transparent boundaries, and negligible absorption unless stated otherwise. Use the given refractive indices as exact modelling values, and round calculated angles to one decimal place where appropriate.

Question 1

A ray travels through air with refractive index \(1.00\) and strikes a flat glass surface with refractive index \(1.60\).

The incidence angle is \(45.0^\circ\) to the normal.

Calculate the refracted angle and explain whether total internal reflection can occur at this boundary.

Solution 1

The ray refracts into the glass at \(26.2^\circ\) to the normal. Total internal reflection cannot occur because light travels from lower to higher refractive index.

Using Snell’s law:

\[ n_a\sin\theta_a=n_g\sin\theta_g \]

Substitute:

\[ \begin{aligned} \sin\theta_g &=\frac{1.00\sin45.0^\circ}{1.60}\\ &=0.4419 \end{aligned} \]

Therefore:

\[ \theta_g=\sin^{-1}(0.4419) \]

\[ \theta_g=26.2^\circ \]

The ray bends towards the normal because light travels more slowly in glass than in air.

The common mistake is to check only whether the incidence angle is large. TIR also requires the ray to travel from higher to lower refractive index, which is not happening here.

Some light can reflect from the surface, but that is partial reflection.

Question 2

A ray enters the flat face of a semicircular glass block from air. The angle between the incident ray and the glass surface is \(12.0^\circ\).

The refractive indices are:

• Air: \(1.00\)
• Glass: \(1.50\)

A strong reflected beam is visible in the air.

A student claims, “The ray is almost parallel to the surface, so this must be total internal reflection.”

a. Calculate the refracted angle inside the glass, measured from the normal.

b. Predict what happens if a ray is sent back through the glass along exactly the reverse of this refracted path.

c. Describe one observation that could distinguish partial reflection from total internal reflection without measuring beam intensity.

Solution 2

The refracted angle is \(40.7^\circ\) inside the glass. The reversed ray refracts back into air at \(78.0^\circ\), and the original reflection is not total internal reflection.

Part a: Calculate the refracted angle.

The given \(12.0^\circ\) is measured from the surface, so the incidence angle from the normal is:

\[ \theta_a=90.0^\circ-12.0^\circ \]

\[ \theta_a=78.0^\circ \]

Apply Snell’s law:

\[ \begin{aligned} \sin\theta_g &=\frac{1.00\sin78.0^\circ}{1.50}\\ &=0.6521 \end{aligned} \]

Therefore:

\[ \theta_g=\sin^{-1}(0.6521) \]

\[ \theta_g=40.7^\circ \]

The ray enters the glass and bends towards the normal.

Part b: Reverse the ray.

A ray following the reversed path reaches the flat face from inside the glass at \(40.7^\circ\).

The critical angle for glass into air is:

\[ \theta_c =\sin^{-1}\left(\frac{1.00}{1.50}\right) =41.8^\circ \]

Because:

\[ 40.7^\circ<41.8^\circ \]

total internal reflection does not occur.

The ray refracts back into the air at \(78.0^\circ\) to the normal.

Part c: Test the claim.

Trace the transmitted ray through the glass and check whether it emerges from the curved surface.

Since the ray travels from the flat-face centre along a radius, its direction isn’t changed at the curved exit surface.

Finding this transmitted ray demonstrates that light crossed the original air-glass boundary.

The student’s prediction feels reasonable because a strong reflected beam resembles the result of TIR. But the existence of a transmitted ray disproves the claim, regardless of how bright the reflection appears.

Question 3

A parallel-sided glass slab has refractive index \(1.50\) and is surrounded by air with refractive index \(1.00\).

A ray enters the first face from air at an incidence angle that can be varied anywhere between \(0^\circ\) and \(90^\circ\), not including \(90^\circ\).

A student argues:

“At the first face, light cannot undergo TIR because it’s entering glass. But at the second face it’s travelling from glass into air, so if I make the original incidence angle large enough, it must undergo TIR.”

a. Determine whether this claim is correct for a slab with two perfectly parallel faces. Prove your answer for every allowed entrance angle.

b. Explain whether changing the thickness of the slab could change this conclusion.

c. A separate light source is embedded inside the same glass and sends a ray towards the second face at \(50.0^\circ\) to its normal. Determine whether this ray undergoes TIR and explain why your answer differs from part a.

Solution 3

The student’s claim is incorrect. A ray that enters a parallel-sided glass slab from air cannot undergo TIR at the opposite face. The slab’s thickness doesn’t change this, but a separate ray originating inside glass at \(50.0^\circ\) does undergo TIR.

Part a: Prove the limit.

Let \(\theta_a\) be the angle of incidence at the first surface and \(\theta_g\) the refracted angle inside the glass.

Snell’s law gives:

\[ 1.00\sin\theta_a =1.50\sin\theta_g \]

Therefore:

\[ \sin\theta_g =\frac{\sin\theta_a}{1.50} \]

For every allowed incident angle below \(90^\circ\):

\[ \sin\theta_a<1 \]

Therefore:

\[ \sin\theta_g<\frac{1}{1.50} \]

The critical angle for glass into air is:

\[ \theta_c =\sin^{-1}\left(\frac{1}{1.50}\right) =41.8^\circ \]

It follows that:

\[ \theta_g<\theta_c \]

Because the slab faces are parallel, their normals are parallel. The ray’s incidence angle at the second face equals \(\theta_g\).

It therefore remains below the critical angle, so TIR cannot occur at that face.

Part b: Change the thickness.

Changing the slab’s thickness alters the distance travelled inside the glass and the lateral displacement of the emerging ray.

It does not change the angle between the ray and either surface normal.

Therefore, changing the thickness cannot make TIR occur in this arrangement.

Part c: Use the internal light source.

This ray already travels inside the glass and strikes the glass-air boundary at:

\[ \theta_i=50.0^\circ \]

Since:

\[ 50.0^\circ>41.8^\circ \]

it undergoes total internal reflection.

The crucial difference is how the ray acquired its incidence angle.

A ray entering from air through a parallel face is constrained by Snell’s law. A source already inside glass can produce a direction that exceeds the glass-air critical angle.

The tempting argument checks the direction at the second boundary but overlooks the restriction imposed by the first.

Question 4

A triangular glass prism has refractive index \(1.50\) and is surrounded by air with refractive index \(1.00\).

Light enters through face A and travels towards face B.

The normal to face B is rotated by \(60.0^\circ\) from the normal to face A, towards the direction of the refracted ray. In the cross-section, the incidence angle at B is therefore:

\[ i_B=60.0^\circ-\theta_g \]

where \(\theta_g\) is the refracted angle inside the glass measured from the normal to face A.

Consider two rays entering at A:

• Ray P has an air incidence angle of \(20.0^\circ\).
• Ray Q has an air incidence angle of \(60.0^\circ\).

a. Determine which ray undergoes TIR at B.

b. Find the entrance incidence angle for which the ray reaches B at exactly the critical angle.

c. State the range of entrance angles that produce TIR at B.

d. Explain why increasing the entrance angle can stop TIR, even though large incidence angles are normally associated with total internal reflection.

Solution 4

Ray P undergoes TIR at face B, while Ray Q refracts out. The change occurs at an entrance angle of approximately \(27.9^\circ\). Increasing the entrance angle actually reduces the incidence angle at B in this particular prism geometry.

Part a: Analyse both rays.

First calculate the glass-air critical angle:

\[ \theta_c =\sin^{-1}\left(\frac{1.00}{1.50}\right) =41.8^\circ \]

For Ray P, Snell’s law at A gives:

\[ \begin{aligned} \sin\theta_g &=\frac{\sin20.0^\circ}{1.50}\\ \theta_g&=13.2^\circ \end{aligned} \]

At face B:

\[ i_B=60.0^\circ-13.2^\circ \]

\[ i_B=46.8^\circ \]

Since \(46.8^\circ>41.8^\circ\), Ray P undergoes TIR.

For Ray Q:

\[ \begin{aligned} \sin\theta_g &=\frac{\sin60.0^\circ}{1.50}\\ \theta_g&=35.3^\circ \end{aligned} \]

At face B:

\[ i_B=60.0^\circ-35.3^\circ \]

\[ i_B=24.7^\circ \]

Since \(24.7^\circ<41.8^\circ\), Ray Q can refract into the air.

Part b: Find the threshold.

At the critical condition:

\[ i_B=\theta_c=41.8^\circ \]

Using the prism geometry:

\[ \theta_g=60.0^\circ-41.8^\circ \]

\[ \theta_g\approx18.2^\circ \]

Now find the corresponding entrance angle using Snell’s law:

\[ \sin\theta_a=1.50\sin\theta_g \]

Using unrounded intermediate values:

\[ \theta_a =\sin^{-1}(1.50\sin18.19^\circ) \]

\[ \theta_a\approx27.9^\circ \]

Part c: State the range.

Within the stated geometry, TIR occurs at B for entrance angles:

\[ 0^\circ\leq\theta_a<27.9^\circ \]

At approximately \(27.9^\circ\), the ray is at the critical condition, not yet undergoing TIR.

For larger entrance angles, it can refract out through B.

Part d: Explain the counterintuitive result.

Increasing the incidence angle at A increases the refracted angle \(\theta_g\) inside the prism.

But because the normal to B is tilted towards that ray, increasing \(\theta_g\) reduces \(i_B\).

The ray becomes more closely aligned with the second normal.

The common mistake is to assume that increasing the incidence angle at one surface must increase it at every later surface.

Different surfaces have different normals. The geometry determines whether an angle increases or decreases.

Question 5

A beam of light in air strikes face A of a specially shaped glass block at \(70.0^\circ\) to the normal.

The refractive indices are:

\[ n_a=1.00,\qquad n_g=1.50 \]

After entering, the beam reaches face B.

The normal to face B is tilted by \(10.0^\circ\) relative to the entrance normal, on the opposite side of the refracted ray. Consequently, the incidence angle at B equals the refracted angle at A plus \(10.0^\circ\).

Consider two designs.

Design 1: Face B is directly exposed to air.

Design 2: Face B is covered by a uniform, transparent layer of water with refractive index \(1.33\). The outer water-air surface is parallel to face B.

Assume the water layer is thick enough for geometrical optics to apply separately at both surfaces, and extends far enough that the ray reaches its outer surface rather than its edge.

A designer claims:

“Adding water must allow light to escape. The glass-water boundary has a higher critical angle than glass-air, so there will no longer be total internal reflection.”

a. Determine what happens at face B in Design 1.

b. In Design 2, determine what happens at the glass-water boundary and calculate the ray’s angle in water.

c. Determine what happens when that ray reaches the water-air boundary. Does a propagating ray emerge into the surrounding air through the coating?

d. Explain precisely why the designer’s reasoning is incomplete.

e. Explain how the results would change if face B and its coating were instead parallel to the entrance face A.

Solution 5

In Design 1, the ray undergoes TIR at the glass-air boundary. In Design 2, it refracts into the water but then undergoes TIR at the outer water-air boundary. No propagating ray escapes through the coated face. If the faces were parallel, the ray could escape in either design.

The surprising result is that adding water moves the location of total internal reflection rather than allowing the light to reach the outside air.

Part a: Analyse the uncoated glass.

First calculate the ray’s direction inside glass.

At face A:

\[ 1.00\sin70.0^\circ =1.50\sin\theta_g \]

Therefore:

\[ \begin{aligned} \sin\theta_g &=\frac{\sin70.0^\circ}{1.50}\\ \theta_g&=38.8^\circ \end{aligned} \]

The normal at B is tilted \(10.0^\circ\) away from this ray.

Therefore:

\[ i_B=38.8^\circ+10.0^\circ \]

\[ i_B=48.8^\circ \]

The glass-air critical angle is:

\[ \theta_c =\sin^{-1}\left(\frac{1.00}{1.50}\right) =41.8^\circ \]

Since:

\[ 48.8^\circ>41.8^\circ \]

the ray undergoes TIR at B in Design 1.

Notice that the ray entered the glass from air successfully. The TIR occurs later, at a differently oriented boundary.

Part b: Add the water layer.

The ray still reaches B inside glass at \(48.8^\circ\). Its direction before B hasn’t changed.

But the second medium is now water, not air.

The glass-water critical angle is:

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

\[ \theta_c=62.5^\circ \]

Since \(48.8^\circ<62.5^\circ\), total internal reflection does not occur at this glass-water boundary.

The transmitted ray enters the water.

Use Snell’s law with unrounded angles:

\[ 1.50\sin48.79^\circ =1.33\sin\theta_w \]

Therefore:

\[ \begin{aligned} \sin\theta_w &=\frac{1.50\sin48.79^\circ}{1.33}\\ &\approx0.8485 \end{aligned} \]

\[ \theta_w\approx58.0^\circ \]

The ray refracts away from the normal into the lower-index water.

Some light can also reflect at the glass-water boundary, but this is partial reflection, not TIR.

Part c: Examine the water-air boundary.

Because the outer water surface is parallel to B, its normal is parallel to B’s normal.

The ray therefore reaches the water-air boundary at \(58.0^\circ\).

The critical angle from water into air is:

\[ \theta_c =\sin^{-1}\left(\frac{1.00}{1.33}\right) \]

\[ \theta_c=48.8^\circ \]

Since:

\[ 58.0^\circ>48.8^\circ \]

the ray undergoes total internal reflection at the water-air boundary.

No propagating transmitted ray emerges into the air through that surface.

The reflected light can subsequently return towards the glass-water boundary, where further reflection and transmission may occur. It may eventually leave through another part of the apparatus, but not directly through the specified outer water-air face.

Part d: Identify the missing constraint.

The designer correctly recognises that adding water increases the critical angle at the glass boundary.

But that only answers what happens at the first interface.

The light still has to cross the outer water-air boundary.

There’s a more powerful way to see the constraint.

Snell’s law at two parallel interfaces preserves the quantity \(n\sin\theta\):

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

Numerically:

\[ n_g\sin i_B\approx1.13 \]

If a refracted ray were to emerge into the final air, it would have to satisfy:

\[ n_a\sin\theta_a\approx1.13 \]

Since \(n_a=1.00\), this requires:

\[ \sin\theta_a\approx1.13 \]

That’s impossible for a real angle.

The intervening parallel water layer changes the ray’s angle and where reflection takes place, but it doesn’t remove the restriction on transmission into air.

This is why considering only one boundary can give a confident but incorrect prediction.

Part e: Make the faces parallel.

If face B is parallel to face A, the internal ray reaches B at the same angle it made with the entrance normal:

\[ i_B=38.8^\circ \]

For the uncoated surface, this is below the glass-air critical angle of \(41.8^\circ\).

Therefore, the ray refracts into air. Snell’s law gives an exit angle of \(70.0^\circ\), parallel to the original incident direction.

For the water-coated surface, the ray first refracts from glass into water.

Using Snell’s law:

\[ 1.50\sin38.79^\circ =1.33\sin\theta_w \]

\[ \theta_w\approx45.0^\circ \]

The water-air critical angle is \(48.8^\circ\), so \(45.0^\circ\) is below the threshold.

The ray can therefore refract into air at the outer surface, again at \(70.0^\circ\) to the normal.

The essential difference between the original problem and this one is the geometry.

A tilted surface can increase the internal incidence angle enough to allow TIR. A parallel surface cannot do that to a ray entering from air through the first face.

The water layer doesn’t magically release trapped light. Whether it can escape depends on the complete sequence of boundaries and the angles at each one.

06The useful distinction to take into your next ray diagram

When an HSC question shows light travelling from air into glass, don’t decide that total internal reflection occurs just because the incident ray is nearly parallel to the surface.

At that entry boundary, it cannot occur. The refracted ray bends towards the normal, although ordinary reflection may also be strong.

But don’t extend that answer automatically to every later surface.

The ray may encounter a differently oriented glass-air boundary, a coating, or a second medium. Each boundary needs its own incident medium, normal, incidence angle, and refractive-index comparison.

If the next problem contains more than one boundary, the natural continuation is How to Decide if a Ray Refracts or Reflects at a Second Boundary. The important move is to follow the actual ray path rather than carrying the first boundary’s answer through the rest of the diagram.