How to Find the Critical Angle Between Two Named Media

A question asks you to find the critical angle between crown glass and water. You calculate \(61.0^\circ\). Another student says there is no critical angle. Who is right?

Either answer could be correct. It depends on which way the light travels.

Light travelling from crown glass into water has a critical angle. Light travelling from water into crown glass does not. The two materials are identical in both cases, but the direction of travel changes the physics.

To find the critical angle between two named media, identify the material containing the incoming ray, then the material it is trying to enter. If the first material has a higher refractive index, use:

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

Here, \(n_1\) is the refractive index of the incident medium, \(n_2\) is the refractive index of the second medium, and \(\theta_c\) is the critical angle, measured in degrees from the normal.

The formula only gives a physical critical angle for total internal reflection when \(n_1>n_2\).

01Identify the two media in the correct order

Suppose a ray travels through acrylic and reaches a surface in contact with water.

The important description is not simply “acrylic and water”. It is:

Acrylic → water

This tells us which refractive index belongs in the numerator and which belongs in the denominator.

A refractive index, \(n\), describes how much light slows down in a material compared with its speed in a vacuum. It has no units.

Here are some approximate values we can use for practice.

These are rounded illustrative values for visible light near room temperature, not exact values for every sample. Refractive index can vary with wavelength, temperature, and material composition. In an HSC question, use the values supplied.

The first decision is straightforward:

• If \(n_1>n_2\), a critical angle exists for light travelling in that direction.
• If \(n_1<n_2\), there is no critical angle for total internal reflection.
• If \(n_1=n_2\), there is no meaningful critical-angle threshold because the two materials have equal refractive indices.

Why does the direction matter?

When light enters a material with a lower refractive index, its speed increases, and its direction bends away from the normal. As the incidence angle increases, the transmitted ray moves closer to the boundary.

Eventually, it reaches a limiting direction: along the boundary itself.

But when light travels into a material with a higher refractive index, it bends towards the normal. Increasing the incidence angle cannot produce the same limiting behaviour.

02Draw the normal before calculating

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

For a horizontal boundary, the normal is vertical. For a tilted boundary, the normal tilts with it.

There are two angles to distinguish:

• 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 are measured in degrees, from \(0^\circ\) to \(90^\circ\) for the usual ray diagrams.

Imagine a ray that makes an angle of \(20^\circ\) with the surface.

Its angle of incidence is not \(20^\circ\). The normal is perpendicular to the surface, so:

\[ i=90^\circ-20^\circ=70^\circ \]

You must use \(70^\circ\) when comparing the ray with a critical angle.

This is especially important near total internal reflection, because a ray that looks almost parallel to the surface actually has a large incidence angle.

What exactly happens at the critical angle?

Picture light travelling from glass into water.

At a small incidence angle, the ray refracts into the water. As the incidence angle increases, the refracted ray moves further away from the normal.

Predict what happens when the refracted angle reaches \(90^\circ\).

The transmitted direction is now parallel to the boundary. This is the limiting case that defines the critical angle.

If the incidence angle increases beyond that limit, the ordinary ray model predicts no propagating refracted ray in the second medium. Instead, total internal reflection occurs.

The distinction is:

A reflected ray can exist even below the critical angle. Seeing reflection does not, by itself, prove total internal reflection.

For the broader distinction between partial reflection and complete trapping of the ray, see Total Internal Reflection: Conditions and Critical Angle.

03Derive the critical-angle formula from Snell’s law

Snell’s law connects the angles of incidence and refraction:

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

Here, \(n_1\) and \(n_2\) are dimensionless refractive indices, while \(i\) and \(r\) are angles measured from the normal.

At the critical angle:

\[ i=\theta_c,\qquad r=90^\circ \]

Substitute these into Snell’s law:

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

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

\[ n_1\sin\theta_c=n_2 \]

Rearrange:

\[ \sin\theta_c=\frac{n_2}{n_1} \]

Then take the inverse sine:

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

The inverse sine, \(\sin^{-1}\), returns an angle. It does not mean dividing by sine.

Set your calculator to degree mode.

Before substituting, check that \(n_2<n_1\). If the refractive index of the second material is larger, the ratio is greater than 1, and no real critical angle exists.

04How could an experiment measure the critical angle?

A calculation predicts where the transition should occur. A ray-box investigation gives us a way to test that prediction.

The main experimental problem is controlling the angle at the boundary we actually want to investigate.

First attempt: a rectangular acrylic block

Imagine placing a rectangular acrylic block on a protractor sheet.

A narrow ray from a ray box enters one face from air, travels through the acrylic, and reaches the opposite face, which is in contact with water.

You increase the angle of the ray entering the block, hoping to find the point where light stops escaping into the water.

What could go wrong?

The ray refracts as it enters the acrylic. Its direction inside the block is different from its direction in air.

Therefore, the incidence angle measured outside the block is not the incidence angle at the acrylic-water boundary.

If the two faces are parallel, the entry refraction also restricts which internal angles are reachable. Simply making the incoming ray more oblique does not guarantee that the critical angle at the exit face can be reached.

A better design needs to control the ray’s internal direction without introducing an unwanted change of direction at the entrance.

Revised design: a semicircular acrylic block

Use a semicircular acrylic block with its flat face in direct contact with water in a transparent cell. There must be no air gap between the acrylic and water.

Place a protractor centred on the midpoint of the flat face.

Direct a narrow beam through the curved surface so that it follows a radius towards that midpoint.

This works because the normal to a circular surface points along the radius. The beam therefore enters the curved surface at \(0^\circ\) incidence.

It can slow down and partially reflect at that surface, but its transmitted direction does not bend.

The beam then travels through the acrylic and strikes the flat acrylic-water boundary.

By moving the source around the curved edge while keeping the beam directed towards the centre, we can vary the incidence angle at the flat face without introducing an initial change of direction.

What should we observe?

Assume:

• Acrylic has refractive index \(1.49\).
• Water has refractive index \(1.33\).
• Both media are transparent, homogeneous, and approximately uniform in refractive index.
• We use light of one wavelength and treat the boundaries as smooth.
• Angles are measured in degrees from the normal.

The predicted critical angle is:

\[ \begin{aligned} \theta_c&=\sin^{-1}\left(\frac{1.33}{1.49}\right)\\ &=63.2^\circ \end{aligned} \]

Now consider the following predictions.

These values are calculated from Snell’s law. They are not claimed experimental measurements.

As the incidence angle increases from \(50^\circ\) to \(63^\circ\), the refracted ray moves towards the boundary.

At approximately \(63.2^\circ\), its limiting direction is along the acrylic-water interface.

Beyond that angle, no ordinary transmitted ray should propagate into the water.

The experimental signature would be a transition from a visible transmitted ray to total internal reflection as the incidence angle increases.

Why the measured value might not be exactly \(63.2^\circ\)

A real beam has a finite width. Its edges strike the boundary at slightly different angles.

Small alignment errors, scattering from imperfect surfaces, uncertainty in the material’s refractive index, and difficulty detecting a nearly grazing transmitted beam can affect the measured threshold.

Also, a faint transmitted ray may be difficult to see even when it exists. The disappearance of one visible spot is not enough to prove total internal reflection.

A stronger investigation tracks both the transmitted direction and the reflected beam as the incidence angle is changed.

Rather than claiming an exact critical angle from one observation, we can bracket it between the largest angle at which transmission is reliably observed and the smallest angle at which total internal reflection is reliably identified.

That bracket can also tell us something about an unknown liquid.

If the critical angle is measured and the first refractive index is known, rearrange the equation:

\[ n_2=n_1\sin\theta_c \]

This turns critical-angle measurement into a method for estimating the refractive index of the second medium.

05What happens if you change the liquid but keep the ray angle?

The critical angle depends on the refractive indices of both materials, not just the material containing the incoming ray.

Imagine an acrylic window with a liquid on one side. Light travels through the acrylic towards that liquid.

You replace water with glycerol, keeping the acrylic and ray direction unchanged.

Which liquid is more likely to let the light escape?

Glycerol has a refractive index closer to acrylic’s. Its critical angle is therefore larger, so refraction remains possible over a wider range of incident angles.

06Questions and solutions

Question 1

A ray travels from glycerol, with refractive index \(1.47\), into water, with refractive index \(1.33\).

Calculate the critical angle to one decimal place. State whether total internal reflection occurs at an incidence angle of \(67.0^\circ\).

Solution 1

The critical angle is \(64.8^\circ\), so a ray incident at \(67.0^\circ\) undergoes total internal reflection.

Light travels from the higher refractive index, \(1.47\), into the lower refractive index, \(1.33\). A critical angle therefore exists.

Using:

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

Substitute:

\[ \begin{aligned} \theta_c&=\sin^{-1}\left(\frac{1.33}{1.47}\right)\\ &=64.8^\circ \end{aligned} \]

Since:

\[ 67.0^\circ>64.8^\circ \]

the incident angle exceeds the critical angle.

No propagating refracted ray enters the water. The ray reflects back into the glycerol.

The result is specific to glycerol-to-water travel. Reversing the two media would remove the possibility of total internal reflection at that boundary.

Question 2

An investigator uses a semicircular acrylic block of refractive index \(1.49\) to identify an unknown transparent liquid.

The liquid could be water, \(n=1.33\), or ethanol, \(n=1.36\).

At the flat acrylic-liquid boundary, the following observations are recorded. The angles are measured between the incoming ray and the surface, not the normal.

Assume the angle readings are accurate and that transmission and total internal reflection have been correctly identified.

1. Find the possible interval for the critical angle.
2. Calculate the corresponding interval for the unknown liquid’s refractive index.
3. Identify which liquid is consistent with the observations.
4. Explain why using the listed angles directly in the critical-angle formula would be wrong.

Solution 2

The unknown liquid is consistent with water, not ethanol. Its critical angle is between \(63.0^\circ\) and \(64.0^\circ\), and its refractive index lies approximately between \(1.328\) and \(1.339\).

First convert both angles from the surface to the normal.

For the transmitted ray:

\[ i_1=90^\circ-27^\circ=63^\circ \]

For the totally internally reflected ray:

\[ i_2=90^\circ-26^\circ=64^\circ \]

Transmission at \(63^\circ\) tells us the critical angle is larger than \(63^\circ\). Total internal reflection at \(64^\circ\) tells us it is smaller than \(64^\circ\).

Therefore:

\[ 63^\circ<\theta_c<64^\circ \]

Now rearrange the critical-angle equation:

\[ n_2=n_1\sin\theta_c \]

At the lower bound:

\[ \begin{aligned} n_{\text{lower}}&=1.49\sin63^\circ\\ &=1.3276 \end{aligned} \]

At the upper bound:

\[ \begin{aligned} n_{\text{upper}}&=1.49\sin64^\circ\\ &=1.3392 \end{aligned} \]

Therefore:

\[ 1.3276<n_2<1.3392 \]

Water, with \(n=1.33\), lies inside this interval.

Ethanol, with \(n=1.36\), lies outside it.

The correct identification is water, under the stated assumptions.

The tempting mistake is using \(27^\circ\) and \(26^\circ\) as incidence angles. They were measured from the surface, whereas Snell’s law uses angles from the normal.

A final limitation matters: these observations establish a range, not an exact refractive index. Greater precision would require more closely spaced measurements and careful alignment.

Question 3

A ray travels from water, \(n=1.33\), into crown glass, \(n=1.52\), at \(75.0^\circ\) to the normal.

A faint reflected ray is observed in the water.

A student claims:

“Since \(75^\circ\) is greater than the critical angle between glass and water, the reflection must be total internal reflection.”

1. Explain whether the student’s claim is correct.
2. Calculate the angle of the refracted ray inside the crown glass.
3. The refracted ray is reflected by a separate mirror inside the glass so that it returns along exactly the same path to the original glass-water boundary. Determine whether total internal reflection occurs on its return, and explain why.

Assume smooth interfaces and ideal geometrical optics.

Solution 3

The student is incorrect. The first ray refracts into glass at \(57.7^\circ\), and the ray returning from glass into water also refracts rather than undergoing total internal reflection.

The student’s mistake is treating the critical angle as a property of two material names without considering direction.

1. First boundary: water into glass

The incident medium is water:

\[ n_1=1.33 \]

The second medium is crown glass:

\[ n_2=1.52 \]

Because \(n_1<n_2\), total internal reflection is impossible for this direction of travel.

A faint reflected ray can still be present because ordinary partial reflection occurs at refractive-index boundaries.

2. Angle inside crown glass

Apply Snell’s law:

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

Substitute:

\[ \begin{aligned} 1.33\sin75^\circ&=1.52\sin r\\ \sin r&=\frac{1.33}{1.52}\sin75^\circ\\ &=0.8453\\ r&=57.7^\circ \end{aligned} \]

The refracted ray travels at \(57.7^\circ\) to the normal, closer to the normal than the original ray.

3. Returning from glass into water

On the return journey, light travels from crown glass into water, so a critical angle exists.

Calculate it:

\[ \begin{aligned} \theta_c&=\sin^{-1}\left(\frac{1.33}{1.52}\right)\\ &=61.0^\circ \end{aligned} \]

The ray returns to the boundary at \(57.7^\circ\) to the normal because the mirror makes it retrace the original transmitted path.

Since:

\[ 57.7^\circ<61.0^\circ \]

total internal reflection does not occur.

The ray can refract back into water at \(75.0^\circ\) to the normal, reversing its original path.

The key constraint is that the angle inside the glass is not the original \(75^\circ\). Refraction on entry changed it.

A ray that entered from water at \(75^\circ\) cannot simply be assigned that same incidence angle when it later approaches the boundary from inside glass.

Question 4

A transparent model glass has slightly different refractive indices for two colours of light:

The glass is in contact with water, which has refractive index \(1.33\) for both colours in this simplified model.

Blue and red rays travel through the glass and reach the same flat glass-water boundary at an incidence angle of \(60.0^\circ\).

1. Calculate the critical angle for each colour.
2. Determine which colour undergoes total internal reflection and which can refract into water.
3. Calculate the angle of refraction for any transmitted colour.
4. Explain why it is misleading to say that a named pair of media must always have one fixed critical angle.

All index values are illustrative. Ignore absorption and assume each colour travels independently.

Solution 4

Blue light undergoes total internal reflection, while red light refracts into water at approximately \(81.8^\circ\) to the normal.

This happens because the two colours have different critical angles, even though the named materials are the same.

1. Calculate the blue critical angle

For blue light:

\[ \begin{aligned} \theta_{c,b}&=\sin^{-1}\left(\frac{1.33}{1.54}\right)\\ &=59.7^\circ \end{aligned} \]

2. Calculate the red critical angle

For red light:

\[ \begin{aligned} \theta_{c,r}&=\sin^{-1}\left(\frac{1.33}{1.52}\right)\\ &=61.0^\circ \end{aligned} \]

3. Compare with the common incidence angle

Both colours arrive at \(60.0^\circ\).

For blue:

\[ 60.0^\circ>59.7^\circ \]

So blue light undergoes total internal reflection.

For red:

\[ 60.0^\circ<61.0^\circ \]

So red light can refract into water.

4. Calculate the red refracted angle

Using Snell’s law:

\[ \begin{aligned} 1.52\sin60^\circ&=1.33\sin r\\ \sin r&=\frac{1.52}{1.33}\sin60^\circ\\ &=0.9898\\ r&=81.8^\circ \end{aligned} \]

The red ray emerges nearly parallel to the interface.

The tempting prediction is that both colours must behave identically because they encounter the same glass and water.

But refractive index can depend on wavelength. This dependence is called dispersion.

The higher refractive index for blue light makes its critical angle smaller in this example.

A critical angle is therefore not always determined by material names alone. It is determined by their refractive indices at the relevant wavelength and conditions.

Question 5

A thick, plane-parallel acrylic slab of refractive index \(1.49\) has glycerol, \(n=1.47\), in contact with its right face.

A ray is launched from another transparent medium into the slab through its left face.

The left and right faces are parallel, so their normals point in the same direction.

A student calculates the acrylic-glycerol critical angle and argues:

“Because acrylic has a higher refractive index than glycerol, I can always make total internal reflection happen at the right face by increasing the angle of the incoming ray.”

Investigate the claim for three possible launch media:

• Air, \(n=1.00\)
• Water, \(n=1.33\)
• Crown glass, \(n=1.52\)

For the crown-glass case, assume perfect optical contact with acrylic, with no intervening air gap.

1. Calculate the critical angle for light travelling from acrylic into glycerol.
2. Determine whether any external incidence angle from air can produce total internal reflection at the acrylic-glycerol boundary.
3. Repeat for an incident ray launched from water.
4. For crown glass, determine the complete range of external incidence angles that allows the ray to enter acrylic and then undergo total internal reflection at the acrylic-glycerol boundary.
5. Explain physically what happens at each endpoint of that range.

Use the ideal geometrical-ray model. Assume the acrylic slab is sufficiently wide and thick for any ray crossing it to reach the opposite face, and neglect wave-tunnelling effects.

Solution 5

The student’s claim is false. Although the acrylic-glycerol boundary has a critical angle of \(80.6^\circ\), neither air nor water can launch a ray through the parallel slab that reaches that angle. Crown glass can, but only for external incidence angles strictly between \(75.3^\circ\) and \(78.6^\circ\).

The central difficulty is that the ray cannot choose its angle inside the acrylic independently of the way it entered.

1. Critical angle for acrylic into glycerol

At the right face:

\[ n_1=1.49,\qquad n_2=1.47 \]

Therefore:

\[ \begin{aligned} \theta_c&=\sin^{-1}\left(\frac{1.47}{1.49}\right)\\ &=80.6^\circ \end{aligned} \]

Total internal reflection at that face requires the internal incidence angle to exceed \(80.6^\circ\).

But can the ray actually reach such an angle?

2. Connect the two boundaries using Snell’s law

Let:

• \(n_s\) be the refractive index of the launch medium.
• \(i\) be the external incidence angle at the left face.
• \(\alpha\) be the refracted angle inside the acrylic.

At the left face:

\[ n_s\sin i=1.49\sin\alpha \]

Because the slab faces are parallel, \(\alpha\) is also the angle of incidence at the right face.

For total internal reflection at the right face:

\[ 1.49\sin\alpha>1.47 \]

Combining the equations gives:

\[ n_s\sin i>1.47 \]

This is the first condition the launch ray must satisfy.

There is another condition: the ray must actually enter the acrylic and travel across the slab.

For a ray entering from a higher-index medium, refraction at the first boundary requires:

\[ n_s\sin i<1.49 \]

Equality would give an internal angle of \(90^\circ\), with the limiting transmitted direction parallel to the entrance face rather than crossing the slab.

Together, the conditions for a crossing ray that is totally internally reflected at the right face are:

\[ 1.47<n_s\sin i<1.49 \]

3. Launching from air

For air:

\[ n_s=1.00 \]

The largest possible value of \(\sin i\) is 1.

Therefore:

\[ n_s\sin i\leq1.00 \]

But total internal reflection at the right face requires:

\[ n_s\sin i>1.47 \]

That is impossible.

No ray launched from air through the left face can produce total internal reflection at the acrylic-glycerol boundary.

4. Launching from water

For water:

\[ n_s=1.33 \]

Again:

\[ n_s\sin i\leq1.33 \]

This is still less than the required value of \(1.47\).

Therefore, no ray entering acrylic from water through the parallel left face can undergo total internal reflection at the right acrylic-glycerol face.

The glass-like acrylic may have a critical angle into glycerol, but the earlier refraction prevents the ray from reaching it.

5. Launching from crown glass

Now the launch medium has:

\[ n_s=1.52 \]

The combined condition becomes:

\[ 1.47<1.52\sin i<1.49 \]

First, find the lower bound:

\[ \begin{aligned} \sin i&>\frac{1.47}{1.52}\\ i&>75.3^\circ \end{aligned} \]

Next, find the upper bound:

\[ \begin{aligned} \sin i&<\frac{1.49}{1.52}\\ i&<78.6^\circ \end{aligned} \]

Therefore, the required external incidence range is:

\[ 75.3^\circ<i<78.6^\circ \]

Within this narrow interval, the ray can enter acrylic from crown glass and reach the acrylic-glycerol boundary at an angle greater than its critical angle.

It then undergoes total internal reflection at the right face.

6. Understand the endpoints

At the lower endpoint, \(i\approx75.3^\circ\), the internal ray reaches the acrylic-glycerol boundary at its critical angle of \(80.6^\circ\).

The limiting transmitted direction is along that boundary. This is not yet total internal reflection.

At the upper endpoint, \(i\approx78.6^\circ\), the crown-glass-acrylic entrance itself reaches its critical angle.

The limiting refracted direction inside acrylic is \(90^\circ\) to the normal, parallel to the entrance face. A ray at exactly this limit does not travel across a slab of finite thickness.

For angles above this upper limit, total internal reflection occurs at the first boundary, before the ray can enter acrylic.

That is why both inequalities are strict.

The surprise is that calculating a critical angle does not guarantee that a particular ray can reach it.

For a named boundary, the formula tells you the local threshold. If the ray has passed through another medium first, Snell’s law at that earlier boundary may restrict the available angles.

This is the next decision to make in a multi-boundary problem: calculate the critical angle where the ray might escape, but also track how the earlier boundary determines its direction.

For a systematic method involving several surfaces, continue with How to Decide if a Ray Refracts or Reflects at a Second Boundary.