What Happens Exactly at the Critical Angle? | HSC Physics

Imagine a light ray travelling through glass towards an air boundary. As the angle of incidence increases, the refracted ray bends further away from the normal. Eventually, it reaches an angle of exactly 90° to the normal and travels along the boundary itself.

This is the critical-angle condition. It is the precise dividing point between ordinary refraction and total internal reflection. At the critical angle, the refracted ray grazes the surface. Only when the angle of incidence becomes greater than the critical angle does total internal reflection occur.

The difference between exactly 90° and just over the critical angle is small numerically, but important physically.

01What happens at exactly the critical angle?

Consider light travelling from glass into air. Glass has a higher refractive index than air, so the light bends away from the normal as it enters the air.

At the critical angle, three things are true:

• The incident ray is travelling in the optically denser medium (glass).
• The angle of incidence is equal to the critical angle, measured from the normal.
• The angle of refraction is exactly 90°, so the refracted ray travels parallel to the boundary.

The key detail is that the 90° angle is measured from the normal, not from the surface. A ray travelling along the surface is therefore at 90° to the normal and 0° to the boundary.

This is easy to mix up in an exam. If you measure the angle from the surface instead of the normal, you’ll reverse the interpretation.

At the critical angle, the relationship between the two angles is:

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

where:

• \(i\) is the angle of incidence in the glass.
• \(\theta_c\) is the critical angle.
• \(r\) is the angle of refraction in the air.

The critical angle is found using Snell’s law:

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

At the critical angle, substitute \(i=\theta_c\) and \(r=90^\circ\):

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

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

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

Here, \(n_1\) is the refractive index of the material the light starts in, and \(n_2\) is the refractive index of the material it approaches. The formula only gives a real critical angle when \(n_1>n_2\).

Notice what the equation tells us: the critical angle is an incidence angle inside the higher-refractive-index material. It is not the angle of the grazing ray.

02The difference between below, at, and above the critical angle

Let’s use a glass-air boundary with:

• Glass: \(n_1=1.50\)
• Air: \(n_2=1.00\)

The critical angle is:

\[ \theta_c=\sin^{-1}\left(\frac{1.00}{1.50}\right) =41.81^\circ \]

Now compare three incident angles.

The middle case is the important one.

At \(40^\circ\), light still escapes into the air. At \(41.81^\circ\), the refracted direction has reached its maximum possible angle. At \(42^\circ\), Snell’s law can no longer produce a real refracted angle.

Why?

Rearrange Snell’s law:

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

For the glass-air boundary:

\[ \sin r=1.50\sin i \]

At \(i=42^\circ\):

\[ \sin r=1.50\sin42^\circ \]

\[ \sin r=1.0037 \]

This is impossible for a real refracted angle because the sine of a real angle cannot exceed 1.

The ray therefore undergoes total internal reflection.

The distinction is:

• \(i<\theta_c\): a refracted ray enters the second medium.
• \(i=\theta_c\): the limiting refracted direction is along the boundary.
• \(i>\theta_c\): total internal reflection occurs.

The inequality matters. Total internal reflection requires an angle strictly greater than the critical angle.

What the graph tells us

As the angle of incidence approaches the critical angle from below, the refracted angle approaches \(90^\circ\).

The relationship is not linear. Small increases in incidence angle near the critical angle produce increasingly large changes in the refracted angle.

This explains why a ray can appear to disappear so suddenly when you increase its incidence angle.

It hasn’t gradually turned into a reflected ray over a wide range of angles. The refracted direction is approaching a limiting value.

03Can an experiment show the exact critical angle?

Imagine investigating the critical angle at a water-air boundary.

The question is straightforward: can you observe the refracted ray gradually approaching the surface, then identify the point where it becomes parallel to that surface?

A useful experimental arrangement would use a transparent water tank, a narrow laser beam, an adjustable underwater laser mount, a protractor, and a screen positioned just above the water.

Step 1: Set up the investigation

Fill the transparent tank with water.

Position the laser below the water surface and aim it towards a fixed point on the surface. The laser must be able to rotate without moving that point of incidence.

Draw a normal at the point where the beam meets the surface.

The angle of incidence must be measured inside the water, between the incoming beam and this normal.

A screen above the surface allows you to observe the transmitted beam.

Step 2: Predict what should happen

For water with refractive index \(1.33\), approaching air with refractive index \(1.00\):

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

\[ \boxed{\theta_c=48.75^\circ} \]

This is the predicted critical angle.

What should happen as the laser’s incidence angle increases?

The refracted ray should move further away from the normal and closer to the water surface.

At the critical angle, its direction should be parallel to the surface. Beyond that angle, the transmitted ray should disappear and total internal reflection should occur.

Step 3: Compare the predicted observations

The following values are illustrative calculations using Snell’s law, not claimed experimental measurements.

Notice how quickly the refracted angle approaches 90 degrees.

The beam at \(48^\circ\) is already travelling at more than 81 degrees to the normal. Increasing the incidence angle by less than one degree brings it to the limiting condition.

Step 4: Understand the experimental limitation

There is a problem with trying to observe the exact critical angle.

Suppose the screen is positioned 1.0 cm above the water surface.

At \(48^\circ\), the refracted ray makes an angle of approximately \(81.3^\circ\) with the normal. Its horizontal displacement before reaching the screen is:

\[ x=h\tan r \]

where \(h=0.010\text{ m}\) is the screen height above the surface.

Substituting:

\[ x=0.010\tan81.3^\circ \]

\[ \boxed{x\approx0.065\text{ m}} \]

The spot appears approximately 6.5 cm from the normal.

As the incidence angle approaches \(48.75^\circ\), the refracted ray becomes increasingly parallel to the surface. Its horizontal displacement increases dramatically.

At exactly \(90^\circ\), it cannot reach a screen at any finite height above the surface.

This is why the transmitted spot becomes difficult to locate near the critical angle.

A real experiment also has limitations:

• The laser beam has a finite width.
• A protractor has limited angular precision.
• Small ripples change the orientation of the water surface.
• The refractive index depends slightly on wavelength and temperature.

You therefore cannot claim to have measured an infinitely precise critical angle just because the spot disappears.

Instead, you can bracket the transition using measurements immediately below and above the predicted value.

There is another design issue worth noticing. If you initially shine the laser through the side of the tank, the beam refracts at that wall before reaching the water surface. The external angle you set is not necessarily the internal angle of incidence you need.

Using an underwater source with a suitably aligned output window avoids that extra refraction and gives better control of the angle being investigated.

06Questions and solutions

Question 1

A light ray travels from glass with refractive index \(1.50\) into air with refractive index \(1.00\).

Calculate the critical angle. Then describe what happens when the angle of incidence is exactly equal to this value.

Solution 1

The critical angle is \(41.8^\circ\). At exactly this angle, the limiting refracted ray travels parallel to the boundary, at \(90^\circ\) to the normal.

Start with the critical-angle equation:

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

Substitute:

\[ \sin\theta_c=\frac{1.00}{1.50} \]

Therefore:

\[ \theta_c=\sin^{-1}(0.6667) \]

\[ \boxed{\theta_c=41.8^\circ} \]

At exactly this incidence angle:

\[ i=\theta_c \]

and:

\[ r=90^\circ \]

The ray reaches the limiting refracted direction and travels along the glass-air boundary.

The tempting mistake is to call this total internal reflection because the ray no longer travels into the air at a finite angle.

However, the distinction is the equality condition. Total internal reflection occurs when the incidence angle is greater than the critical angle, not equal to it.

Question 2

A ray travels through water with refractive index \(1.33\) towards an air boundary with refractive index \(1.00\).

Its angle of incidence is \(48.0^\circ\), measured from the normal.

Calculate:

a) The critical angle.

b) The angle of refraction.

c) Whether total internal reflection occurs.

Solution 2

The critical angle is approximately \(48.8^\circ\), and the refracted angle is approximately \(81.3^\circ\). Total internal reflection does not occur.

Part a: Calculate the critical angle.

The ray travels from water into air:

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

Therefore:

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

\[ \boxed{\theta_c=48.8^\circ} \]

Part b: Calculate the refracted angle.

Use Snell’s law:

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

Substitute:

\[ 1.33\sin48.0^\circ=1.00\sin r \]

\[ \sin r=0.9884 \]

Taking the inverse sine:

\[ \boxed{r=81.3^\circ} \]

Part c: Determine whether total internal reflection occurs.

Compare the angles:

\[ 48.0^\circ<48.8^\circ \]

Therefore, the ray has not reached the critical angle.

\[ \boxed{\text{Refraction occurs, not total internal reflection.}} \]

Notice that the refracted angle is much larger than the incident angle. The ray is travelling very close to the surface, but it still enters the air.

A ray can look almost parallel to a boundary without having reached the exact critical-angle condition.

Question 3

A light ray travels from a transparent solid with refractive index \(1.52\) into a liquid with refractive index \(1.40\).

The incident ray makes an angle of \(24.0^\circ\) with the boundary surface.

Calculate the critical angle and determine the refracted angle. Explain why the ray does not undergo total internal reflection.

Solution 3

The ray refracts into the liquid at approximately \(82.7^\circ\) to the normal. Total internal reflection does not occur because the actual incidence angle is \(66.0^\circ\), which is below the critical angle of approximately \(67.1^\circ\).

Step 1: Convert the supplied angle into an angle of incidence.

The question gives the angle from the surface, but Snell’s law requires the angle from the normal.

Since the normal is perpendicular to the surface:

\[ i=90.0^\circ-24.0^\circ \]

\[ \boxed{i=66.0^\circ} \]

Step 2: Calculate the critical angle.

\[ \theta_c=\sin^{-1}\left(\frac{1.40}{1.52}\right) \]

\[ \boxed{\theta_c=67.1^\circ} \]

Step 3: Compare the angles.

\[ 66.0^\circ<67.1^\circ \]

Therefore, refraction occurs.

Step 4: Calculate the refracted angle.

Using Snell’s law:

\[ 1.52\sin66.0^\circ=1.40\sin r \]

Rearrange:

\[ \sin r=\frac{1.52\sin66.0^\circ}{1.40} \]

\[ \sin r\approx0.9918 \]

Therefore:

\[ \boxed{r\approx82.7^\circ} \]

The ray bends away from the normal and travels close to the boundary.

The tempting mistake is to use \(24^\circ\) as the angle of incidence. That would produce an incorrect comparison with the critical angle and an incorrect prediction of the ray’s path.

Always identify the normal before using an angle in Snell’s law.

Question 4

A student investigates the critical angle at a water-air boundary.

The water has refractive index \(1.33\), and the air has refractive index \(1.00\).

A horizontal screen is positioned \(1.0\text{ cm}\) above the water surface. It extends \(30\text{ cm}\) horizontally from the normal.

The student records the following illustrative observations:

The angle measurements have an uncertainty of approximately \( \pm0.05^\circ \).

Answer the following:

a) Calculate the theoretical critical angle.

b) Explain why the transmitted spot moves further from the normal as the incidence angle increases.

c) Explain what the disappearance of the transmitted spot at \(49.0^\circ\) indicates.

d) Can the student claim that the critical angle has been experimentally measured as exactly \(48.7^\circ\)? Explain.

Solution 4

The theoretical critical angle is approximately \(48.75^\circ\). The observations support a transition near this value, but they do not establish that \(48.7^\circ\) is the exact critical angle.

Part a: Calculate the theoretical critical angle.

Using:

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

Substitute:

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

\[ \boxed{\theta_c=48.75^\circ} \]

Part b: Explain the movement of the spot.

As the incidence angle increases towards the critical angle, the refracted angle also increases.

The ray therefore travels closer to the horizontal water surface.

The screen is only 1.0 cm above the surface, so a ray travelling almost horizontally must travel a much greater horizontal distance before reaching it.

This explains why the spot moves from 6.5 cm to 24.7 cm as the incidence angle increases.

Part c: Explain the missing transmitted spot.

At \(49.0^\circ\):

\[ 49.0^\circ>48.75^\circ \]

The incidence angle is greater than the critical angle.

The ray therefore undergoes total internal reflection instead of producing a propagating refracted ray in the air.

The reflected spot is consistent with this prediction.

Part d: Assess the experimental conclusion.

No. The student cannot claim that \(48.7^\circ\) is the exact critical angle.

There are several reasons:

• The calculated critical angle is approximately \(48.75^\circ\).
• The measured angle has an uncertainty of \( \pm0.05^\circ \).
• The laser has a finite beam width, so the transition is not a perfectly sharp visual event.
• The screen has a finite width, so a refracted ray could travel beyond its edge without being detected.

The experiment supports a transition close to \(48.75^\circ\), but the exact value is established more reliably from the refractive indices using Snell’s law.

This is an important experimental distinction: observing that a beam disappears is evidence of a transition, not proof of an infinitely precise angle.

Question 5

Challenge question

A light ray travels through glass with refractive index \(1.50\) towards an air boundary.

Its angle of incidence is exactly the critical angle for a glass-air boundary.

Now imagine replacing the air with a uniform water layer of refractive index \(1.33\). The water layer has a finite thickness, and its upper surface is parallel to the glass-water boundary. Air is above the water.

Assume:

• Both boundaries are perfectly parallel.
• The refractive indices remain constant.
• The ray follows geometrical optics.
• Absorption and interference effects can be ignored.

Answer the following:

a) Calculate the critical angle for the glass-water boundary. Does total internal reflection occur at this boundary?

b) Calculate the angle of refraction inside the water.

c) Determine what happens when the ray reaches the upper water-air boundary.

d) Now increase the original angle of incidence in the glass to \(42.0^\circ\). Will the ray escape into the air, or will it undergo total internal reflection somewhere in the layered system?

Solution 5

At the original incidence angle of \(41.81^\circ\), the ray refracts into the water and then reaches the water-air boundary at exactly its critical angle. It therefore travels along the upper boundary in the limiting ray model.

When the original incidence angle increases to \(42.0^\circ\), the ray enters the water but undergoes total internal reflection at the upper water-air boundary.

The important detail is that the critical angle must be recalculated for each boundary. The refractive index of the intermediate water layer changes the first refraction, but the second boundary introduces another critical-angle condition.

Part a: Calculate the glass-water critical angle.

At the first boundary:

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

Therefore:

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

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

The original incidence angle is \(41.81^\circ\).

Compare:

\[ 41.81^\circ<62.5^\circ \]

Therefore, total internal reflection does not occur at the glass-water boundary.

The ray refracts into the water.

Part b: Calculate the angle inside the water.

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

\[ 1.50\sin41.81^\circ=1.33\sin r_w \]

Rearrange:

\[ \sin r_w= \frac{1.50\sin41.81^\circ}{1.33} \]

Remember that \(41.81^\circ\) is the glass-air critical angle.

Therefore:

\[ 1.50\sin41.81^\circ\approx1.00 \]

So:

\[ \sin r_w=\frac{1.00}{1.33} \]

Taking the inverse sine:

\[ \boxed{r_w=48.75^\circ} \]

Part c: Determine what happens at the water-air boundary.

This is where the problem becomes interesting.

Calculate the critical angle for water-air:

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

\[ \boxed{\theta_{c,wa}=48.75^\circ} \]

The ray enters the water at \(48.75^\circ\).

Because the two surfaces are parallel, their normals are also parallel. The ray therefore reaches the upper boundary at the same angle:

\[ i_{wa}=48.75^\circ \]

Thus:

\[ i_{wa}=\theta_{c,wa} \]

The ray reaches exactly the critical-angle condition at the upper boundary.

In the geometrical ray model, the limiting refracted direction is parallel to the water-air surface.

It does not emerge into the air at a finite angle.

Part d: Increase the original incidence angle to \(42.0^\circ\).

The first boundary is still glass-water.

Its critical angle remains \(62.5^\circ\), so:

\[ 42.0^\circ<62.5^\circ \]

The ray still refracts into the water.

Calculate its new angle:

\[ 1.50\sin42.0^\circ=1.33\sin r_w \]

Rearrange:

\[ \sin r_w= \frac{1.50\sin42.0^\circ}{1.33} \]

\[ \sin r_w\approx0.7547 \]

Therefore:

\[ \boxed{r_w\approx49.0^\circ} \]

The two surfaces are parallel, so the incidence angle at the upper water-air boundary is also approximately \(49.0^\circ\).

Compare this with the water-air critical angle:

\[ 49.0^\circ>48.75^\circ \]

Therefore:

\[ \boxed{\text{Total internal reflection occurs at the water-air boundary.}} \]

The ray does not escape into the air.

The tempting first answer is that replacing air with water at the original glass boundary raises the critical angle, allowing the ray to escape.

That reasoning only considers the first boundary.

The ray still has to cross the second boundary. The angle it acquires inside the water determines whether it can enter the air.

There is a deeper relationship here. For parallel interfaces, Snell’s law preserves the relationship between the refractive indices and the ray’s direction across the layers. At the original glass-air critical angle, the ray arrives at the final water-air boundary at its own critical angle.

The intermediate layer changes the direction at each individual boundary, but it does not remove the overall limiting condition.

07Where to go next

The most useful habit to take from this is to treat the critical angle as an exact boundary condition, not as a synonym for total internal reflection.

Remember the three cases:

\[ \begin{aligned} i<\theta_c &: \text{refraction}\\ i=\theta_c &: \text{grazing refracted direction}\\ i>\theta_c &: \text{total internal reflection} \end{aligned} \]

Once a ray reaches another surface, draw a new normal and assess the new pair of materials. The original critical angle does not automatically apply to the next boundary.

For the broader derivation and conditions for total internal reflection, read Critical Angle and Total Internal Reflection for HSC Physics.

For more complicated paths involving multiple surfaces, continue with How to Decide if a Ray Refracts or Reflects at a Second Boundary.