Why a Ray Can Change Speed Without Changing Direction at a Boundary

A laser enters a glass block at right angles to its surface. It continues travelling in a straight line, but its speed decreases. How can a boundary change the speed of light without changing its direction?

The key is that speed and direction are separate properties of motion. At normal incidence, light travels along the normal to the boundary. Its speed changes when it enters the new medium, but there is no sideways deflection to make the ray turn.

Predict what happens if the same laser approaches the glass at an angle instead. It will generally bend towards the normal as it slows down. The difference comes from the angle at which it meets the boundary, not from whether its speed changes.

01Start with the boundary and the normal

Before calculating anything, get the geometry right.

A boundary is the surface separating two different optical media. For example, when light travels from air into glass, the air-glass interface is the boundary.

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

Three terms describe the ray’s path:

• Incident ray: the incoming ray approaching the boundary.
• Refracted ray: the transmitted ray travelling into the second medium.
• Angle of incidence, \(\theta_1\): the angle between the incident ray and the normal.

The angle between the refracted ray and the normal is called the angle of refraction, \(\theta_2\).

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

This is particularly important at normal incidence.

A ray travelling at \(90^\circ\) to the surface is travelling at \(0^\circ\) to the normal.

Therefore:

\[
\theta_1=0^\circ
\]

At normal incidence, the incident ray and the normal occupy the same line.

02Why does the ray stay straight when its speed changes?

Imagine light travelling from air into glass.

For this example, use the following illustrative refractive indices:

The refractive index tells us how the speed of light in a medium compares with its speed in a vacuum.

The relationship is:

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

Where:

• \(v\) is the speed of light in the medium, measured in \(\mathrm{m\,s^{-1}}\).
• \(c\) is the speed of light in a vacuum, approximately \(3.00\times10^8\ \mathrm{m\,s^{-1}}\).
• \(n\) is the dimensionless refractive index of the medium.

For air:

\[
v_{\text{air}}\approx 3.00\times10^8
\ \mathrm{m\,s^{-1}}
\]

For glass:

\[
v_{\text{glass}}
=\frac{3.00\times10^8}{1.50}
\]

\[
=2.00\times10^8\ \mathrm{m\,s^{-1}}
\]

The light travels one-third slower in glass than in air.

Yet its direction does not change.

Why?

The answer becomes clearer when we consider the wavefront.

The wavefront explanation

A wavefront is a line joining points on a wave that are at the same stage of oscillation.

The direction of wave travel is perpendicular to the wavefront.

When light approaches a boundary at normal incidence, its wavefronts are parallel to that boundary.

This means the entire wavefront reaches the new medium at the same time.

Every point along the wavefront experiences the change in speed together. There is no part of the wavefront that enters first and gets a head start on the rest.

Consequently, the wavefront does not rotate.

Its spacing changes because the speed changes, but its orientation stays the same. Since the ray remains perpendicular to the wavefront, its direction also stays the same.

A useful way to picture this is a row of cyclists crossing from a smooth path onto grass.

If the entire row reaches the grass at the same time, everyone slows together. The row keeps its orientation.

If the row approaches diagonally, one cyclist reaches the grass first and slows while the others are still moving faster. The row begins to rotate.

The cyclists represent points along a wavefront, and the change in their forward speed represents the change in wave speed.

The analogy has limits: a wavefront is not a rigid row of people. It is a line of equal phase. However, the analogy captures the important difference between simultaneous and staggered entry.

What does Snell’s law predict?

We can express the same result mathematically using Snell’s law:

\[
n_1\sin\theta_1=n_2\sin\theta_2
\]

Where:

• \(n_1\) is the refractive index of the initial medium.
• \(n_2\) is the refractive index of the second medium.
• \(\theta_1\) is the angle of incidence.
• \(\theta_2\) is the angle of refraction.

At normal incidence:

\[
n_1\sin0^\circ=n_2\sin\theta_2
\]

Since \(\sin0^\circ=0\):

\[
0=n_2\sin\theta_2
\]

Therefore:

\[
\boxed{\theta_2=0^\circ}
\]

The refracted ray remains along the normal, regardless of whether the second medium has a higher or lower refractive index.

Notice what Snell’s law does not say: it does not say the speed stays constant.

The refractive indices can be different, so the speed can change even though both angles are zero.

03Speed, wavelength and direction are not the same thing

There is another useful detail to understand.

When light crosses a stationary boundary, its frequency remains constant.

The wave equation is:

\[
v=f\lambda
\]

Where:

• \(v\) is wave speed in \(\mathrm{m\,s^{-1}}\).
• \(f\) is frequency in hertz (Hz).
• \(\lambda\) is wavelength in metres (m).

If the speed decreases while frequency stays constant, the wavelength must also decrease.

For example, suppose light travels from air into glass with:

\[
v_1=3.00\times10^8\ \mathrm{m\,s^{-1}}
\]

\[
v_2=2.00\times10^8\ \mathrm{m\,s^{-1}}
\]

The wavelength in glass becomes two-thirds of its original value.

So three things are happening:

1. The speed changes.
2. The wavelength changes.
3. The direction stays the same.

This is perfectly consistent with wave behaviour.

One common mistake is assuming that a wavelength change must produce a direction change. It doesn’t. The wavefront can become more closely spaced without rotating.

04How could you investigate this experimentally?

Suppose you wanted to demonstrate that light changes speed without changing direction.

There are two separate things to establish:

• The ray does not change direction at normal incidence.
• The speed inside the material is different from its speed in air.

A simple ray-tracing experiment can establish the first. It cannot establish the second by itself.

The initial experiment: trace the ray

A suitable setup uses a ray box, a narrow slit, a sheet of paper, and a rectangular transparent glass block.

First, place the glass block on the paper and trace its outline.

Draw a normal at the midpoint of one flat face. Position the ray box so that its narrow beam travels directly along this normal.

The beam approaches the air-glass boundary at \(0^\circ\) to the normal.

Predict what happens.

The transmitted ray should continue along the same straight line inside the glass. It should not bend towards or away from the normal.

Tracing the incoming and transmitted paths allows you to compare their directions.

However, there is a problem.

A straight ray does not prove that its speed is unchanged.

The experiment measures direction, not travel time.

A ray can travel at different speeds along the same straight path.

Revising the experiment: measure refractive index

To establish the speed difference, we need another measurement.

One practical approach uses a semicircular glass block.

Direct a narrow beam into the curved surface along a radius passing through the centre of curvature. The ray meets the curved surface normally, so it enters without changing direction.

It then reaches the flat glass-air boundary at an angle.

Measure the angle inside the glass and the angle outside in air. Both must be measured from the normal to the flat face.

Using Snell’s law:

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

Since \(n_a\approx1.00\), we can determine the refractive index of the glass.

Once we know \(n_g\), we calculate its light speed using:

\[
v_g=\frac{c}{n_g}
\]

This is an indirect measurement of speed: we determine refractive index from angles, then calculate speed.

Why not simply time the light through the glass?

The difficulty is the extremely short travel time.

Consider light travelling through 10 cm of air compared with 10 cm of glass, using \(n_{\text{glass}}=1.50\).

The time taken is:

\[
t=\frac{d}{v}
\]

For air:

\[
t_{\text{air}}
=\frac{0.10}{3.00\times10^8}
=3.33\times10^{-10}\ \mathrm{s}
\]

For glass:

\[
t_{\text{glass}}
=\frac{0.10}{2.00\times10^8}
=5.00\times10^{-10}\ \mathrm{s}
\]

The difference is only:

\[
\Delta t=1.67\times10^{-10}\ \mathrm{s}
\]

That is approximately \(0.17\) nanoseconds.

A normal school stopwatch cannot measure this difference. The angle-based method is much more practical for a school investigation.

The important experimental conclusion is that the straight-line observation and the refractive-index measurement answer different questions. You need both to establish the complete result.

05Worked examples

06The contrasting case: when the ray does bend

Now consider the same air-glass boundary, but with an incident angle of \(40^\circ\).

The refractive indices remain:

\[
n_{\text{air}}=1.00
\]

\[
n_{\text{glass}}=1.50
\]

This time:

\[
1.00\sin40^\circ
=1.50\sin\theta_2
\]

Rearranging:

\[
\sin\theta_2
=\frac{\sin40^\circ}{1.50}
\]

\[
\theta_2\approx25.4^\circ
\]

The ray bends towards the normal.

Compare the two situations:

The important distinction is that a change in speed does not automatically produce a change in direction.

For more detail on the general relationship between wave speed, wavelength, and direction, see Wave Refraction: Speed, Wavelength and Direction Changes.

07Questions and solutions

Question 1

A laser travels from air into water at normal incidence.

The refractive index of air is 1.00, and the refractive index of water is 1.33. The laser has a frequency of \(4.50\times10^{14}\ \mathrm{Hz}\).

Calculate:

a) The speed of light in water.

b) The wavelength of light in water.

c) The angle of refraction.

Solution 1

The transmitted ray remains straight, travels at approximately \(2.26\times10^8\ \mathrm{m\,s^{-1}}\), and has a wavelength of approximately 502 nm.

Part a: Calculate the speed.

Use:

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

Substitute:

\[
v=\frac{3.00\times10^8}{1.33}
\]

\[
\boxed{v=2.26\times10^8\ \mathrm{m\,s^{-1}}}
\]

Part b: Calculate the wavelength.

Use:

\[
\lambda=\frac{v}{f}
\]

Substitute:

\[
\lambda=
\frac{2.26\times10^8}{4.50\times10^{14}}
\]

\[
\boxed{\lambda=5.02\times10^{-7}\ \mathrm{m}}
\]

Or:

\[
\boxed{\lambda=502\ \mathrm{nm}}
\]

The wavelength decreases because the speed decreases while the frequency remains constant.

Part c: Calculate the angle of refraction.

Using Snell’s law:

\[
1.00\sin0^\circ
=1.33\sin\theta_2
\]

Therefore:

\[
\boxed{\theta_2=0^\circ}
\]

The ray changes speed and wavelength, but not direction.

Question 2

A ray travels from water into acrylic glass.

The refractive indices are:

\[
n_{\text{water}}=1.33
\]

\[
n_{\text{acrylic}}=1.49
\]

The incident angle is \(40^\circ\), measured from the normal.

Calculate:

a) The angle of refraction.

b) The speed of light in each medium.

c) Whether the ray bends towards or away from the normal.

d) What the refracted angle would be if the ray instead approached the same boundary at normal incidence.

Solution 2

The ray refracts at approximately \(35.0^\circ\), bends towards the normal, and slows down from approximately \(2.26\times10^8\) to \(2.01\times10^8\ \mathrm{m\,s^{-1}}\).

Part a: Calculate the angle of refraction.

Use Snell’s law:

\[
n_1\sin\theta_1=n_2\sin\theta_2
\]

Substitute:

\[
1.33\sin40^\circ
=1.49\sin\theta_2
\]

Rearrange:

\[
\sin\theta_2
=\frac{1.33\sin40^\circ}{1.49}
\]

\[
\theta_2=\sin^{-1}(0.5738)
\]

\[
\boxed{\theta_2=35.0^\circ}
\]

Part b: Calculate the speed in each medium.

For water:

\[
v_w=\frac{3.00\times10^8}{1.33}
\]

\[
\boxed{v_w=2.26\times10^8\ \mathrm{m\,s^{-1}}}
\]

For acrylic:

\[
v_a=\frac{3.00\times10^8}{1.49}
\]

\[
\boxed{v_a=2.01\times10^8\ \mathrm{m\,s^{-1}}}
\]

Part c: Determine the direction of bending.

The ray enters a medium with a higher refractive index.

Its speed decreases, and its refracted angle decreases from \(40^\circ\) to \(35.0^\circ\).

Therefore, it bends towards the normal.

Part d: Consider normal incidence.

If the incident angle were \(0^\circ\), Snell’s law would give:

\[
1.33\sin0^\circ
=1.49\sin\theta_2
\]

Therefore:

\[
\boxed{\theta_2=0^\circ}
\]

The speed would still change to \(2.01\times10^8\ \mathrm{m\,s^{-1}}\).

This demonstrates that the same pair of media can produce a speed change with or without a change in direction, depending on the incident angle.

Question 3

Two unknown transparent materials, X and Y, are investigated using the same monochromatic light source.

Light travels from air into each material at an incident angle of \(50^\circ\).

The following illustrative experimental measurements are obtained:

The refractive index of air is 1.00.

The light source has a frequency of \(5.00\times10^{14}\ \mathrm{Hz}\).

a) Calculate the refractive index of each material.

b) Calculate the speed of light in each material.

c) Calculate the wavelength in each material.

d) If the light entered both materials at normal incidence, would the two transmitted rays travel in different directions? Explain.

Solution 3

Material X has the higher refractive index and the lower light speed. Both materials would nevertheless transmit a normally incident ray without changing its direction.

Part a: Calculate the refractive indices.

For material X:

\[
n_X\sin50^\circ
=1.00\sin30^\circ
\]

Rearranging:

\[
n_X=\frac{\sin50^\circ}{\sin30^\circ}
\]

\[
n_X=\frac{0.7660}{0.5000}
\]

\[
\boxed{n_X=1.53}
\]

For material Y:

\[
n_Y\sin50^\circ
=1.00\sin36^\circ
\]

Rearranging:

\[
n_Y=\frac{\sin50^\circ}{\sin36^\circ}
\]

\[
n_Y=\frac{0.7660}{0.5878}
\]

\[
\boxed{n_Y=1.30}
\]

Material X has the higher refractive index.

Part b: Calculate the speeds.

For material X:

\[
v_X=\frac{3.00\times10^8}{1.53}
\]

\[
\boxed{v_X\approx1.96\times10^8\ \mathrm{m\,s^{-1}}}
\]

For material Y:

\[
v_Y=\frac{3.00\times10^8}{1.30}
\]

\[
\boxed{v_Y\approx2.30\times10^8\ \mathrm{m\,s^{-1}}}
\]

Light travels more slowly in material X.

Part c: Calculate the wavelengths.

Use:

\[
\lambda=\frac{v}{f}
\]

For material X:

\[
\lambda_X=
\frac{1.96\times10^8}{5.00\times10^{14}}
\]

\[
\boxed{\lambda_X=392\ \mathrm{nm}}
\]

For material Y:

\[
\lambda_Y=
\frac{2.30\times10^8}{5.00\times10^{14}}
\]

\[
\boxed{\lambda_Y=460\ \mathrm{nm}}
\]

The frequency is the same in both materials, but the wavelength is shorter in material X because its light speed is lower.

Part d: Consider normal incidence.

Both transmitted rays would travel along their respective normals.

Therefore, neither ray would change direction.

However, their speeds would still be different:

• Material X: approximately \(1.96\times10^8\ \mathrm{m\,s^{-1}}\).
• Material Y: approximately \(2.30\times10^8\ \mathrm{m\,s^{-1}}\).

A straight ray does not tell you that two materials have the same refractive index.

Question 4

A student directs a laser at normal incidence onto a transparent plastic block.

They observe that the transmitted ray continues in a straight line and conclude:

“The refractive index of the plastic must be 1.00 because the ray does not bend.”

A second investigation uses a semicircular block made from the same plastic.

The ray enters the curved surface along a radius and reaches the flat surface from inside the plastic. At the flat surface:

• The angle of incidence inside the plastic is \(25^\circ\).
• The angle of refraction into air is \(40^\circ\).

Assume the refractive index of air is 1.00.

a) Explain why the student’s original conclusion is incorrect.

b) Calculate the refractive index of the plastic.

c) Calculate the speed of light in the plastic.

d) Explain why the two experiments are consistent with each other.

Solution 4

The student’s conclusion is incorrect because zero angular deviation does not mean the refractive indices are equal. The second investigation gives a plastic refractive index of approximately 1.52 and a light speed of approximately \(1.97\times10^8\ \mathrm{m\,s^{-1}}\).

Part a: Evaluate the original conclusion.

At normal incidence:

\[
\theta_1=\theta_2=0^\circ
\]

Snell’s law becomes:

\[
n_1\sin0^\circ=n_2\sin0^\circ
\]

This simplifies to:

\[
0=0
\]

The equation is true for any positive refractive index.

It therefore provides no information about the value of the plastic’s refractive index.

The student has confused unchanged direction with unchanged speed.

Part b: Calculate the refractive index.

In the second experiment, light travels from plastic into air.

Apply Snell’s law:

\[
n_p\sin25^\circ
=1.00\sin40^\circ
\]

Rearrange:

\[
n_p=\frac{\sin40^\circ}{\sin25^\circ}
\]

\[
n_p=\frac{0.6428}{0.4226}
\]

\[
\boxed{n_p=1.52}
\]

Part c: Calculate the speed.

\[
v_p=\frac{c}{n_p}
\]

\[
v_p=\frac{3.00\times10^8}{1.52}
\]

\[
\boxed{v_p=1.97\times10^8\ \mathrm{m\,s^{-1}}}
\]

Part d: Explain why the results agree.

The first experiment establishes that light does not change direction when it enters the plastic at normal incidence.

The second experiment establishes that the plastic has a refractive index greater than 1.00 and therefore reduces the speed of light compared with air.

These results are entirely consistent.

The first experiment measures the geometry of the transmitted ray. The second provides information about the material’s refractive index.

Question 5

A ray of monochromatic light travels from air into a transparent glass prism.

The prism is positioned so that the ray strikes its first face at normal incidence. Its first face is therefore perpendicular to the incoming ray.

The ray then reaches a second face separating the glass from a layer of water.

The geometry and material properties are:

• Refractive index of air: \(n_a=1.00\).
• Refractive index of glass: \(n_g=1.60\).
• Refractive index of water: \(n_w=1.33\).
• The normal to the second face is tilted by \(40^\circ\) relative to the ray travelling inside the glass.
• The outer water-air boundary is parallel to the glass-water boundary.

Assume stationary, smooth boundaries, monochromatic light, and negligible partial reflections at the first two boundaries. Consider the main transmitted ray.

Determine:

a) The speed of light inside the glass immediately after the first boundary.

b) The angle of incidence at the glass-water boundary.

c) The angle of refraction into water.

d) Whether the ray emerges from the water into air or undergoes total internal reflection.

e) Whether the speed or direction changes at the water-air boundary.

Solution 5

The ray initially changes speed without changing direction. It subsequently refracts into water, but undergoes total internal reflection at the water-air boundary. At that final boundary, its direction changes while its speed remains constant.

This problem tests whether you can track the normal at each boundary rather than assuming that the original direction remains normal to every surface.

Part a: Calculate the speed in glass.

At the first boundary, the ray travels from air into glass.

Use:

\[
v_g=\frac{c}{n_g}
\]

Substitute:

\[
v_g=\frac{3.00\times10^8}{1.60}
\]

\[
\boxed{v_g=1.875\times10^8\ \mathrm{m\,s^{-1}}}
\]

The speed decreases as the ray enters glass.

However, because the ray enters at normal incidence, it continues in the same direction.

Part b: Determine the incidence angle at the glass-water boundary.

The second boundary has a different orientation from the first.

Its normal is tilted by \(40^\circ\) relative to the ray travelling inside the glass.

Therefore:

\[
\boxed{\theta_g=40^\circ}
\]

The important point is that normal incidence at the first face does not guarantee normal incidence at the second face.

The normal is defined separately at each boundary.

Part c: Calculate the refracted angle in water.

Apply Snell’s law:

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

Substitute:

\[
1.60\sin40^\circ
=1.33\sin\theta_w
\]

Rearrange:

\[
\sin\theta_w
=\frac{1.60\sin40^\circ}{1.33}
\]

\[
\sin\theta_w\approx0.773
\]

Therefore:

\[
\boxed{\theta_w\approx50.6^\circ}
\]

The ray bends away from the normal because it enters a medium with a lower refractive index.

Part d: Determine whether total internal reflection occurs.

The ray now travels from water towards air.

We must compare its incidence angle with the critical angle for the water-air boundary.

The critical angle is:

\[
\sin\theta_c=\frac{n_{\text{air}}}{n_{\text{water}}}
\]

Substitute:

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

\[
\theta_c=\sin^{-1}(0.7519)
\]

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

The water-air boundary is parallel to the glass-water boundary.

Therefore, its normal is also parallel to the previous normal, and the ray reaches it at the same angle:

\[
\theta_i=50.6^\circ
\]

Compare:

\[
50.6^\circ>48.8^\circ
\]

The incidence angle exceeds the critical angle.

Therefore:

\[
\boxed{\text{Total internal reflection occurs.}}
\]

The ray does not emerge into air.

Part e: Explain what happens to speed and direction.

At the water-air boundary, total internal reflection occurs.

The reflected ray remains inside water.

Its speed therefore remains:

\[
v_w=\frac{3.00\times10^8}{1.33}
\]

\[
\boxed{v_w=2.26\times10^8\ \mathrm{m\,s^{-1}}}
\]

However, its direction changes because it reflects from the boundary.

This is the reverse of what happened at the first boundary:

The first boundary changes the speed without changing direction. The final boundary changes direction without changing the speed of the reflected ray.

That is why you should never assume that a change in speed and a change in direction must happen together.

The next useful step is learning how to decide whether a ray refracts or undergoes total internal reflection when it reaches another boundary. That requires identifying the new normal, the new medium pair, and the critical angle.

See How to Find the Critical Angle Between Two Named Media.