Why Light Slows in a Medium but Not in a Vacuum

In 1850, physicists faced a useful test of two competing models of light: when electromagnetic radiation enters water, does it speed up or slow down? The answer would help distinguish the wave model from the simple particle model of light being debated at the time.

The experiment showed that light travels more slowly in water than in air. We now understand why: the electric field of an electromagnetic wave interacts with charged particles in the material, changing how quickly the wave’s phase propagates. In a vacuum, there are no material particles to respond to the field, so electromagnetic radiation travels at its vacuum speed, \(c=3.00\times10^8\ \mathrm{m\,s^{-1}}\).

There is an important catch. Light slowing down in glass does not mean that its photons necessarily lose energy, and it does not mean that fewer photons must emerge. Speed, wavelength, photon energy, and photon count are separate quantities.

01Foucault’s experiment: does light travel slower in water?

Before we explain the modern physics, consider how scientists established that light actually travels at different speeds in different materials.

During the nineteenth century, scientists were still debating two descriptions of light:

• The wave model predicted that light would travel more slowly in water than in air.
• The simple Newtonian corpuscular model predicted that light would travel faster in water than in air.

The disagreement gave scientists a clear experimental question.

Could they compare the travel times of light through air and water?

How the experiment worked

French physicist Léon Foucault developed an experiment in 1850 using a rapidly rotating mirror.

The arrangement worked as follows:

1. A light source produced a narrow beam, which travelled towards a small rotating mirror.
2. The rotating mirror directed light along two comparable optical paths. One passed through air, while the other passed through a water-filled tube.
3. At the far end of each path, fixed mirrors reflected the light back towards the rotating mirror.
4. The returning light reflected from the rotating mirror again and travelled towards an observation point.
5. The observer compared the positions of the returned images.

The important detail is that the light had to travel to the fixed mirrors and back. During that time, the rotating mirror continued turning.

What did Foucault predict?

Imagine the rotating mirror turning at a constant rate.

If light takes longer to complete its journey, the mirror has more time to rotate before the light returns.

The returning light therefore strikes the mirror at a slightly different orientation, changing the direction of the reflected beam. The longer the travel time, the greater the displacement of the returned image.

Predict which image should move further: the air-path image or the water-path image?

If the wave model is correct, the water-path image should be displaced further because light travels more slowly through water.

What did he observe?

Foucault observed that the water-path image was displaced further than the air-path image.

The result was consistent with light taking longer to complete the journey through water. At the same mirror rotation rate, the water-path image showed a greater displacement.

Foucault successfully observed the two images simultaneously on 27 April 1850. The recorded displacements were very small, reaching at most approximately \(0.3\ \mathrm{mm}\) at mirror rotation rates between 600 and 800 revolutions per second.

This supported the prediction that light travels more slowly in water than in air. It contradicted the simple Newtonian corpuscular prediction being tested. \(\cite{turn467615view1}\)

Why did the apparatus need improving?

Getting a clear comparison was not straightforward.

The rotating mirror had to remain stable at extremely high speeds. An earlier geared driving arrangement proved unreliable, so Foucault developed a steam-turbine drive. He also worked on balancing the mirror and reducing irregularities in its rotation.

There was another problem: the air-path image was brighter than the water-path image. Its surrounding light made the weaker water-path image difficult to distinguish.

Foucault introduced a narrow slit into the air-path optical arrangement to separate the images more clearly.

These changes allowed him to compare the two images at the same mirror rotation rate.

However, his 1850 experiment was primarily comparative. His estimate of the mirror’s rotation rate was not accurate enough to produce reliable absolute values for the speeds in air and water.

In 1862, Foucault improved the rotating-mirror technique, extending the optical path and using a more reliable method of measuring the mirror’s rotation. He obtained an improved absolute measurement of the speed of light in air. This was a refinement of the measurement technique, not a precise measurement of the water speed in his original comparison. \(\cite{turn440043view2}\)

What did the experiment actually establish?

The historical result established that light travels more slowly in water than in air.

At the time, this was strong evidence in favour of the wave model over the simple Newtonian particle model.

Modern hindsight adds an important qualification: the experiment did not establish that light has no particle properties. Modern physics describes electromagnetic radiation using both wave behaviour and photons.

Foucault measured the propagation of light. The experiment did not directly investigate the energy of individual photons, which became a separate question in the development of quantum physics.

Worked scale estimate: how much longer does light take in water?

Suppose light travels through a \(3.00\ \mathrm{m}\) section of water.

Use an illustrative refractive index of \(n=1.333\), and assume the medium is stationary and transparent.

First, calculate the speed of light in water:

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

\[ v=\frac{3.00\times10^8}{1.333} \]

\[ v=2.25\times10^8\ \mathrm{m\,s^{-1}} \]

Now calculate the travel time through the water:

\[ t_{\mathrm{water}}=\frac{d}{v} \]

\[ t_{\mathrm{water}} =\frac{3.00}{2.25\times10^8} \]

\[ t_{\mathrm{water}}=1.33\times10^{-8}\ \mathrm{s} \]

Therefore:

\[ t_{\mathrm{water}}=13.3\ \mathrm{ns} \]

In a vacuum, the same distance takes:

\[ t_{\mathrm{vacuum}} =\frac{3.00}{3.00\times10^8} =10.0\ \mathrm{ns} \]

The additional travel time is:

\[ \Delta t=13.3-10.0 \]

\[ \boxed{\Delta t=3.3\ \mathrm{ns}} \]

The light takes approximately \(3.3\ \mathrm{ns}\) longer to travel through the water.

This is a scale estimate using modern refractive-index data, not a numerical result obtained directly by Foucault.

02Why does electromagnetic radiation slow down in a material?

The historical experiment established what happens. We now need to understand why.

The key is the interaction between the electromagnetic wave and the charged particles inside the material.

1. The electric field interacts with bound electrons

Glass and water contain atoms and molecules with charged particles.

The electrons are negatively charged, while the nuclei contain positively charged protons.

When an electromagnetic wave enters the material, its oscillating electric field exerts a force on these charges.

The electrons are bound to their atoms, so they cannot simply travel away. Instead, their electron clouds undergo small oscillations relative to the positive nuclei.

This creates tiny electric dipoles.

The process is called electric polarisation of the material.

Be careful: this is not the same as the polarisation direction of an electromagnetic wave.

2. The material produces its own electromagnetic response

The oscillating electric dipoles produce secondary electromagnetic fields.

These fields combine with the original electromagnetic wave through superposition.

This is the crucial part.

The combined electromagnetic field propagates through the material with a different phase progression from a wave travelling through a vacuum.

For ordinary transparent materials at visible-light frequencies, the resulting wave has a lower phase velocity.

In other words, its crests and troughs advance more slowly through the material.

The material’s electromagnetic response is described by its refractive index.

A useful mental model is to imagine each bound electron cloud as a tiny spring-mounted object. The incoming electric field drives its oscillation, and that response contributes to the overall electromagnetic field.

The analogy has limits. Light is not being passed from atom to atom like a message in a queue. The incoming field and the induced responses combine coherently to produce the propagating wave.

3. Why doesn’t the same thing happen in a vacuum?

A vacuum contains no material atoms or bound electrons to polarise.

An electromagnetic wave can still propagate because changing electric and magnetic fields are linked by Maxwell’s equations. The fields sustain the electromagnetic disturbance without requiring a material medium.

There is therefore no material polarisation response to modify the wave’s propagation.

Its speed is:

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

This is true for all electromagnetic frequencies in a vacuum.

Radio waves, microwaves, visible light, ultraviolet radiation, and X-rays all travel at the same vacuum speed.

Their frequencies, wavelengths, and photon energies differ, but their vacuum speeds do not.

A common misconception: photons repeatedly stop at atoms

You might picture a photon travelling at \(c\), hitting an atom, being absorbed, and then being re-emitted. Repeat this millions of times, and the average speed becomes lower.

It sounds reasonable, but it is not an accurate explanation of ordinary refraction in a transparent material.

The reduction in wave speed is a consequence of the coherent interaction between the electromagnetic field and the material’s charged particles.

It is not simply a series of separate absorption-and-release events.

A transparent material can interact with light without significantly absorbing its energy.

That distinction will become important when we examine photon energy and photon count.

03Refractive index: putting a number on the speed reduction

We describe the effect of a material on light using its refractive index, \(n\).

The refractive index is defined as:

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

where:

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

Rearranging:

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

For ordinary transparent materials at visible wavelengths, \(n\) is greater than 1.

This means the speed of light in the material is lower than its vacuum speed.

Comparing different materials

The following are representative refractive indices for light with a vacuum wavelength close to \(589\ \mathrm{nm}\).

These values are approximate. The refractive index depends slightly on wavelength, temperature, and the particular material. \(\cite{turn467615view2}\)

Notice that air has a refractive index extremely close to 1. Light therefore travels through air at almost its vacuum speed.

Glass has a higher refractive index, so its phase velocity is lower.

The important point is that the speed reduction depends on the material’s electromagnetic response, not simply how physically dense it is.

04What happens to photon energy and photon count?

This is where students often make an incorrect prediction.

If light travels more slowly in glass, does each photon lose energy?

Does the material automatically reduce the number of photons?

Neither conclusion follows from the reduction in speed.

We need to consider photon energy and photon count separately.

Photon energy stays the same

The energy of an individual photon is:

\[ \boxed{E_\gamma=hf} \]

where:

• \(E_\gamma\) is the energy of one photon, in joules (J).
• \(h\) is Planck’s constant, \(6.63\times10^{-34}\ \mathrm{J\,s}\).
• \(f\) is the frequency, in hertz (Hz).

When light enters an ordinary stationary transparent medium, its frequency remains unchanged.

Therefore:

\[ E_{\gamma,\mathrm{before}} = E_{\gamma,\mathrm{after}} \]

The energy of each transmitted photon stays the same.

This is true even though the wavelength decreases.

To see why, start with the vacuum relationship:

\[ c=f\lambda_0 \]

Substitute into the photon energy equation:

\[ E_\gamma=hf \]

\[ \boxed{E_\gamma=\frac{hc}{\lambda_0}} \]

Here, \(\lambda_0\) is the wavelength in a vacuum.

Notice that this equation uses the vacuum wavelength, not the wavelength inside the material.

A common mistake is to substitute the shorter in-medium wavelength into \(E=hc/\lambda\). That incorrectly predicts an increase in photon energy.

Inside a medium, the correct relationships are:

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

\[ \lambda_{\mathrm{medium}}=\frac{\lambda_0}{n} \]

\[ E_\gamma=hf \]

The photon energy has not increased. The wave’s phase velocity and wavelength have changed.

Photon count is a separate quantity

Now consider the number of photons.

Imagine a laser emitting \(1.0\times10^{18}\) photons every second into a perfectly transparent, lossless material.

If we ignore reflection at the boundary, the same number of photons per second passes through the material and emerges from the other side.

The slower propagation speed does not automatically destroy photons.

However, real materials can introduce losses.

For example:

• Reflection at the entrance surface sends some photons away from the transmitted beam.
• Absorption transfers energy from photons to the material.
• Scattering can redirect photons away from the detector.

These processes can reduce the number of photons reaching a detector.

They are separate from the change in propagation speed.

The quantities can be summarised as follows:

The key distinction is that photon energy describes the energy of each photon, while photon count describes how many photons are present or detected.

If you need to calculate the optical power from a photon emission rate, see How to Calculate Light Power from Photon Emission Rate.

05Why the in-medium wavelength matters in real applications

A shorter wavelength inside a material is not just a mathematical consequence of the wave equation.

It affects how electromagnetic radiation behaves in optical systems.

Two useful examples are optical fibres and prisms.

Optical fibres

Optical fibres guide light through a transparent core surrounded by cladding with a lower refractive index.

The difference in refractive index allows light to be guided by total internal reflection.

The wavelength inside the fibre core is shorter than the wavelength in a vacuum.

For example, suppose light has a vacuum wavelength of \(1550\ \mathrm{nm}\) and travels through a fibre core with an illustrative refractive index of \(1.45\).

Its wavelength inside the core is:

\[ \lambda_{\mathrm{core}}=\frac{1550}{1.45} \]

\[ \boxed{\lambda_{\mathrm{core}}\approx1069\ \mathrm{nm}} \]

The photon energy is still determined by its frequency, or equivalently its vacuum wavelength.

The in-medium wavelength matters when considering the phase accumulated along the fibre and the electromagnetic modes that can propagate through its core.

Prisms and camera lenses

Different wavelengths can experience slightly different refractive indices in the same material.

This effect is called dispersion.

For ordinary glass in the visible range, blue light generally has a slightly higher refractive index than red light.

Since:

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

blue light travels more slowly through the glass than red light.

This difference in speed causes different colours to refract by different amounts.

A prism can therefore separate white light into its component colours.

The same effect is important in camera lenses, where different colours can focus at slightly different positions. This is one cause of chromatic aberration.

The important application rule is simple:

• Use the vacuum wavelength when calculating photon energy using \(E_\gamma=hc/\lambda_0\).
• Use the wavelength inside the material when considering wave spacing and phase propagation within that material.
• Remember that the refractive index can vary with wavelength, so different colours need not travel at exactly the same speed.

06Questions and solutions

Question 1

A monochromatic laser produces light with a vacuum wavelength of \(650\ \mathrm{nm}\).

The light enters a transparent liquid with a refractive index of \(1.25\), travelling at normal incidence.

Calculate:

(a) The speed of light in the liquid.

(b) The wavelength inside the liquid.

(c) State whether the frequency changes.

Solution 1

The speed is \(2.40\times10^8\ \mathrm{m\,s^{-1}}\), the wavelength is \(520\ \mathrm{nm}\), and the frequency remains unchanged.

(a) Speed

Use:

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

Substitute:

\[ v=\frac{3.00\times10^8}{1.25} \]

\[ \boxed{v=2.40\times10^8\ \mathrm{m\,s^{-1}}} \]

(b) Wavelength

The frequency remains unchanged when light enters a stationary medium.

Therefore:

\[ \lambda_{\mathrm{medium}}=\frac{\lambda_0}{n} \]

\[ \lambda_{\mathrm{medium}} =\frac{650}{1.25} \]

\[ \boxed{\lambda_{\mathrm{medium}}=520\ \mathrm{nm}} \]

(c) Frequency

The frequency stays the same.

The source determines the oscillation frequency, and a stationary boundary does not change that frequency during ordinary refraction.

The material changes the propagation speed, so the wavelength must decrease to maintain \(v=f\lambda\).

Question 2

Two monochromatic light sources emit different colours.

• Source A emits red light with a vacuum wavelength of \(660\ \mathrm{nm}\).
• Source B emits blue light with a vacuum wavelength of \(440\ \mathrm{nm}\).

Both sources emit \(4.00\times10^{17}\) photons per second.

The light from both sources enters the same transparent glass, which has a refractive index of \(1.50\) for both wavelengths.

Assume the glass is lossless, and ignore dispersion.

Calculate:

(a) The speed of each colour inside the glass.

(b) The wavelength of each colour inside the glass.

(c) The energy of each photon.

(d) The optical power of each source.

Explain why the two sources can have equal photon emission rates but different optical powers.

Solution 2

Both colours travel at \(2.00\times10^8\ \mathrm{m\,s^{-1}}\) in the glass. The blue photons have 1.50 times the energy of the red photons, so the blue source has 1.50 times the optical power despite emitting the same number of photons each second.

(a) Speed

Both colours have the same refractive index.

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

\[ v=\frac{3.00\times10^8}{1.50} \]

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

(b) Wavelengths

For red light:

\[ \lambda_{\mathrm{red}} =\frac{660}{1.50} \]

\[ \boxed{\lambda_{\mathrm{red}}=440\ \mathrm{nm}} \]

For blue light:

\[ \lambda_{\mathrm{blue}} =\frac{440}{1.50} \]

\[ \boxed{\lambda_{\mathrm{blue}}=293\ \mathrm{nm}} \]

Both wavelengths decrease by the same factor because the refractive index is assumed to be identical.

(c) Photon energies

Use:

\[ E_\gamma=\frac{hc}{\lambda_0} \]

For red light:

\[ E_{\mathrm{red}} = \frac{(6.63\times10^{-34})(3.00\times10^8)} {660\times10^{-9}} \]

\[ \boxed{E_{\mathrm{red}}=3.01\times10^{-19}\ \mathrm{J}} \]

For blue light:

\[ E_{\mathrm{blue}} = \frac{(6.63\times10^{-34})(3.00\times10^8)} {440\times10^{-9}} \]

\[ \boxed{E_{\mathrm{blue}}=4.52\times10^{-19}\ \mathrm{J}} \]

Blue photons have greater energy because they have a higher frequency and shorter vacuum wavelength.

Notice that we used vacuum wavelengths, not the wavelengths inside the glass.

(d) Optical power

Power is the energy transferred per second.

For a monochromatic source:

\[ P=RE_\gamma \]

where \(R\) is the photon emission rate in photons per second.

For red light:

\[ P_{\mathrm{red}} =(4.00\times10^{17})(3.01\times10^{-19}) \]

\[ \boxed{P_{\mathrm{red}}=0.120\ \mathrm{W}} \]

For blue light:

\[ P_{\mathrm{blue}} =(4.00\times10^{17})(4.52\times10^{-19}) \]

\[ \boxed{P_{\mathrm{blue}}=0.181\ \mathrm{W}} \]

The blue source has greater power because each photon carries more energy.

The photon emission rates are identical, but the energy transferred per photon is different.

This is why photon count and photon energy must be treated as separate quantities.

Question 3

A monochromatic laser has a vacuum wavelength of \(500\ \mathrm{nm}\).

It enters crown glass with a refractive index of \(1.52\), striking the surface at an angle of \(40^\circ\) to the normal.

Assume the refractive index of air is \(1.00\).

Calculate:

(a) The speed of light in the glass.

(b) The wavelength inside the glass.

(c) The angle of refraction.

(d) Explain why the light bends towards the normal.

Solution 3

The speed is \(1.97\times10^8\ \mathrm{m\,s^{-1}}\), the wavelength is approximately \(329\ \mathrm{nm}\), and the refracted angle is approximately \(25.0^\circ\).

(a) Speed

Use:

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

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

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

(b) Wavelength

\[ \lambda_{\mathrm{glass}} =\frac{\lambda_0}{n} \]

\[ \lambda_{\mathrm{glass}} =\frac{500}{1.52} \]

\[ \boxed{\lambda_{\mathrm{glass}}=329\ \mathrm{nm}} \]

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

(c) Angle of refraction

Use Snell’s law:

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

where:

• \(n_1\) is the refractive index of air.
• \(n_2\) is the refractive index of glass.
• \(\theta_1\) is the angle of incidence.
• \(\theta_2\) is the angle of refraction.

All angles are measured from the normal.

Substitute:

\[ 1.00\sin40^\circ =1.52\sin\theta_2 \]

Rearrange:

\[ \sin\theta_2 =\frac{\sin40^\circ}{1.52} \]

\[ \theta_2 =\sin^{-1}(0.423) \]

\[ \boxed{\theta_2=25.0^\circ} \]

(d) Explanation

The light bends towards the normal because it enters a material with a higher refractive index and lower propagation speed.

The frequency remains constant, but the wavelength decreases.

At an angled boundary, the change in wave speed causes the wavefronts to change direction. This produces refraction.

The tempting mistake is to assume that the light bends because its frequency changes. It does not. The change in propagation speed is responsible for the change in direction.

Question 4

A student constructs a simplified version of Foucault’s rotating-mirror experiment.

Two light paths have equal one-way lengths of \(3.00\ \mathrm{m}\).

One path passes through air, and the other passes through water.

The rotating mirror operates at a constant speed. The returned image displacement is proportional to the total time taken for the light to complete its round trip.

The student measures the following illustrative image displacements:

Assume the speed of light in air is \(3.00\times10^8\ \mathrm{m\,s^{-1}}\).

(a) Calculate the ratio of the travel times through water and air.

(b) Determine the speed of light in water.

(c) Calculate the refractive index of water using your result.

(d) Explain why an inaccurate measurement of the rotating mirror’s speed would affect an absolute speed determination.

Solution 4

The water travel time is \(1.33\) times the air travel time, giving a water speed of \(2.25\times10^8\ \mathrm{m\,s^{-1}}\) and a refractive index of approximately \(1.33\).

(a) Ratio of travel times

At constant mirror rotation speed and identical optical geometry, image displacement is proportional to travel time.

Therefore:

\[ \frac{t_{\mathrm{water}}}{t_{\mathrm{air}}} = \frac{0.24}{0.18} \]

\[ \boxed{ \frac{t_{\mathrm{water}}}{t_{\mathrm{air}}} =1.33 } \]

The water path takes approximately 1.33 times as long.

(b) Speed in water

Both paths have the same one-way distance, so they also have the same round-trip distance.

The speed of light in air is:

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

Since:

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

For equal distances, speed is inversely proportional to travel time.

Therefore:

\[ \frac{v_{\mathrm{water}}}{v_{\mathrm{air}}} = \frac{t_{\mathrm{air}}}{t_{\mathrm{water}}} \]

Substitute:

\[ v_{\mathrm{water}} =\frac{3.00\times10^8}{1.33} \]

\[ \boxed{ v_{\mathrm{water}} =2.25\times10^8\ \mathrm{m\,s^{-1}} } \]

(c) Refractive index

Use:

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

\[ n=\frac{3.00\times10^8}{2.25\times10^8} \]

\[ \boxed{n=1.33} \]

This is consistent with the expected refractive index of water for visible light.

(d) Experimental limitation

An inaccurate mirror rotation rate makes the calculated travel time inaccurate.

The experiment relies on knowing how far the mirror turns while the light completes its round trip.

If the actual rotation rate differs from the assumed rate, the calculated time and therefore the calculated speed will be incorrect.

Comparing the two paths simultaneously helps because both measurements use the same mirror rotation. However, accurate absolute measurements still require reliable calibration of the rotation rate.

This is why Foucault’s 1850 result was particularly useful as a comparison, even though it did not provide precise absolute speeds for both media.

Question 5

An idealised Mach-Zehnder interferometer splits a monochromatic laser beam into two coherent beams.

The laser has a vacuum wavelength of \(600\ \mathrm{nm}\).

The two beams travel along separate arms with equal geometric lengths. A glass plate of thickness \(600\ \mathrm{nm}\) is inserted into one arm, replacing an equal length of vacuum.

The glass has a refractive index of \(1.50\).

The beams are then recombined at a second beam splitter and directed towards two detectors, D1 and D2.

Initially, with no glass plate inserted, the interferometer is adjusted so that D1 receives constructive interference and D2 receives destructive interference.

Assume:

• Both beams have equal amplitudes.
• The glass is perfectly transparent and lossless.
• The glass surfaces have antireflection coatings.
• The coatings introduce no additional relative phase shift.
• The two beams contain equal photon numbers before recombination.
• All other optical path lengths and beam-splitter phase effects remain fixed.

Answer the following:

(a) Calculate the speed of light inside the glass.

(b) Calculate the wavelength inside the glass.

(c) Calculate the energy of one photon before entering the glass.

(d) Determine the additional number of wave cycles accumulated in the glass arm compared with the same distance in a vacuum.

(e) Predict what happens at detectors D1 and D2 when the glass plate is inserted.

(f) Explain how the detector photon counts can change even though the photons have not lost energy and the interferometer has no absorption.

Solution 5

The glass arm accumulates an additional half-cycle of phase, corresponding to a phase difference of \(\pi\) radians. This reverses the interference pattern: D1 becomes dark and D2 becomes bright. The energy of each photon remains unchanged, and the total photon count across both detectors is conserved.

This question combines refractive index, wavelength, photon energy, and interference.

The important idea is that unchanged photon energy does not mean unchanged phase.

(a) Speed inside glass

Use:

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

Substitute:

\[ v=\frac{3.00\times10^8}{1.50} \]

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

(b) Wavelength inside glass

The vacuum wavelength is:

\[ \lambda_0=600\ \mathrm{nm} \]

Therefore:

\[ \lambda_{\mathrm{glass}} =\frac{\lambda_0}{n} \]

\[ \lambda_{\mathrm{glass}} =\frac{600}{1.50} \]

\[ \boxed{\lambda_{\mathrm{glass}}=400\ \mathrm{nm}} \]

The wavelength is shorter inside the glass.

(c) Photon energy

First, calculate the frequency.

\[ f=\frac{c}{\lambda_0} \]

\[ f=\frac{3.00\times10^8}{600\times10^{-9}} \]

\[ f=5.00\times10^{14}\ \mathrm{Hz} \]

Now calculate the photon energy:

\[ E_\gamma=hf \]

\[ E_\gamma =(6.63\times10^{-34})(5.00\times10^{14}) \]

\[ \boxed{E_\gamma=3.32\times10^{-19}\ \mathrm{J}} \]

The photon energy is unchanged inside the glass because the frequency is unchanged.

(d) Additional wave cycles

This is the most important calculation.

The glass plate replaces a \(600\ \mathrm{nm}\) section of vacuum.

In the vacuum section, the number of complete wavelengths fitting into that distance is:

\[ N_{\mathrm{vacuum}} =\frac{600}{600} =1.00 \]

Inside the glass, the number of wavelengths fitting into the same distance is:

\[ N_{\mathrm{glass}} =\frac{600}{400} =1.50 \]

The additional number of cycles is:

\[ \Delta N=1.50-1.00 \]

\[ \boxed{\Delta N=0.50\ \mathrm{cycles}} \]

Half a cycle corresponds to a phase difference of:

\[ \Delta\phi=2\pi\Delta N \]

\[ \Delta\phi=2\pi(0.50) \]

\[ \boxed{\Delta\phi=\pi\ \mathrm{rad}} \]

The glass arm therefore accumulates an additional half-cycle of phase compared with the vacuum arm.

(e) What happens at the detectors?

The original arrangement produces constructive interference at D1 and destructive interference at D2.

Inserting the glass plate introduces an additional phase difference of \(\pi\) radians.

This reverses the interference pattern.

Therefore:

• D1 changes from constructive to destructive interference.
• D2 changes from destructive to constructive interference.

For the ideal equal-amplitude arrangement, D1 becomes dark and D2 becomes bright.

(f) Why does this not violate energy conservation?

The tempting prediction is that if the photons have unchanged energy and the same number enter the interferometer, the detector counts must remain unchanged.

That reasoning misses the effect of interference.

The two coherent beams combine when they reach the second beam splitter. Their relative phase determines how the resulting light is distributed between the two output detectors.

The glass changes the phase accumulated along one path, even though it does not change the photon energy.

Photons that would previously have been detected at D1 are now redirected towards D2 through the interference process.

The total number of photons detected across both outputs remains unchanged in this ideal lossless arrangement.

This is the crucial distinction:

• Photon energy is determined by frequency.
• Photon count is affected by transmission and by how interference distributes light between detectors.
• Phase depends on the wave’s propagation through each optical path.

A slower wave can therefore accumulate a different phase without losing energy.

07Where this takes you next

The next useful step is connecting refractive index to two different wave behaviours.

Refraction explains how a change in propagation speed changes the direction of light at a boundary. Interference explains how differences in accumulated phase determine whether two waves reinforce or cancel one another.

Keep the quantities separate when approaching HSC questions:

• Use \(v=c/n\) to calculate phase speed.
• Use \(v=f\lambda\) to calculate wavelength inside a medium.
• Use \(E_\gamma=hf\) to calculate photon energy.
• Use photon emission rate and transmission information to calculate photon count.
• Use phase differences when comparing coherent beams.

Once these relationships are clear, questions about glass, water, optical fibres, prisms, and interferometers become much easier to reason through.