How to Compare Radio, Microwave and Infrared Wavelengths

A 100 MHz radio signal, a 2.45 GHz microwave signal, and infrared radiation at 30 THz are all electromagnetic waves. They travel at essentially the same speed through a vacuum, yet their wavelengths are dramatically different.

Before calculating anything, predict which has the longest wavelength. Which carries the most energy per photon?

The answer is:

• Radio waves have the longest wavelengths.
• Microwaves have shorter wavelengths than radio waves.
• Infrared waves have the shortest wavelengths of these three.

However, photon energy follows the opposite order. Infrared photons have the most energy, while radio photons have the least.

The key to comparing them is understanding three separate quantities: wavelength, energy per photon, and the number of photons in a beam. Mixing these up is a common source of mistakes in HSC Physics.

01A quick comparison of radio, microwave, and infrared wavelengths

Consider three representative electromagnetic signals:

These are illustrative frequencies, not exact boundaries between the electromagnetic spectrum regions. The precise classification of radio and microwave radiation varies between conventions. Microwaves are sometimes treated as a subset of radio-frequency radiation.

The important comparison is the order:

\[ \boxed{\lambda_{\text{radio}}> \lambda_{\text{microwave}}> \lambda_{\text{infrared}}} \]

For photon energy, the order reverses:

\[ \boxed{E_{\text{radio}}< E_{\text{microwave}}< E_{\text{infrared}}} \]

Let’s work out why these relationships hold.

01Why do shorter wavelengths have higher frequencies?

All electromagnetic waves travel at the speed of light in a vacuum:

\[ c=3.00\times10^8\ \text{m s}^{-1} \]

Their speed, frequency, and wavelength are connected by:

\[ \boxed{c=f\lambda} \]

where:

• \(c\) is the speed of light, measured in metres per second (\(\text{m s}^{-1}\)).
• \(f\) is frequency, measured in hertz (Hz).
• \(\lambda\) is wavelength, measured in metres (m).

Frequency tells us how many complete wave cycles pass a fixed point each second. Wavelength is the distance between corresponding points on successive cycles.

Picture runners moving along a track at the same speed. If more runners pass a particular point each second, they must be more closely spaced.

The analogy captures the relationship between frequency and spacing, although electromagnetic waves are oscillations of electric and magnetic fields rather than objects travelling in a queue.

The same principle applies to electromagnetic waves.

Because their speed in a vacuum is constant, increasing frequency means decreasing wavelength.

Rearrange the wave equation to find wavelength:

\[ \boxed{\lambda=\frac{c}{f}} \]

What happens when electromagnetic waves enter a material?

There is an important limitation to the equation \(c=f\lambda\).

It applies when the wave travels through a vacuum. Inside a material, electromagnetic waves generally travel more slowly.

We then use:

\[ v=f\lambda \]

where \(v\) is the wave speed in the material.

At a stationary boundary, the frequency remains unchanged. The wave speed and wavelength can change.

This distinction becomes important when comparing photon energies, particularly if a question involves electromagnetic radiation travelling through glass or another transparent material.

If you need to practise identifying wavelength from wave diagrams before tackling these calculations, see Wavelength in HSC Physics: How to Measure It Correctly.

02How does wavelength affect photon energy?

So far, we have compared electromagnetic waves using their wave properties.

However, electromagnetic radiation also transfers energy in discrete packets called photons.

The energy of an individual photon is given by Planck’s equation:

\[ \boxed{E=hf} \]

where:

• \(E\) is the energy of one photon, measured in joules (J).
• \(h\) is Planck’s constant, \(6.63\times10^{-34}\ \text{J s}\).
• \(f\) is the frequency of the electromagnetic radiation, measured in hertz (Hz).

Since \(c=f\lambda\), we can substitute \(f=c/\lambda\):

\[ \boxed{E=\frac{hc}{\lambda}} \]

Here, \(\lambda\) represents the wavelength in a vacuum, or approximately in air.

This equation reveals an important relationship:

The shorter the wavelength, the greater the energy carried by each photon.

For example, an infrared photon at \(10\ \mu\text{m}\) has substantially more energy than a radio photon with a wavelength of \(3.0\ \text{m}\).

The relationship is inverse. If the wavelength decreases by a factor of 10, the energy per photon increases by a factor of 10.

The important distinction: photon energy versus total energy

Here is where students often make a mistake.

Suppose an infrared source and a radio transmitter both deliver exactly \(1.0\ \text{J}\) of electromagnetic energy.

Does the infrared source deliver more total energy because its individual photons are more energetic?

No. Both deliver the same total energy.

The difference is the number of photons needed to deliver that energy.

Think about exchanging a fixed amount of money using either one-dollar coins or ten-cent coins. You can exchange the same total value using fewer large coins or many smaller coins.

The analogy helps distinguish the value of each unit from the total value. Photons, of course, are quantised energy packets rather than physical coins.

The number of photons in a beam is:

\[ \boxed{N=\frac{U}{E}} \]

where:

• \(N\) is the number of photons.
• \(U\) is the total energy of the radiation, measured in joules.
• \(E\) is the energy of one photon, measured in joules.

Substituting Planck’s equation gives:

\[ N=\frac{U}{hf} \]

This means that, for equal total energy, longer-wavelength radiation contains more photons.

There is another useful version of this relationship.

If a source produces radiation with power \(P\), the number of photons emitted per second is:

\[ \boxed{\dot N=\frac{P}{hf}} \]

Here, \(\dot N\) represents the photon emission rate, measured in photons per second.

03How do wavelength differences affect real applications?

The different wavelengths of radio, microwave, and infrared radiation make them useful for different purposes.

However, applications are not determined by photon energy alone. The way radiation interacts with matter, the dimensions of equipment, and the properties of the detector also matter.

Radio waves: Long wavelengths and communication

Radio waves have relatively long wavelengths.

For example, a \(100\ \text{MHz}\) radio signal has a wavelength of approximately \(3.0\ \text{m}\).

This affects the dimensions of the antennas used to transmit and receive radio signals. Antennas are commonly designed using dimensions that are a fraction of the wavelength.

Longer wavelengths can also diffract around obstacles more effectively when the obstacle dimensions are comparable to the wavelength.

Common applications include:

• Radio broadcasting.
• Television broadcasting.
• Wireless communication.
• Long-distance communication systems.

Radio photons have very little energy individually, but enormous numbers of them can transfer energy and information.

Microwaves: Shorter wavelengths and directional transmission

Microwaves have shorter wavelengths than the radio examples considered here.

A microwave signal at \(2.45\ \text{GHz}\), for example, has a wavelength of approximately \(12.2\ \text{cm}\).

This makes it practical to construct relatively compact antennas and dishes that direct electromagnetic radiation into narrow beams.

Microwaves are commonly used in:

• Radar systems.
• Satellite communication.
• Wireless networking.
• Microwave ovens.

In a microwave oven, the alternating electric field transfers energy to polar molecules and other charge-containing components in food. Molecular reorientation and dielectric losses help convert electromagnetic energy into thermal energy.

The heating process involves many photons and collective interactions with matter. It is not simply a matter of one microwave photon having enough energy to heat an entire molecule.

Infrared radiation: Thermal detection and molecular vibrations

Infrared radiation has shorter wavelengths than microwaves.

A wavelength of \(10\ \mu\text{m}\), for example, is commonly associated with thermal infrared applications.

Warm objects emit infrared radiation, making it useful for thermal imaging. A thermal camera detects infrared radiation and uses the measured signal to produce an image related to the temperatures of different surfaces.

Infrared radiation is also important in chemistry.

Molecular bonds can absorb particular infrared frequencies, causing changes in their vibrational energy. These absorption patterns provide information about molecular structure.

This is the basis of infrared spectroscopy, which is used to identify functional groups in organic compounds.

For a more detailed explanation of this application, see Infrared Spectroscopy of Organic Compounds: HSC Chemistry Guide.

04How did scientists measure invisible radio wavelengths?

Knowing that radio waves have long wavelengths is useful. But how can you actually measure a wavelength when you cannot see the wave?

This was part of the experimental challenge faced by Heinrich Hertz during his investigations of electromagnetic waves in the late nineteenth century.

Hertz needed experimental evidence that electromagnetic disturbances could travel through space and behave like waves.

His apparatus provided a way to generate, detect, and investigate these invisible waves.

The original apparatus

The transmitter consisted of metal conductors separated by a small spark gap. An induction coil supplied a high voltage, causing a spark to jump across the gap.

This rapid electrical discharge produced electromagnetic radiation.

Some distance away, Hertz positioned a receiving loop of wire with another tiny spark gap. When electromagnetic radiation reached the loop, it induced an electrical response, producing a small spark across the receiver gap.

The receiving spark provided visible evidence that energy had travelled from the transmitter to the receiver.

What should happen if the waves behave like waves?

Hertz investigated what happened when electromagnetic radiation reflected from a conducting metal surface.

Imagine placing a large metal reflector opposite the transmitter.

Some of the outgoing electromagnetic radiation travels towards the reflector and is reflected backwards. The outgoing and reflected waves overlap.

If the waves interfere, they should produce a repeating pattern of strong and weak signals.

This pattern is called a standing wave.

At some positions, the waves interfere destructively, producing minima in the detector’s response. These positions are called nodes.

At other positions, constructive interference produces maxima called antinodes.

The important observation is that adjacent nodes are separated by half a wavelength:

\[ d=\frac{\lambda}{2} \]

where \(d\) is the distance between adjacent nodes.

Therefore:

\[ \boxed{\lambda=2d} \]

By moving the receiver along the path and recording the positions of successive minima, the wavelength could be estimated.

The repeating pattern supported the wave model because the detector response depended on the relative phase of the outgoing and reflected waves.

What was the limitation of the original design?

The tiny spark detector was useful for showing that electromagnetic energy had arrived, but it was not a precise measuring instrument.

Its response depended on factors such as the detector’s orientation, the spark-gap separation, and how strongly it coupled to the incoming field.

This made it difficult to identify the exact positions of nodes and measure their spacing accurately.

A more controlled modern experiment would use:

• A stable microwave transmitter with a known frequency.
• A fixed metal reflector.
• A detector probe that can move along a measured track.
• A detector with a calibrated electrical readout.

The detector’s orientation would remain fixed, and measurements could be taken across several minima to reduce uncertainty.

This is a useful example of experimental improvement: the original apparatus demonstrated the phenomenon, while a more controlled arrangement makes quantitative measurements more reliable.

05A reliable method for comparing the three types of radiation

When an HSC question asks you to compare radio, microwave, and infrared radiation, follow this sequence.

1. Identify the wavelength or frequency.

Check whether the question gives you wavelength, frequency, or both. Convert all values into SI units before calculating.

2. Compare wavelengths.

If the frequencies are given, use:

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

For electromagnetic waves travelling in a vacuum, the highest frequency corresponds to the shortest wavelength.

3. Compare energy per photon.

Use:

\[ E=hf \]

Or, using the vacuum wavelength:

\[ E=\frac{hc}{\lambda} \]

A higher-frequency photon carries more energy.

4. Calculate photon number separately.

If the question gives total energy:

\[ N=\frac{U}{hf} \]

If it gives power:

\[ \dot N=\frac{P}{hf} \]

Do not assume that higher-energy photons mean a greater total number of photons.

5. Connect the wavelength to an application.

Consider the physical properties that make the radiation useful:

• Radio: communication and antenna design.
• Microwave: radar, wireless communication, and heating.
• Infrared: thermal imaging and molecular spectroscopy.

07Questions and solutions

The following questions move from straightforward wavelength calculations to more demanding problems involving photon energy, experimental measurements, and changes in the medium through which electromagnetic radiation travels.

Question 1

A radio transmitter operates at a frequency of \(150\ \text{MHz}\), while a microwave transmitter operates at \(3.00\ \text{GHz}\).

Both signals travel through air.

Calculate the wavelength of each signal and determine how many times longer the radio wavelength is than the microwave wavelength.

Use \(c=3.00\times10^8\ \text{m s}^{-1}\).

Solution 1

The radio wavelength is \(2.00\ \text{m}\), while the microwave wavelength is \(0.100\ \text{m}\). The radio wavelength is therefore 20 times longer.

Use the electromagnetic wave equation:

\[ c=f\lambda \]

Rearranging:

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

For the radio signal:

\[ f=150\times10^6\ \text{Hz} \]

\[ \lambda_{\text{radio}} = \frac{3.00\times10^8} {150\times10^6} \]

\[ \lambda_{\text{radio}}=2.00\ \text{m} \]

For the microwave signal:

\[ f=3.00\times10^9\ \text{Hz} \]

\[ \lambda_{\text{microwave}} = \frac{3.00\times10^8} {3.00\times10^9} \]

\[ \lambda_{\text{microwave}}=0.100\ \text{m} \]

Calculate the ratio:

\[ \frac{\lambda_{\text{radio}}} {\lambda_{\text{microwave}}} = \frac{2.00}{0.100} \]

\[ =20 \]

The radio wavelength is 20 times longer.

The important principle is that both signals travel at approximately the same speed in air. The radio signal has a lower frequency, so its wavelength must be longer.

Question 2

Two monochromatic electromagnetic sources operate continuously.

• Source A emits radio waves at \(100\ \text{MHz}\) with a power of \(2.00\ \text{W}\).
• Source B emits microwaves at \(2.00\ \text{GHz}\) with a power of \(0.500\ \text{W}\).

Calculate the number of photons emitted per second by each source.

Determine which source emits more photons per second and calculate the ratio of their emission rates.

Use \(h=6.63\times10^{-34}\ \text{J s}\).

Solution 2

Source A emits approximately \(3.02\times10^{25}\) photons per second, while Source B emits approximately \(3.77\times10^{23}\) photons per second.

Source A therefore emits 80 times as many photons per second, despite having only four times the power.

The important distinction is that each microwave photon carries more energy than each radio photon.

Step 1: Calculate the energy of an individual photon from each source.

Use:

\[ E=hf \]

For Source A:

\[ E_A=(6.63\times10^{-34})(1.00\times10^8) \]

\[ E_A=6.63\times10^{-26}\ \text{J} \]

For Source B:

\[ E_B=(6.63\times10^{-34})(2.00\times10^9) \]

\[ E_B=1.326\times10^{-24}\ \text{J} \]

Step 2: Calculate the photon emission rates.

The number of photons emitted per second is:

\[ \dot N=\frac{P}{E} \]

For Source A:

\[ \dot N_A= \frac{2.00}{6.63\times10^{-26}} \]

\[ \dot N_A=3.02\times10^{25} \ \text{photons s}^{-1} \]

For Source B:

\[ \dot N_B= \frac{0.500}{1.326\times10^{-24}} \]

\[ \dot N_B=3.77\times10^{23} \ \text{photons s}^{-1} \]

Step 3: Calculate the ratio.

\[ \frac{\dot N_A}{\dot N_B} = \frac{3.02\times10^{25}} {3.77\times10^{23}} \]

\[ =80.1\approx80 \]

The radio source emits approximately 80 times as many photons per second.

The tempting approach is to assume that the source with greater power must produce more photons. However, power represents energy transferred per second, not the number of photons transferred per second.

Each microwave photon carries 20 times the energy of each radio photon. Source A produces four times the total power, but its much lower photon energy means its photon emission rate is 80 times greater.

Question 3

Two monochromatic radiation pulses are completely absorbed by separate detectors.

• Pulse A contains \(2.00\times10^{18}\) radio photons, each with a frequency of \(100\ \text{MHz}\).
• Pulse B contains \(2.00\times10^{18}\) infrared photons, each with a frequency of \(30.0\ \text{THz}\).

Calculate the total energy absorbed by each detector.

Determine the ratio of the absorbed energies and explain why equal photon numbers do not produce equal energy transfers.

Use \(h=6.63\times10^{-34}\ \text{J s}\).

Solution 3

The infrared pulse transfers \(3.00\times10^5\) times as much energy as the radio pulse.

Although both pulses contain the same number of photons, their individual photon energies are very different.

Step 1: Calculate the energy of one radio photon.

\[ E_A=hf \]

\[ E_A=(6.63\times10^{-34})(1.00\times10^8) \]

\[ E_A=6.63\times10^{-26}\ \text{J} \]

Step 2: Calculate the total energy of the radio pulse.

\[ U_A=NE_A \]

\[ U_A=(2.00\times10^{18}) (6.63\times10^{-26}) \]

\[ U_A=1.326\times10^{-7}\ \text{J} \]

Step 3: Calculate the energy of one infrared photon.

\[ E_B=hf \]

\[ E_B=(6.63\times10^{-34})(3.00\times10^{13}) \]

\[ E_B=1.989\times10^{-20}\ \text{J} \]

Step 4: Calculate the total energy of the infrared pulse.

\[ U_B=NE_B \]

\[ U_B=(2.00\times10^{18}) (1.989\times10^{-20}) \]

\[ U_B=3.978\times10^{-2}\ \text{J} \]

Step 5: Compare the total energies.

\[ \frac{U_B}{U_A} = \frac{3.978\times10^{-2}} {1.326\times10^{-7}} \]

\[ =3.00\times10^5 \]

The infrared pulse transfers 300,000 times more energy.

The misconception is assuming that equal photon numbers mean equal total energy. The correct relationship is:

\[ U=N E_{\text{photon}} \]

Both pulses have the same \(N\), but each infrared photon carries 300,000 times the energy of each radio photon.

Question 4

A student investigates the wavelength of microwave radiation using a transmitter, a flat metal reflector, and a movable detector.

The transmitter operates at a frequency of \(1.25\ \text{GHz}\).

The detector records eight consecutive minima in signal strength over a total distance of \(0.84\ \text{m}\), measured from the first minimum to the eighth.

(a) Explain why the detector records repeated minima.

(b) Calculate the wavelength of the microwave radiation.

(c) Calculate the speed of the radiation and comment on whether the result is consistent with electromagnetic waves travelling in air.

(d) Explain why dividing the measured distance by eight would produce an incorrect wavelength.

Solution 4

(a) The repeated minima occur because the outgoing and reflected microwave waves interfere destructively at particular positions.

The metal reflector sends part of the incoming radiation backwards. Where the two waves have the appropriate phase relationship, their amplitudes cancel, producing nodes.

At these positions, the detector records a minimum signal.

Adjacent minima are separated by half a wavelength.

(b) The wavelength is \(0.240\ \text{m}\).

There are eight measured minima, which means there are seven intervals between the first and eighth minima.

Step 1: Calculate the distance between adjacent minima.

\[ d=\frac{0.84}{7} \]

\[ d=0.120\ \text{m} \]

Step 2: Calculate the wavelength.

Since adjacent minima are separated by half a wavelength:

\[ \lambda=2d \]

\[ \lambda=2(0.120) \]

\[ \boxed{\lambda=0.240\ \text{m}} \]

(c) The wave speed is \(3.00\times10^8\ \text{m s}^{-1}\), consistent with electromagnetic radiation travelling through air.

Use:

\[ v=f\lambda \]

Substitute:

\[ v=(1.25\times10^9)(0.240) \]

\[ \boxed{v=3.00\times10^8\ \text{m s}^{-1}} \]

This is the expected approximate speed of electromagnetic radiation in air.

(d) Dividing by eight would incorrectly treat the eight measured minima as eight separate intervals.

There are only seven intervals between eight positions.

This is the same counting principle as the gaps between fence posts: eight posts create seven gaps.

The best way to improve the reliability of this experiment is to measure across as many consecutive minima as possible and divide by the correct number of intervals. This reduces the influence of uncertainty in locating any individual minimum.

Question 5

A physics student compares two electromagnetic pulses travelling through a transparent material.

Both pulses have a total energy of \(1.00\ \text{mJ}\).

• Pulse A is radio radiation with a vacuum wavelength of \(3.00\ \text{m}\).
• Pulse B is infrared radiation with a vacuum wavelength of \(10.0\ \mu\text{m}\).

The pulses enter a stationary, transparent, non-reflecting, non-absorbing material.

The material is dispersive, with the following refractive indices:

• Radio radiation: \(n_A=1.20\).
• Infrared radiation: \(n_B=1.50\).

Assume that the refractive indices remain constant for each pulse, no photons are absorbed or converted, and both pulses emerge back into a vacuum.

(a) Calculate the frequency and energy of an individual photon for each pulse.

(b) Calculate the number of photons contained in each pulse. Which pulse contains more photons, and by what factor?

(c) Calculate the wavelength of each pulse inside the material.

(d) A student calculates the ratio of photon energies inside the material using \(E=hc/\lambda\), substituting the wavelengths measured inside the material. They conclude that the energy ratio has changed.

Explain precisely why this conclusion is incorrect. Your explanation must account for the different refractive indices, the unchanged frequencies, and what happens when the pulses emerge back into a vacuum.

Solution 5

The radio pulse contains \(1.51\times10^{22}\) photons, while the infrared pulse contains \(5.03\times10^{16}\) photons. The radio pulse therefore contains 300,000 times as many photons.

Inside the material, both wavelengths decrease, but neither photon energy changes. The key is that photon energy depends on frequency, not directly on the wavelength measured inside a material.

(a) Calculate the frequencies and individual photon energies.

First, use:

\[ c=f\lambda_0 \]

where \(\lambda_0\) is the vacuum wavelength.

For the radio pulse:

\[ f_A=\frac{3.00\times10^8}{3.00} \]

\[ f_A=1.00\times10^8\ \text{Hz} \]

Its photon energy is:

\[ E_A=hf_A \]

\[ E_A=(6.63\times10^{-34})(1.00\times10^8) \]

\[ \boxed{E_A=6.63\times10^{-26}\ \text{J}} \]

For the infrared pulse, first convert the wavelength:

\[ 10.0\ \mu\text{m}=1.00\times10^{-5}\ \text{m} \]

Therefore:

\[ f_B=\frac{3.00\times10^8}{1.00\times10^{-5}} \]

\[ f_B=3.00\times10^{13}\ \text{Hz} \]

Its photon energy is:

\[ E_B=hf_B \]

\[ E_B=(6.63\times10^{-34})(3.00\times10^{13}) \]

\[ \boxed{E_B=1.989\times10^{-20}\ \text{J}} \]

(b) Calculate the photon numbers.

Use:

\[ N=\frac{U}{E} \]

Both pulses contain:

\[ U=1.00\ \text{mJ} =1.00\times10^{-3}\ \text{J} \]

For the radio pulse:

\[ N_A= \frac{1.00\times10^{-3}} {6.63\times10^{-26}} \]

\[ \boxed{N_A=1.51\times10^{22}} \]

For the infrared pulse:

\[ N_B= \frac{1.00\times10^{-3}} {1.989\times10^{-20}} \]

\[ \boxed{N_B=5.03\times10^{16}} \]

Calculate the ratio:

\[ \frac{N_A}{N_B} = \frac{1.51\times10^{22}} {5.03\times10^{16}} \]

\[ \boxed{\frac{N_A}{N_B}=3.00\times10^5} \]

The radio pulse contains 300,000 times as many photons.

This makes sense because both pulses have equal total energy, but each infrared photon carries 300,000 times as much energy as each radio photon.

(c) Calculate the wavelengths inside the material.

The refractive index is defined by:

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

Rearranging the wave equation \(v=f\lambda\), and using the fact that frequency remains unchanged at a stationary boundary, gives:

\[ \boxed{\lambda_{\text{material}} =\frac{\lambda_0}{n}} \]

For the radio pulse:

\[ \lambda_A=\frac{3.00}{1.20} \]

\[ \boxed{\lambda_A=2.50\ \text{m}} \]

For the infrared pulse:

\[ \lambda_B=\frac{10.0}{1.50} \]

\[ \boxed{\lambda_B=6.67\ \mu\text{m}} \]

Both wavelengths decrease inside the material because the waves travel more slowly.

Notice that they decrease by different factors because the refractive indices are different.

(d) Explain why the photon energies do not change.

The student’s conclusion is incorrect because they have used the vacuum speed of light, \(c\), with wavelengths measured inside the material.

The equation:

\[ E=\frac{hc}{\lambda} \]

uses the vacuum wavelength when \(c\) represents the vacuum speed of light.

Inside the material, the correct relationship is:

\[ E=hf \]

The frequency does not change when electromagnetic radiation crosses a stationary boundary.

Therefore, the energy of each photon remains constant.

We can also demonstrate this mathematically.

Inside the material:

\[ E= h\frac{v}{\lambda} \]

Since:

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

and:

\[ \lambda=\frac{\lambda_0}{n} \]

we obtain:

\[ E=h\frac{c/n}{\lambda_0/n} \]

The refractive index cancels:

\[ \boxed{E=\frac{hc}{\lambda_0}} \]

This is exactly the original photon energy.

The ratio of photon energies therefore remains:

\[ \frac{E_B}{E_A} = \frac{f_B}{f_A} \]

\[ \boxed{\frac{E_B}{E_A}=3.00\times10^5} \]

There is one final subtlety.

Because the material is dispersive, the two radiation frequencies experience different refractive indices. Consequently, the ratio of their wavelengths inside the material is:

\[ \frac{\lambda_A}{\lambda_B} = \frac{2.50}{6.67\times10^{-6}} \]

\[ =3.75\times10^5 \]

This is not the same as the photon energy ratio of \(3.00\times10^5\).

The wavelength ratio changes because the two frequencies experience different wave speeds inside the material. The photon energy ratio does not change because neither frequency changes.

When both pulses emerge back into a vacuum, their wavelengths return to \(3.00\ \text{m}\) and \(10.0\ \mu\text{m}\), respectively.

Their photon energies and photon numbers also remain unchanged under the stated lossless conditions.

This is the important distinction: wavelength depends on the medium, but photon energy depends on frequency.

Once you can compare wavelength, frequency, photon energy, and photon number independently, you can start predicting how electromagnetic radiation interacts with matter.

That is the next useful step: understanding why antennas respond to radio waves, why microwave radiation transfers energy to materials, and why molecules absorb particular infrared frequencies. The last of these leads directly into infrared spectroscopy and the identification of organic functional groups.