How to Find an Electromagnetic Wave’s Frequency from Its Period

An antenna receives a radio signal, and an oscilloscope shows that consecutive positive peaks are \(4.0\ \text{ns}\) apart. What is the frequency of the electromagnetic wave?

The answer is \(250\ \text{MHz}\). The \(4.0\ \text{ns}\) is the time taken for one complete cycle, so we take its reciprocal:

\[ f=\frac{1}{T} =\frac{1}{4.0\times10^{-9}} =2.5\times10^8\ \text{Hz} \]

That’s the essential method for finding an electromagnetic wave’s frequency from its period: measure the time for one complete cycle, convert that time into seconds, and calculate its reciprocal.

The interesting part is making sure you’ve measured the correct time interval. An electromagnetic signal can contain several repeating features, and not all of them represent the period of the wave itself.

01Measuring the period of an electromagnetic wave

You can’t watch a radio wave move through the air in the same way you can watch a ripple travel across water. However, you can measure its oscillating electric field indirectly.

Consider a radio-frequency transmitter sending a signal to a receiving antenna.

The experimental arrangement works in this order:

1. A signal generator produces an alternating electrical signal.
2. The signal travels through a cable to a transmitting antenna.
3. The transmitting antenna produces an electromagnetic wave that propagates through the surrounding air.
4. A receiving antenna detects the changing electric field and produces an alternating voltage.
5. The receiving antenna connects to an oscilloscope, which displays voltage against time.

The oscilloscope therefore gives us a way to measure the period of the received electromagnetic signal.

What should the oscilloscope show?

Suppose the transmitter produces a steady electromagnetic signal.

The receiving antenna experiences a repeating electric field, so its output voltage also repeats. The oscilloscope displays this as a periodic waveform.

If the transmitter produces a frequency of \(100\ \text{MHz}\), the received signal should repeat every:

\[ T=\frac{1}{100\times10^6} =1.0\times10^{-8}\ \text{s} \]

Therefore, consecutive equivalent points on the waveform should be separated by \(10\ \text{ns}\).

Before continuing, make a prediction.

If the transmitter frequency increases, will the peaks on the oscilloscope move closer together or further apart?

They move closer together. A higher frequency means more complete cycles occur each second, so each cycle must take less time.

Measuring one cycle accurately

The period, \(T\), is the time between two consecutive points in the same stage of a cycle.

For example:

• Positive crest to the next positive crest: one period.
• Negative crest to the next negative crest: one period.
• Upward equilibrium crossing to the next upward equilibrium crossing: one period.
• Positive crest to negative crest: half a period.

This last distinction matters. Measuring from a positive crest to the next negative crest gives you only half the time required for a complete cycle.

In an experiment, measuring just one cycle can also introduce a relatively large timing uncertainty.

A better approach is to measure several complete cycles and divide the total time by the number of intervals.

\[ T=\frac{\text{total measured time}} {\text{number of complete cycles}} \]

For example, if six positive crests span \(50\ \text{ns}\) from the first crest to the sixth, there are five complete intervals, not six.

The improved method reduces the effect of uncertainty in reading the oscilloscope’s time scale.

This is the key measurement decision: count the gaps between equivalent points, rather than simply counting the peaks.

02Calculating frequency from period

Once you have measured the period, calculating frequency is straightforward.

The relationship is:

\[ \boxed{f=\frac{1}{T}} \]

where:

• \(f\) is frequency, measured in hertz (Hz).
• \(T\) is period, measured in seconds (s).

One hertz represents one complete cycle per second.

The equation works because frequency counts how many cycles fit into one second, while period measures how long one cycle takes.

Convert the period into seconds first

Electromagnetic waves often have extremely short periods, so questions commonly use nanoseconds, microseconds, or femtoseconds.

You should recognise these conversions:

For example:

\[ 5.0\ \text{ns}=5.0\times10^{-9}\ \text{s} \]

Never substitute \(5.0\) directly into the reciprocal equation if the original period is given in nanoseconds.

03From frequency to photon energy

Once you know the frequency, you can also calculate the energy carried by an individual photon.

This is particularly useful when a question connects electromagnetic radiation with the photon model of light.

The relationship is:

\[ E_{\gamma}=hf \]

where:

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

Since frequency is the reciprocal of period, we can substitute \(f=1/T\):

\[ \boxed{E_{\gamma}=\frac{h}{T}} \]

This gives us a useful connection: a shorter electromagnetic wave period means a higher frequency and greater energy per photon.

For example, consider two electromagnetic waves:

• Wave A has a period of \(2.0\ \text{ns}\).
• Wave B has a period of \(4.0\ \text{ns}\).

Wave A has twice the frequency of Wave B. Therefore, each photon in Wave A carries twice the energy of each photon in Wave B.

But this does not mean Wave A necessarily carries twice the total energy.

That depends on how many photons are present.

04Photon energy and photon count are different

This is an important distinction when working with electromagnetic radiation.

Photon energy tells you how much energy one individual photon carries.

Photon count tells you how many photons are present in the beam or pulse.

To calculate the number of photons, you need the total electromagnetic energy as well as the energy of one photon.

\[ \boxed{N=\frac{U}{E_{\gamma}}} \]

where:

• \(N\) is the number of photons.
• \(U\) is the total energy of the electromagnetic radiation, in joules.
• \(E_{\gamma}\) is the energy of one photon, in joules.

Substituting the photon energy relationship:

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

The period determines the energy per photon, but it cannot tell you the photon count on its own.

You must also know the total energy.

A useful contrasting case

Imagine two monochromatic light beams with exactly the same frequency.

Beam A contains \(1.0\times10^{15}\) photons, while Beam B contains \(3.0\times10^{15}\) photons.

Because their frequencies are identical, their individual photons have identical energies.

However, Beam B contains three times as many photons, so it carries three times the total energy.

Increasing the number of photons does not increase the energy of each photon.

Increasing the frequency does.

This distinction becomes important when calculating photon emission rates and electromagnetic power. For a further worked explanation, see How to Calculate Light Power from Photon Emission Rate.

05Connecting period, frequency, and wavelength to real applications

Finding frequency from period is useful because frequency connects directly to wavelength.

For electromagnetic waves travelling through a vacuum:

\[ c=f\lambda \]

Since \(f=1/T\), we can also write:

\[ \boxed{\lambda=cT} \]

This means that a longer period corresponds to a longer wavelength, provided the wave travels at the same speed.

The relationship helps explain why different parts of the electromagnetic spectrum are useful for different applications.

Consider three representative electromagnetic signals.

These are illustrative operating frequencies, not exact boundaries between regions of the electromagnetic spectrum.

Radio waves: Longer wavelengths and communication

A radio signal at \(100\ \text{MHz}\) has a wavelength of approximately \(3.0\ \text{m}\).

This matters when designing antennas. Their dimensions are often selected as fractions of the wavelength.

Radio waves are useful for broadcasting and wireless communication because electromagnetic signals can transport information over considerable distances.

Microwaves: Shorter wavelengths and heating

A typical microwave oven operates at approximately \(2.45\ \text{GHz}\).

Its wavelength is approximately \(12.2\ \text{cm}\).

The alternating electric field transfers energy to polar molecules and other charge-containing components in food. Their interactions with the changing field contribute to heating.

It is worth remembering that microwave heating is not simply a matter of individual microwave photons carrying enough energy to heat an entire molecule. The process involves collective interactions with electromagnetic radiation.

Infrared: Thermal radiation and detection

Thermal imaging commonly uses long-wave infrared radiation around \(10\ \mu\text{m}\).

Objects emit electromagnetic radiation as a result of their thermal energy. A thermal camera detects infrared radiation and uses the signal to construct an image related to the temperatures of different surfaces.

The much shorter period of infrared radiation corresponds to a much higher frequency than the radio and microwave examples.

06What happens to the period when an electromagnetic wave enters another medium?

There is one more important distinction to understand before tackling harder questions.

The equations above use the speed of light in a vacuum, or approximately in air. Electromagnetic waves can also travel through materials such as glass.

When light enters glass, its speed decreases.

You might predict that its frequency also decreases because the wave is travelling more slowly.

However, that prediction is incorrect.

The frequency of the electromagnetic wave remains unchanged when it crosses an ordinary stationary boundary between two media. Its period therefore remains unchanged as well.

What changes is its wavelength.

The wave equation inside a material is:

\[ v=f\lambda \]

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

The refractive index, \(n\), relates the speed in a vacuum to the speed in the material:

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

Rearranging:

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

Because frequency stays constant while speed decreases, wavelength must decrease.

This gives us three relationships to remember:

Photon energy remains unchanged because \(E_{\gamma}=hf\), and frequency has not changed.

For a more detailed explanation of electromagnetic wave behaviour, see Mechanical vs Electromagnetic Waves for HSC Physics.

07A final measurement trap: Carrier period versus pulse repetition

Some electromagnetic signals are transmitted in short bursts rather than as a continuous wave.

This creates two different time intervals that are easy to confuse.

Consider a radio transmitter that sends a short burst containing many rapid oscillations.

There are two things you could measure:

• Carrier period: the time between consecutive oscillations within one burst.
• Pulse repetition period: the time between the starts of consecutive bursts.

These are not the same quantity.

If a question asks for the frequency of the electromagnetic carrier, you must use the rapid oscillations within each burst.

If it asks how frequently the transmitter sends bursts, use the time between successive bursts.

For a narrow-band carrier, the energy of an individual photon is determined by the carrier frequency:

\[ E_{\gamma}=hf_{\text{carrier}} \]

The burst repetition rate instead helps determine how much energy is transmitted per second, provided the energy of each burst is known.

A finite burst technically contains a small spread of frequencies. In the calculations below, we use the usual narrow-band approximation and treat the photons as having the carrier frequency.

08Choosing the correct equation

Before attempting practice questions, use this table to decide which calculation is actually required.

The most important decision is to identify what the measured time interval actually represents.

A period, a travel time, and a pulse repetition interval are three different things. Taking the reciprocal of the wrong one can give a perfectly calculated but physically incorrect answer.

09Questions and solutions

Question 1

An electromagnetic radio signal is detected by a receiving antenna. An oscilloscope shows that consecutive positive peaks are separated by \(25.0\ \text{ns}\).

Calculate:

a) The period of the wave.

b) Its frequency in hertz.

c) Its wavelength in air.

Solution 1

The frequency is \(4.00\times10^7\ \text{Hz}\), and the wavelength is \(7.50\ \text{m}\).

Step 1: Identify the period.

The time between consecutive positive peaks represents one complete cycle.

\[ T=25.0\ \text{ns} =2.50\times10^{-8}\ \text{s} \]

Step 2: Calculate frequency.

\[ \begin{aligned} f&=\frac{1}{T}\\ &=\frac{1}{2.50\times10^{-8}}\\ &=4.00\times10^7\ \text{Hz} \end{aligned} \]

Step 3: Calculate wavelength.

Using \(c=f\lambda\):

\[ \begin{aligned} \lambda&=\frac{c}{f}\\ &=\frac{3.00\times10^8}{4.00\times10^7}\\ &=7.50\ \text{m} \end{aligned} \]

The signal has a frequency of \(40.0\ \text{MHz}\) and a wavelength of \(7.50\ \text{m}\).

The result is physically sensible: a relatively long period corresponds to a relatively low frequency and a long wavelength in air.

Question 2

An oscilloscope records an electromagnetic signal and displays seven consecutive positive peaks.

The first positive peak and the seventh positive peak are separated by \(48.0\ \text{ns}\).

Determine:

a) The period of the electromagnetic wave.

b) Its frequency.

c) Its wavelength in air.

A student calculates the period by dividing \(48.0\ \text{ns}\) by seven. Explain why this approach is incorrect.

Solution 2

The correct period is \(8.00\ \text{ns}\), giving a frequency of \(125\ \text{MHz}\) and a wavelength of \(2.40\ \text{m}\).

Step 1: Count the complete intervals.

There are seven peaks, but only six intervals between the first and seventh peaks.

This is the important distinction.

\[ T=\frac{48.0\ \text{ns}}{6} \]

\[ T=8.00\ \text{ns} \]

Convert into seconds:

\[ T=8.00\times10^{-9}\ \text{s} \]

Step 2: Calculate frequency.

\[ \begin{aligned} f&=\frac{1}{T}\\ &=\frac{1}{8.00\times10^{-9}}\\ &=1.25\times10^8\ \text{Hz} \end{aligned} \]

Step 3: Calculate wavelength.

\[ \begin{aligned} \lambda&=\frac{c}{f}\\ &=\frac{3.00\times10^8}{1.25\times10^8}\\ &=2.40\ \text{m} \end{aligned} \]

The student’s mistake is counting the peaks rather than the intervals between them.

Seven peaks represent six complete periods when measuring from the first peak to the seventh.

This is a common experimental measurement error. The equation is correct, but the quantity being substituted into it is wrong.

Question 3

Two monochromatic electromagnetic beams, A and B, each carry a total energy of \(2.00\times10^{-3}\ \text{J}\).

Their periods are:

• Beam A: \(2.00\ \text{fs}\).
• Beam B: \(4.00\ \text{fs}\).

Assume both beams consist of photons with the same frequency within each beam.

Calculate:

a) The frequency of each beam.

b) The energy of an individual photon in each beam.

c) The number of photons in each beam.

d) Which beam contains more photons, and by what factor?

Solution 3

Beam B contains twice as many photons as Beam A, even though each photon in Beam A carries twice as much energy.

Step 1: Calculate the frequency of each beam.

For Beam A:

\[ \begin{aligned} f_A&=\frac{1}{2.00\times10^{-15}}\\ &=5.00\times10^{14}\ \text{Hz} \end{aligned} \]

For Beam B:

\[ \begin{aligned} f_B&=\frac{1}{4.00\times10^{-15}}\\ &=2.50\times10^{14}\ \text{Hz} \end{aligned} \]

Beam A has twice the frequency of Beam B.

Step 2: Calculate the energy of each photon.

Use:

\[ E_{\gamma}=hf \]

For Beam A:

\[ \begin{aligned} E_A&=(6.63\times10^{-34}) (5.00\times10^{14})\\ &=3.315\times10^{-19}\ \text{J} \end{aligned} \]

For Beam B:

\[ \begin{aligned} E_B&=(6.63\times10^{-34}) (2.50\times10^{14})\\ &=1.6575\times10^{-19}\ \text{J} \end{aligned} \]

Each photon in Beam A carries twice the energy of each photon in Beam B.

Step 3: Calculate the number of photons.

The total energy of both beams is:

\[ U=2.00\times10^{-3}\ \text{J} \]

Use:

\[ N=\frac{U}{E_{\gamma}} \]

For Beam A:

\[ \begin{aligned} N_A&=\frac{2.00\times10^{-3}} {3.315\times10^{-19}}\\ &=6.03\times10^{15} \end{aligned} \]

For Beam B:

\[ \begin{aligned} N_B&=\frac{2.00\times10^{-3}} {1.6575\times10^{-19}}\\ &=1.21\times10^{16} \end{aligned} \]

Step 4: Compare the photon numbers.

\[ \frac{N_B}{N_A} =\frac{1.21\times10^{16}} {6.03\times10^{15}} =2.00 \]

Beam B contains twice as many photons.

The tempting prediction is that the higher-frequency beam must contain more photons because its photons have greater energy.

However, both beams have the same total energy. Each photon in Beam A uses twice as much of that energy, so only half as many photons are required.

This demonstrates why photon energy and photon count must be calculated separately.

Question 4

A monochromatic visible electromagnetic wave travels through air with a period of \(2.00\ \text{fs}\).

It then enters a glass block with a refractive index of \(1.50\), striking the surface at normal incidence.

Assume the glass is transparent and non-dispersive at this frequency, and ignore reflection and absorption.

Calculate:

a) The frequency of the wave.

b) Its speed inside the glass.

c) Its wavelength inside the glass.

d) The energy of one photon.

e) Explain whether the period and frequency change when the wave enters the glass.

Solution 4

The period remains \(2.00\ \text{fs}\), the frequency remains \(5.00\times10^{14}\ \text{Hz}\), and the wavelength inside the glass decreases to \(400\ \text{nm}\). Each photon retains its original energy.

Step 1: Calculate the frequency in air.

\[ \begin{aligned} f&=\frac{1}{T}\\ &=\frac{1}{2.00\times10^{-15}}\\ &=5.00\times10^{14}\ \text{Hz} \end{aligned} \]

Step 2: Calculate the speed inside the glass.

The refractive index is:

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

Rearranging:

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

Substitute:

\[ \begin{aligned} v&=\frac{3.00\times10^8}{1.50}\\ &=2.00\times10^8\ \text{m s}^{-1} \end{aligned} \]

Step 3: Calculate the wavelength inside the glass.

Use the wave equation:

\[ v=f\lambda \]

Rearrange:

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

Substitute:

\[ \begin{aligned} \lambda&=\frac{2.00\times10^8} {5.00\times10^{14}}\\ &=4.00\times10^{-7}\ \text{m} \end{aligned} \]

Therefore:

\[ \lambda=400\ \text{nm} \]

Step 4: Calculate the energy of one photon.

Photon energy depends on frequency, not directly on the speed of light in the material.

\[ \begin{aligned} E_{\gamma}&=hf\\ &=(6.63\times10^{-34}) (5.00\times10^{14})\\ &=3.315\times10^{-19}\ \text{J} \end{aligned} \]

Step 5: Explain what happens to period and frequency.

The period remains:

\[ T=2.00\ \text{fs} \]

The frequency remains:

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

The student’s tempting prediction might be that frequency decreases because the wave slows down.

However, the oscillation rate must remain consistent across the stationary boundary. The wave slows down by a factor of \(1.50\), so its wavelength decreases by the same factor.

The period and frequency remain unchanged, and therefore the energy per photon also remains unchanged.

Question 5

A radio transmitter sends short bursts of a narrow-band electromagnetic carrier.

An oscilloscope is used to examine the transmitted signal. A zoomed trace shows that consecutive carrier peaks are separated by \(10.0\ \text{ns}\).

A second, slower trace shows that the starts of consecutive bursts are separated by \(2.00\ \mu\text{s}\).

Each burst:

• Contains 20 complete carrier cycles.
• Carries a total electromagnetic energy of \(1.00\ \mu\text{J}\).

Assume that each burst can be treated as monochromatic at the carrier frequency, ignoring small sidebands.

Calculate:

a) The carrier frequency.

b) The wavelength of the carrier in air.

c) The energy of one photon.

d) The number of photons in each burst.

e) The burst repetition frequency.

f) The average power transmitted by the source.

The transmitter is then adjusted so that the starts of consecutive bursts are separated by \(1.00\ \mu\text{s}\), while the carrier waveform and energy per burst remain unchanged.

g) Explain what happens to the carrier frequency, photon energy, photon count per burst, and average transmitted power.

Solution 5

The carrier frequency is \(100\ \text{MHz}\), and each burst contains approximately \(1.51\times10^{19}\) photons. Initially, the average transmitted power is \(0.50\ \text{W}\). Doubling the burst repetition frequency doubles the average power to \(1.00\ \text{W}\), without changing the energy of an individual photon or the photon count per burst.

Step 1: Calculate the carrier frequency.

The carrier period is the time between consecutive peaks within a burst.

\[ T_c=10.0\ \text{ns} =1.00\times10^{-8}\ \text{s} \]

Therefore:

\[ \begin{aligned} f_c&=\frac{1}{T_c}\\ &=\frac{1}{1.00\times10^{-8}}\\ &=1.00\times10^8\ \text{Hz} \end{aligned} \]

The carrier frequency is \(100\ \text{MHz}\).

Step 2: Calculate the wavelength.

Using the speed of light in air:

\[ \begin{aligned} \lambda&=\frac{c}{f_c}\\ &=\frac{3.00\times10^8}{1.00\times10^8}\\ &=3.00\ \text{m} \end{aligned} \]

Step 3: Calculate the energy of one photon.

\[ \begin{aligned} E_{\gamma}&=hf_c\\ &=(6.63\times10^{-34}) (1.00\times10^8)\\ &=6.63\times10^{-26}\ \text{J} \end{aligned} \]

Each photon carries \(6.63\times10^{-26}\ \text{J}\).

Step 4: Calculate the number of photons in each burst.

The total energy per burst is:

\[ U=1.00\ \mu\text{J} =1.00\times10^{-6}\ \text{J} \]

Use:

\[ N=\frac{U}{E_{\gamma}} \]

Substitute:

\[ \begin{aligned} N&=\frac{1.00\times10^{-6}} {6.63\times10^{-26}}\\ &=1.51\times10^{19} \end{aligned} \]

Each burst contains approximately \(1.51\times10^{19}\) photons.

Step 5: Calculate the burst repetition frequency.

This is where we must distinguish the carrier period from the time between bursts.

The repetition period is:

\[ T_{\text{rep}}=2.00\ \mu\text{s} =2.00\times10^{-6}\ \text{s} \]

Therefore:

\[ \begin{aligned} f_{\text{rep}}&=\frac{1}{T_{\text{rep}}}\\ &=\frac{1}{2.00\times10^{-6}}\\ &=5.00\times10^5\ \text{Hz} \end{aligned} \]

The transmitter sends \(5.00\times10^5\) bursts per second.

This is the burst repetition frequency, not the electromagnetic carrier frequency.

Step 6: Calculate average transmitted power.

Power is the total energy transferred per second.

Each burst carries \(1.00\times10^{-6}\ \text{J}\), and the transmitter sends \(5.00\times10^5\) bursts per second.

Therefore:

\[ P=U_{\text{burst}}f_{\text{rep}} \]

\[ \begin{aligned} P&=(1.00\times10^{-6})(5.00\times10^5)\\ &=0.50\ \text{W} \end{aligned} \]

The average transmitted power is \(0.50\ \text{W}\).

Step 7: Analyse the effect of changing the burst repetition period.

The new repetition period is:

\[ T_{\text{rep,new}}=1.00\ \mu\text{s} \]

The new repetition frequency is:

\[ f_{\text{rep,new}} =\frac{1}{1.00\times10^{-6}} =1.00\times10^6\ \text{Hz} \]

The transmitter now sends twice as many bursts each second.

However, the carrier waveform itself has not changed.

Therefore:

• Carrier period remains \(10.0\ \text{ns}\).
• Carrier frequency remains \(100\ \text{MHz}\).
• Energy per photon remains \(6.63\times10^{-26}\ \text{J}\).
• Photon count per burst remains \(1.51\times10^{19}\).
• Number of photons transmitted per second doubles.

The new average power is:

\[ \begin{aligned} P_{\text{new}} &=(1.00\times10^{-6})(1.00\times10^6)\\ &=1.00\ \text{W} \end{aligned} \]

The average power doubles to \(1.00\ \text{W}\).

Why is \(500\ \text{kHz}\) not the carrier frequency?

The tempting approach is to take the reciprocal of the \(2.00\ \mu\text{s}\) interval and call the result the electromagnetic wave’s frequency.

The arithmetic is correct, but the interpretation is wrong.

That interval measures how frequently bursts are transmitted. It does not measure the time taken for one oscillation of the electromagnetic carrier.

The carrier completes 20 cycles within each burst, with each cycle taking \(10.0\ \text{ns}\).

Consequently, the carrier frequency is \(100\ \text{MHz}\), while the burst repetition frequency is \(500\ \text{kHz}\).

Changing the burst repetition rate changes how much energy is transmitted each second. It does not change the energy carried by an individual photon when the carrier frequency remains constant.

Once you can reliably find frequency from period, the next useful step is deciding whether a question is asking about wavelength, individual photon energy, total photon count, or power. Each requires a different piece of information, even though they all connect back to the same electromagnetic wave.