Why Red and Blue Beams with Equal Power Have Different Photon Rates

Two laser beams each deliver exactly \(1.00\ \text{mW}\) of optical power. One produces red light with a wavelength of \(650\ \text{nm}\). The other produces blue light with a wavelength of \(450\ \text{nm}\).

Which beam sends more photons towards a target every second?

The red beam does. Each blue photon carries more energy than each red photon, so the red beam needs to send more photons to deliver the same total energy each second. In fact, it sends approximately \(1.44\) times as many photons.

This is the key distinction: photon energy and photon rate are two different quantities. A beam can contain fewer, more energetic photons and still deliver exactly the same power as a beam containing more, less energetic photons.

01Why equal power does not mean equal photon numbers

Think about paying for something using different denominations of banknotes.

If you have to pay exactly $100, you could use twenty $5 notes or ten $10 notes. The total value is identical, but the number of notes is different.

Photons work in a similar way when we compare energy.

Each photon carries a particular amount of energy. If a beam delivers a fixed amount of energy every second, using more energetic photons means fewer photons are needed.

The analogy is only about counting energy packets. Photons are not physical banknotes, and their energy is determined by their electromagnetic frequency.

Three quantities you must keep separate

Power is energy transferred per unit time:

\[ P=\frac{E_{\text{total}}}{t} \]

One watt is equivalent to one joule transferred every second.

If a beam contains \(N\) identical photons, its total energy is:

\[ E_{\text{total}}=NE_\gamma \]

The photon rate is:

\[ R=\frac{N}{t} \]

Combining these relationships gives:

\[ P=\frac{NE_\gamma}{t} \]

Therefore:

\[ \boxed{P=RE_\gamma} \]

This equation is the starting point for understanding why equal-power beams can have different photon rates.

02Why blue photons carry more energy than red photons

The energy of an individual photon depends on its frequency.

Planck’s equation gives:

\[ E_\gamma=hf \]

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 frequency, measured in hertz (Hz).

Frequency and wavelength are connected through the wave equation:

\[ c=f\lambda \]

Here, \(c=3.00\times10^8\ \text{m s}^{-1}\) is the speed of light in a vacuum, and \(\lambda\) is wavelength in metres.

Rearranging:

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

Substituting into Planck’s equation:

\[ E_\gamma=h\frac{c}{\lambda} \]

Therefore:

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

Notice the inverse relationship.

• Shorter wavelength means higher photon energy.
• Longer wavelength means lower photon energy.

Blue light has a shorter wavelength than red light. Therefore, an individual blue photon carries more energy.

Comparing red and blue photon energies

Use the following illustrative wavelengths:

The energy ratio is:

\[ \frac{E_{\text{blue}}}{E_{\text{red}}} = \frac{4.42\times10^{-19}} {3.06\times10^{-19}} \]

\[ \frac{E_{\text{blue}}}{E_{\text{red}}} \approx1.44 \]

Each blue photon carries approximately 1.44 times as much energy as each red photon.

But this still doesn’t tell us how many photons either beam contains.

For that, we need to bring optical power back into the calculation.

03The relationship between power and photon rate

We already established that:

\[ P=RE_\gamma \]

Substitute the photon energy equation:

\[ P=R\frac{hc}{\lambda} \]

Rearrange to make photon rate the subject:

\[ \boxed{R=\frac{P\lambda}{hc}} \]

This is the important equation for comparing photon rates when optical power and wavelength are known.

Notice what happens when power is held constant:

\[ \boxed{R\propto\lambda} \]

The photon rate is directly proportional to wavelength when optical power is constant.

This means:

• Increasing wavelength increases photon rate.
• Decreasing wavelength decreases photon rate.

So, at equal optical power, a red beam sends more photons each second than a blue beam.

Comparing two beams directly

For two monochromatic beams with equal optical power:

\[ P_{\text{red}}=P_{\text{blue}} \]

Using the photon-rate equation:

\[ \frac{R_{\text{red}}}{R_{\text{blue}}} = \frac{P_{\text{red}}\lambda_{\text{red}}} {P_{\text{blue}}\lambda_{\text{blue}}} \]

Since their powers are equal, they cancel:

\[ \boxed{ \frac{R_{\text{red}}}{R_{\text{blue}}} = \frac{\lambda_{\text{red}}}{\lambda_{\text{blue}}} } \]

For our original example:

\[ \frac{R_{\text{red}}}{R_{\text{blue}}} = \frac{650}{450} \]

\[ \boxed{ \frac{R_{\text{red}}}{R_{\text{blue}}}=1.44 } \]

The red beam delivers 1.44 times as many photons every second.

What if the photon rates are equal instead?

Now reverse the situation.

Suppose both beams emit exactly the same number of photons every second.

The blue beam has more power because each blue photon carries more energy.

In this case:

\[ P\propto\frac{1}{\lambda} \]

when photon rate is constant.

This gives us an important comparison:

Be careful not to swap these two situations in an exam question.

Optical power is not the same as brightness

There is another distinction worth making.

Optical power measures the total energy delivered each second. It does not tell you how concentrated the beam is.

Suppose two beams have equal optical power, but one is focused onto a much smaller area. That beam has greater intensity at the target.

Intensity is power per unit area:

\[ I=\frac{P}{A} \]

where \(I\) is intensity in \(\text{W m}^{-2}\), and \(A\) is the beam’s cross-sectional area in square metres.

Similarly, photon flux density describes the number of photons arriving per unit area each second.

Equal optical power does not automatically mean equal photon flux density or equal perceived brightness.

Human vision also responds differently to different wavelengths, so equal optical power does not necessarily make red and blue beams look equally bright.

04Why different wavelengths have different applications

The energy carried by each photon helps explain why different wavelengths are useful for different technologies.

Consider these examples.

The wavelength is not the only factor determining an application’s suitability. Material absorption, detector sensitivity, transmission through different materials, and safety all matter.

The important connection is that changing wavelength changes the energy carried by each photon. That can change how light interacts with matter.

For a deeper explanation of the relationship between photon energy and atomic energy levels, see Photon energy and atomic transitions: HSC Physics guide.

05How could an experiment test this prediction?

So far, we’ve used equations to predict that a red beam should have a greater photon rate than a blue beam when their optical powers are equal.

How could an experiment check this?

We need to measure optical power and photon rate separately.

Setting up the experiment

Imagine two stable, narrow-band laser sources:

• A red laser with a wavelength of \(650\ \text{nm}\).
• A blue laser with a wavelength of \(450\ \text{nm}\).

Each laser sends light along a separate optical path.

Place a calibrated variable attenuator in each path. These devices allow us to reduce the optical power without changing the laser’s wavelength.

Direct the beams, one at a time, onto the same calibrated optical power meter. Position the meter at the same target plane for both measurements.

Adjust the attenuators until the power meter reads:

\[ P_{\text{red}}=P_{\text{blue}}=1.00\ \text{mW} \]

The two beams now deliver the same optical energy each second at the measurement position.

What should the experiment show?

If we could count every photon arriving at the measurement position, the red beam should produce more detection events each second.

Using:

\[ R=\frac{P\lambda}{hc} \]

For the red beam:

\[ R_{\text{red}} = \frac{(1.00\times10^{-3})(650\times10^{-9})} {(6.63\times10^{-34})(3.00\times10^8)} \]

\[ R_{\text{red}}\approx3.27\times10^{15} \ \text{photons s}^{-1} \]

For the blue beam:

\[ R_{\text{blue}} = \frac{(1.00\times10^{-3})(450\times10^{-9})} {(6.63\times10^{-34})(3.00\times10^8)} \]

\[ R_{\text{blue}}\approx2.26\times10^{15} \ \text{photons s}^{-1} \]

The predicted ratio is:

\[ \frac{R_{\text{red}}}{R_{\text{blue}}} = \frac{3.27}{2.26} \approx1.44 \]

The red beam should deliver approximately 44% more photons each second.

These are calculated predictions, not reported experimental measurements.

Why can’t we simply count every photon?

There is a practical problem.

At \(1.00\ \text{mW}\), the beams deliver billions of billions of photons every second. A typical photon-counting detector cannot register every photon individually at that rate.

We need to reduce the optical power before attempting direct photon counting.

For example, suppose calibrated attenuation reduces both beams to:

\[ P=1.00\times10^{-15}\ \text{W} \]

This is one femtowatt.

The expected photon rates become:

• Red: approximately \(3.27\times10^3\) photons per second.
• Blue: approximately \(2.26\times10^3\) photons per second.

These rates are much more suitable for a photon-counting measurement.

However, there is another experimental limitation.

A detector might be more sensitive to one wavelength than another. For example, it might register a greater fraction of incoming red photons than blue photons.

If we simply compare the raw detector counts, we could mistake differences in detector sensitivity for differences in photon rate.

Revising the measurement

A more reliable experiment calibrates the detector’s efficiency at both wavelengths.

Let:

• \(C\) be the number of recorded detection pulses.
• \(t\) be the measurement time in seconds.
• \(g\) be the fraction of the beam intercepted by the detector.
• \(\eta\) be the detector’s efficiency, expressed as a decimal.

The actual photon emission rate is:

\[ R=\frac{C}{g\eta t} \]

The detector efficiency and collection fraction correct for photons that were not recorded.

We must also account for background counts and ensure that the detector is operating below its saturation limit.

The revised experiment should produce a corrected red-to-blue photon-rate ratio of approximately 1.44 when the optical powers are equal.

The conclusion is that the difference is a property of the photon energies and the energy delivered per second, not simply a difference in detector response.

06Worked examples

07Questions and solutions

Question 1

Two monochromatic beams emit exactly \(2.0\times10^{15}\) photons per second.

The red beam has a wavelength of \(640\ \text{nm}\), while the blue beam has a wavelength of \(480\ \text{nm}\).

Calculate the ratio of their optical powers and identify which beam has greater power.

Solution 1

The blue beam has 1.33 times the optical power of the red beam.

The important information is that the photon rates are equal. Therefore, the beam containing more energetic photons must deliver more energy each second.

Start with:

\[ P=RE_\gamma \]

Since both beams have the same photon rate:

\[ \frac{P_{\text{blue}}}{P_{\text{red}}} = \frac{E_{\text{blue}}}{E_{\text{red}}} \]

Using the wavelength relationship:

\[ \frac{P_{\text{blue}}}{P_{\text{red}}} = \frac{\lambda_{\text{red}}}{\lambda_{\text{blue}}} \]

Substitute:

\[ \frac{P_{\text{blue}}}{P_{\text{red}}} = \frac{640}{480} \]

\[ \boxed{ \frac{P_{\text{blue}}}{P_{\text{red}}}=1.33 } \]

The blue photons have shorter wavelengths and therefore greater individual energies.

Since both beams contain the same number of photons each second, the blue beam transfers more energy per second.

Question 2

Two monochromatic laser beams each deliver \(2.0\ \text{mW}\) of optical power onto a screen.

The red beam has a wavelength of \(640\ \text{nm}\) and illuminates an area of \(8.0\ \text{mm}^2\).

The blue beam has a wavelength of \(480\ \text{nm}\) and illuminates an area of \(2.0\ \text{mm}^2\).

Assume both beams have uniform intensity across their illuminated areas.

Calculate:

(a) The photon rate of each beam.

(b) The photon flux density of each beam.

(c) Which beam produces the greater photon flux density, and by what factor?

Solution 2

The red beam has the greater total photon rate, but the blue beam produces three times the photon flux density.

This is because total photon rate depends on wavelength and optical power, while photon flux density also depends on the area receiving the light.

Part (a): Calculate the photon rates.

Use:

\[ R=\frac{P\lambda}{hc} \]

For the red beam:

\[ R_{\text{red}} = \frac{(2.0\times10^{-3})(640\times10^{-9})} {(6.63\times10^{-34})(3.00\times10^8)} \]

\[ R_{\text{red}} = 6.44\times10^{15}\ \text{photons s}^{-1} \]

For the blue beam:

\[ R_{\text{blue}} = \frac{(2.0\times10^{-3})(480\times10^{-9})} {(6.63\times10^{-34})(3.00\times10^8)} \]

\[ R_{\text{blue}} = 4.83\times10^{15}\ \text{photons s}^{-1} \]

The red beam has the greater total photon rate.

Part (b): Calculate photon flux density.

Photon flux density is the number of photons arriving per unit area each second.

We use:

\[ \Phi=\frac{R}{A} \]

where \(\Phi\) is photon flux density in photons \(\text{m}^{-2}\text{s}^{-1}\).

Convert the areas into square metres:

\[ A_{\text{red}}=8.0\times10^{-6}\ \text{m}^2 \]

\[ A_{\text{blue}}=2.0\times10^{-6}\ \text{m}^2 \]

For red:

\[ \Phi_{\text{red}} = \frac{6.44\times10^{15}}{8.0\times10^{-6}} \]

\[ \Phi_{\text{red}} = 8.05\times10^{20} \ \text{photons m}^{-2}\text{s}^{-1} \]

For blue:

\[ \Phi_{\text{blue}} = \frac{4.83\times10^{15}}{2.0\times10^{-6}} \]

\[ \Phi_{\text{blue}} = 2.42\times10^{21} \ \text{photons m}^{-2}\text{s}^{-1} \]

Part (c): Compare the flux densities.

\[ \frac{\Phi_{\text{blue}}}{\Phi_{\text{red}}} = \frac{2.42\times10^{21}} {8.05\times10^{20}} \]

\[ \boxed{ \frac{\Phi_{\text{blue}}}{\Phi_{\text{red}}}=3.00 } \]

The blue beam produces three times the photon flux density of the red beam.

The tempting mistake is to assume that the beam with more photons overall must also have more photons arriving per square metre.

However, the blue beam concentrates its photons onto a much smaller area. That changes the local photon flux density without changing the total photon rate.

Question 3

A photocathode has a work function of \(2.30\ \text{eV}\).

It is illuminated by two monochromatic beams of equal optical power, each delivering \(0.50\ \mu\text{W}\).

• Red beam: \(650\ \text{nm}\).
• Blue beam: \(450\ \text{nm}\).

Assume the ordinary one-photon photoelectric effect applies, with no multiphoton processes.

(a) Determine whether each beam can eject photoelectrons.

(b) Calculate the maximum kinetic energy of the emitted electrons.

(c) Assuming every incident blue photon is absorbed and ejects one electron, calculate the resulting photocurrent.

(d) Explain why the beam with the greater photon rate does not necessarily produce the greater photocurrent.

Solution 3

Only the blue beam ejects photoelectrons. Its maximum electron kinetic energy is approximately \(0.46\ \text{eV}\), and the idealised photocurrent is approximately \(0.18\ \mu\text{A}\).

The reason is that photoelectric emission depends on the energy of each individual photon, not simply the number of photons arriving each second.

Part (a): Determine whether photoelectrons are emitted.

Convert the photon energies into electronvolts.

Using:

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

and:

\[ 1\ \text{eV}=1.60\times10^{-19}\ \text{J} \]

For the red beam:

\[ E_{\text{red}} = \frac{(6.63\times10^{-34})(3.00\times10^8)} {650\times10^{-9}} \]

\[ E_{\text{red}}=3.06\times10^{-19}\ \text{J} \]

Converting to electronvolts:

\[ E_{\text{red}}= \frac{3.06\times10^{-19}} {1.60\times10^{-19}} \]

\[ E_{\text{red}}=1.91\ \text{eV} \]

For the blue beam:

\[ E_{\text{blue}} = \frac{(6.63\times10^{-34})(3.00\times10^8)} {450\times10^{-9}} \]

\[ E_{\text{blue}}=4.42\times10^{-19}\ \text{J} \]

Converting:

\[ E_{\text{blue}}=2.76\ \text{eV} \]

Compare these values with the work function:

\[ \phi=2.30\ \text{eV} \]

The red photon energy is below the work function:

\[ 1.91<2.30 \]

Therefore, red photons cannot eject electrons through the ordinary one-photon photoelectric effect.

The blue photon energy exceeds the work function:

\[ 2.76>2.30 \]

Therefore, blue photons can eject electrons.

Part (b): Calculate the maximum kinetic energy.

Einstein’s photoelectric equation is:

\[ K_{\max}=E_\gamma-\phi \]

For the blue photons:

\[ K_{\max}=2.76-2.30 \]

\[ \boxed{K_{\max}=0.46\ \text{eV}} \]

The emitted electrons can have a maximum kinetic energy of \(0.46\ \text{eV}\).

Part (c): Calculate the photocurrent.

First, calculate the blue photon rate.

Convert the optical power:

\[ P=0.50\times10^{-6}\ \text{W} \]

Use:

\[ R=\frac{P\lambda}{hc} \]

\[ R_{\text{blue}} = \frac{(0.50\times10^{-6})(450\times10^{-9})} {(6.63\times10^{-34})(3.00\times10^8)} \]

\[ R_{\text{blue}} = 1.13\times10^{12}\ \text{photons s}^{-1} \]

Under the stated ideal assumption, each photon produces one electron.

The current is:

\[ I=\frac{Q}{t} \]

Since each electron carries a charge of magnitude \(e=1.60\times10^{-19}\ \text{C}\):

\[ I=eR \]

Substitute:

\[ I=(1.60\times10^{-19})(1.13\times10^{12}) \]

\[ \boxed{I=1.81\times10^{-7}\ \text{A}} \]

Or:

\[ \boxed{I=0.181\ \mu\text{A}} \]

Part (d): Explain the difference.

The red beam actually has the greater photon rate because it has the longer wavelength and both beams have equal optical power.

Its photon rate is approximately 1.44 times the blue photon rate.

However, red photons do not have enough energy to overcome the work function. Therefore, they cannot produce photoelectrons in this model.

This demonstrates an important distinction: having more photons does not guarantee greater photocurrent. Each photon must also have sufficient energy to release an electron.

In a real photocathode, reflection, absorption, and quantum efficiency would also affect the measured current.

Question 4

An experiment compares two monochromatic beams with equal optical power.

The red beam has a wavelength of \(600\ \text{nm}\), while the blue beam has a wavelength of \(400\ \text{nm}\).

A photon-counting detector is used to measure the incoming photons.

The detector has the following efficiencies:

During separate measurement intervals of \(1.0\ \text{s}\), the detector records:

Assume that the detector intercepts the entire beam, background counts are negligible, and the detector does not become saturated.

(a) Calculate the actual photon rate of each beam.

(b) Determine whether the results are consistent with the beams having equal optical power.

(c) Explain why comparing the raw detector counts would give an incorrect photon-rate ratio.

Solution 4

The corrected photon rates are \(1.2\times10^5\) photons per second for red and \(8.0\times10^4\) photons per second for blue.

The results are consistent with equal optical power. The raw detector counts exaggerate the difference because the detector is more efficient at detecting red photons.

Part (a): Correct the measured counts.

The detector records only a fraction of the photons reaching it.

Therefore:

\[ R=\frac{C}{\eta t} \]

where \(C\) is the recorded count, \(\eta\) is the detector efficiency, and \(t\) is the measurement time.

For red:

\[ R_{\text{red}} = \frac{9.0\times10^4}{0.75\times1.0} \]

\[ \boxed{ R_{\text{red}}=1.2\times10^5\ \text{photons s}^{-1} } \]

For blue:

\[ R_{\text{blue}} = \frac{4.0\times10^4}{0.50\times1.0} \]

\[ \boxed{ R_{\text{blue}}=8.0\times10^4\ \text{photons s}^{-1} } \]

Part (b): Check whether the beams have equal power.

Calculate the corrected photon-rate ratio:

\[ \frac{R_{\text{red}}}{R_{\text{blue}}} = \frac{1.2\times10^5}{8.0\times10^4} \]

\[ \frac{R_{\text{red}}}{R_{\text{blue}}}=1.50 \]

For equal optical power, the expected ratio is:

\[ \frac{R_{\text{red}}}{R_{\text{blue}}} = \frac{\lambda_{\text{red}}}{\lambda_{\text{blue}}} \]

\[ =\frac{600}{400} \]

\[ =1.50 \]

The measured ratio agrees with the theoretical prediction.

We can also verify this by calculating the power of each beam.

For red:

\[ P_{\text{red}} = \frac{R_{\text{red}}hc}{\lambda_{\text{red}}} \]

\[ P_{\text{red}} = \frac{(1.2\times10^5)(6.63\times10^{-34})(3.00\times10^8)} {600\times10^{-9}} \]

\[ P_{\text{red}}=3.98\times10^{-14}\ \text{W} \]

For blue:

\[ P_{\text{blue}} = \frac{(8.0\times10^4)(6.63\times10^{-34})(3.00\times10^8)} {400\times10^{-9}} \]

\[ P_{\text{blue}}=3.98\times10^{-14}\ \text{W} \]

Therefore:

\[ \boxed{P_{\text{red}}=P_{\text{blue}}} \]

Part (c): Explain the raw-count error.

The raw count ratio is:

\[ \frac{9.0\times10^4}{4.0\times10^4}=2.25 \]

If we assumed every incoming photon was detected, we would incorrectly conclude that the red beam has 2.25 times the photon rate.

However, the detector registers 75% of red photons but only 50% of blue photons.

Correcting for these different efficiencies reduces the actual ratio to 1.50.

This is why experimental measurements must account for detector sensitivity before drawing conclusions about photon rates.

Question 5

Two light sources, A and B, each deliver exactly \(1.00\ \text{mW}\) of optical power to a measurement plane.

Both sources contain only two narrow spectral bands:

• Blue light at \(450\ \text{nm}\).
• Red light at \(750\ \text{nm}\).

Their optical power is distributed differently.

A student argues:

“Both sources contain the same two wavelengths, and the average of 450 nm and 750 nm is 600 nm. They have equal total power, so they must have the same photon rate.”

(a) Evaluate this claim by calculating the total photon rate of each source.

(b) A filter is then placed in the path of source B. It transmits 25% of the blue optical power and 100% of the red optical power. Calculate the total transmitted optical power and photon rate.

(c) Compare the filtered photon rate of source B with the original photon rate of source A.

(d) Explain why using the arithmetic average of the two wavelengths produces an incorrect prediction for source B.

Solution 5

Source A has the greater original photon rate, emitting approximately \(3.02\times10^{15}\) photons per second compared with \(2.57\times10^{15}\) for source B.

After filtering, source B emits approximately \(1.21\times10^{15}\) photons per second. The important mistake in the student’s argument is treating the two wavelengths as equally weighted when calculating the average.

Part (a): Calculate the original photon rates.

For a single wavelength:

\[ R=\frac{P\lambda}{hc} \]

For a source containing multiple wavelengths, calculate the photon rate of each component separately and add them:

\[ R_{\text{total}} = \sum_i\frac{P_i\lambda_i}{hc} \]

This is important because different wavelengths carry different amounts of energy per photon.

Using \(P\) in milliwatts and \(\lambda\) in nanometres, the conversion factor is:

\[ \frac{(1.0\times10^{-3})(1.0\times10^{-9})} {1.989\times10^{-25}} = 5.03\times10^{12} \]

Thus, each numerical value of \(P_i\lambda_i\), expressed in \(\text{mW nm}\), is multiplied by \(5.03\times10^{12}\) to obtain the photon rate in photons per second.

For source A:

\[ P_{\text{blue}}\lambda_{\text{blue}} = 0.50\times450=225 \]

\[ P_{\text{red}}\lambda_{\text{red}} = 0.50\times750=375 \]

Adding the contributions:

\[ R_A=(225+375)(5.03\times10^{12}) \]

\[ \boxed{ R_A=3.02\times10^{15}\ \text{photons s}^{-1} } \]

For source B:

\[ P_{\text{blue}}\lambda_{\text{blue}} = 0.80\times450=360 \]

\[ P_{\text{red}}\lambda_{\text{red}} = 0.20\times750=150 \]

Therefore:

\[ R_B=(360+150)(5.03\times10^{12}) \]

\[ \boxed{ R_B=2.57\times10^{15}\ \text{photons s}^{-1} } \]

Compare the two rates:

\[ \frac{R_A}{R_B} = \frac{600}{510} \]

\[ \boxed{ \frac{R_A}{R_B}=1.18 } \]

Source A emits approximately 18% more photons each second.

Both sources have equal total power, but source A allocates more of its power to the longer-wavelength red photons.

Those photons carry less energy individually, so more photons are required to carry the same amount of energy.

Part (b): Calculate the transmitted power and photon rate of source B.

The filter transmits 25% of the blue power and all of the red power.

The transmitted blue power is:

\[ P_{\text{blue,out}}=0.80\times0.25 \]

\[ P_{\text{blue,out}}=0.20\ \text{mW} \]

The transmitted red power is:

\[ P_{\text{red,out}}=0.20\times1.00 \]

\[ P_{\text{red,out}}=0.20\ \text{mW} \]

Therefore, the total transmitted optical power is:

\[ P_{\text{out}}=0.20+0.20 \]

\[ \boxed{P_{\text{out}}=0.40\ \text{mW}} \]

Now calculate the photon rate of each transmitted component.

Using the same conversion factor:

\[ R_{\text{out}} = (0.20\times450+0.20\times750) (5.03\times10^{12}) \]

\[ R_{\text{out}} = (90+150)(5.03\times10^{12}) \]

\[ \boxed{ R_{\text{out}}=1.21\times10^{15}\ \text{photons s}^{-1} } \]

Part (c): Compare the filtered source B with the original source A.

We have:

\[ R_A=3.02\times10^{15}\ \text{photons s}^{-1} \]

\[ R_{B,\text{out}}=1.21\times10^{15}\ \text{photons s}^{-1} \]

The ratio is:

\[ \frac{R_A}{R_{B,\text{out}}} = \frac{3.02}{1.21} \]

\[ \boxed{ \frac{R_A}{R_{B,\text{out}}}=2.50 } \]

Source A now delivers 2.50 times as many photons each second as filtered source B.

Notice that the filter has also reduced source B’s optical power from \(1.00\ \text{mW}\) to \(0.40\ \text{mW}\).

We cannot continue treating the two sources as equal-power beams after this change.

Part (d): Explain the average-wavelength misconception.

The student’s arithmetic average is:

\[ \lambda_{\text{average}} = \frac{450+750}{2} = 600\ \text{nm} \]

But this assumes the two wavelengths contribute equally to the relevant average.

Source A distributes its optical power equally between the two wavelengths. Source B does not.

A more useful quantity for this calculation is the power-weighted average wavelength:

\[ \lambda_{\text{effective}} = \frac{\sum_i P_i\lambda_i}{P_{\text{total}}} \]

For source A:

\[ \lambda_{\text{effective,A}} = \frac{0.50(450)+0.50(750)}{1.00} \]

\[ \lambda_{\text{effective,A}}=600\ \text{nm} \]

For source B:

\[ \lambda_{\text{effective,B}} = \frac{0.80(450)+0.20(750)}{1.00} \]

\[ \lambda_{\text{effective,B}}=510\ \text{nm} \]

The effective wavelengths are different because the optical power distributions are different.

This explains why the two sources have different photon rates despite having the same total power and the same two spectral components.

The deeper lesson is that a mixed-wavelength beam cannot generally be treated as though every photon has the energy associated with the arithmetic average wavelength.

You must account for how much optical power is carried at each wavelength.

08Where to take this concept next

The most useful habit to carry into other HSC Physics questions is to identify what the question holds constant before choosing an equation.

• Equal optical power: longer wavelength means more photons per second.
• Equal photon rate: shorter wavelength means greater optical power.
• Equal electrical input: calculate optical efficiency before comparing photon rates.
• Multiple wavelengths: calculate each component separately and add the photon rates.

Once you understand the difference between photon energy and photon number, the next step is to examine what happens when those photons interact with matter.

In particular, atomic transitions depend on the energy of individual photons, while the total number of photons determines how many energy-transfer events may occur.

The reverse calculation is also useful in exam questions. If you are given a photon emission rate and asked to calculate optical power, see How to Calculate Light Power from Photon Emission Rate.