How to Estimate Temperature Change from a Radiation Peak Shift

An infrared spectrometer shows a heated cavity’s radiation peak moving from \(4.00\,\mu\mathrm{m}\) to \(3.20\,\mu\mathrm{m}\). The peak wavelength has fallen by 20%. Has the temperature risen by 20%?

Not quite. Assuming the cavity emits ideal black-body radiation, its absolute temperature has risen by 25%:

\[
\frac{T_2}{T_1}
=\frac{\lambda_{\mathrm{peak},1}}
{\lambda_{\mathrm{peak},2}}
=\frac{4.00}{3.20}
=1.25
\]

The shorter peak wavelength tells you the object has become hotter. The inverse ratio tells you how much hotter.

01How to calculate temperature change from two peak wavelengths

Picture an electrically heated cavity with a small opening. Radiation escapes at many wavelengths, but one wavelength has the greatest emitted power per unit wavelength interval.

Increasing the cavity’s temperature changes the distribution. The peak moves towards shorter wavelengths, while the cavity continues emitting at wavelengths on both sides of that peak.

For an ideal black body, Wien’s displacement law describes this relationship:

\[
\lambda_{\mathrm{peak}}T=b
\]

Here:

• \(\lambda_{\mathrm{peak}}\) is the wavelength of maximum spectral intensity, measured in metres (m).
• \(T\) is the absolute temperature, measured in kelvin (K).
• \(b=2.90\times10^{-3}\ \mathrm{m\,K}\) is Wien’s displacement constant.

For two measurements of the same source, the constant \(b\) cancels:

\[
\frac{T_2}{T_1}
=\frac{\lambda_{\mathrm{peak},1}}
{\lambda_{\mathrm{peak},2}}
\]

Subscript 1 means before the change, and subscript 2 means after the change.

This is the quickest method when a question asks for the temperature ratio or percentage change.

Percentage change versus actual temperature change

A temperature ratio of \(1.25\) means the final temperature is 125% of the initial temperature, so the increase is 25%.

In general:

\[
\%\Delta T
=\left(\frac{T_2}{T_1}-1\right)100\%
\]

The percentage must be calculated using kelvin, not degrees Celsius.

If the question asks for the actual temperature change in kelvin, first find both temperatures:

\[
T_1=\frac{b}{\lambda_{\mathrm{peak},1}}
\]

\[
T_2=\frac{b}{\lambda_{\mathrm{peak},2}}
\]

Then subtract:

\[
\Delta T=T_2-T_1
\]

With this sign convention, a positive \(\Delta T\) means heating, while a negative value means cooling.

For ratios, both wavelengths simply need the same units. For absolute temperatures using the stated value of \(b\), convert wavelengths into metres:

• \(1\ \mu\mathrm{m}=10^{-6}\ \mathrm{m}\)
• \(1\ \mathrm{nm}=10^{-9}\ \mathrm{m}\)

If you need a refresher on finding a single temperature from one peak, see Wien’s Displacement Law: Finding Temperature from Peak Wavelength. Here, the extra challenge is comparing two measurements correctly and deciding whether their peaks are trustworthy.

02How the shift is measured in an experiment

Suppose you want to find out how much a hot object has warmed without putting a thermometer against it.

Pointing an infrared detector at the object is a reasonable first attempt. The detector might report a stronger signal after heating. But that alone cannot tell you where the radiation spectrum peaks.

The object could be hotter, closer to the detector, or simply presenting a larger emitting area.

To isolate temperature, we need to measure how radiation is distributed across wavelengths.

The apparatus: from the heated cavity to a spectrum

A suitable experimental arrangement works like this:

1. An electrically heated cavity is brought to a stable temperature. Its small opening provides radiation that closely approximates an ideal black-body spectrum.
2. Radiation leaves the opening and enters a narrow slit.
3. A diffraction grating separates the radiation into different directions according to wavelength.
4. An infrared detector is positioned to receive a narrow range of wavelengths. Its electrical output provides a measure of the radiation reaching it.
5. The selected wavelength is changed, and the signal is recorded repeatedly to build a spectrum.
6. The heater setting is increased, the cavity reaches a new stable temperature, and the scan is repeated.

The controlled variables include the aperture, optical alignment, detector distance, and wavelength calibration.

What should the second spectrum show?

For an ideal black body, the hotter spectrum should have its peak at a shorter wavelength. Its emitted spectral intensity per unit area should also be higher at every fixed wavelength.

That second prediction is useful. Heating doesn’t simply move a fixed-shaped hump to the left. It changes both the position and the height of the spectral distribution.

Why the first spectrum might be misleading

A detector doesn’t necessarily respond equally strongly to every wavelength.

Suppose it is much more sensitive at \(4\ \mu\mathrm{m}\) than at \(2\ \mu\mathrm{m}\). Its largest raw signal might occur near \(4\ \mu\mathrm{m}\), even though the source’s actual spectral maximum lies somewhere else.

The original measurement therefore needs another correction.

A more reliable procedure is to:

• Measure and subtract the background signal, including radiation from the apparatus.
• Calibrate which wavelength reaches the detector at each setting.
• Correct for the detector’s wavelength-dependent sensitivity.
• Compare signals per equal wavelength interval, rather than comparing unequal wavelength bands.
• Ensure the scan covers both sides of the peak.

The resulting calibrated spectrum can be used to estimate the peak wavelength.

This is an experimental measurement method, not a claim that the numerical examples below were recorded in a laboratory. All numerical scan data in this article are illustrative.

04Three different readings from the same radiation spectrum

A thermal radiation curve contains more information than its peak wavelength. Before using any value to estimate temperature change, make sure you’re reading the correct feature.

For example, spectral radiant exitance can be measured in \(\mathrm{W\,m^{-2}\,\mu m^{-1}}\). This is power per square metre of emitting surface per micrometre of wavelength.

When you include the entire wavelength range, the area under that curve gives emitted power per square metre, in \(\mathrm{W\,m^{-2}}\).

To find the total power emitted by a surface in watts, you must also account for its emitting area.

Extension: Stefan-Boltzmann checks the total emitted power

The Stefan-Boltzmann law concerns the total radiation emitted per unit surface area, rather than the position of the spectral peak.

For an ideal black body:

\[
M=\sigma T^4
\]

Here, \(M\) is total emitted radiant power per unit surface area in \(\mathrm{W\,m^{-2}}\), and \(\sigma=5.67\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}\) is the Stefan-Boltzmann constant.

For the same emitting area before and after heating, total emitted power \(P\), in watts, obeys:

\[
\frac{P_2}{P_1}
=\left(\frac{T_2}{T_1}\right)^4
\]

Consider the opening example, in which the temperature increased by 25%.

\[
\begin{aligned}
\frac{P_2}{P_1}&=(1.25)^4\\
&\approx2.44
\end{aligned}
\]

The ideal black body emits about 2.44 times as much total power from the same area, an increase of approximately 144%.

This does not mean the spectral intensity at every individual wavelength increases by 144%. The Stefan-Boltzmann law describes the whole spectrum combined.

Also, total power emitted by the source is not automatically the same as the power collected by a detector. Distance, collecting area, and optical transmission can change how much radiation reaches the instrument.

For a fuller comparison of peak position, spectral intensity, and curve area, see How to Compare Black-Body Curves at Different Temperatures.

05When a shift does not mean what you first expect

A longer peak means cooling, but watch the percentage

Suppose a cavity’s peak increases from \(3.00\ \mu\mathrm{m}\) to \(3.75\ \mu\mathrm{m}\).

The wavelength has increased by 25%. Does the temperature decrease by 25%?

No. Calculate the inverse ratio:

\[
\begin{aligned}
\frac{T_2}{T_1}&=\frac{3.00}{3.75}\\
&=0.80
\end{aligned}
\]

The final temperature is 80% of the initial absolute temperature. The temperature has decreased by 20%.

A 25% increase in wavelength and a 25% decrease in temperature are not inverse operations.

A dimmer spectrum does not necessarily mean a cooler source

Imagine placing a wavelength-independent attenuator between a hot cavity and its spectrometer.

The attenuator transmits the same fraction of radiation at every wavelength. Every part of the measured spectrum becomes lower by the same factor.

The peak stays at the same wavelength because the relative proportions of the different wavelengths are unchanged.

Now replace that attenuator with a filter that absorbs some wavelengths more strongly than others.

The measured spectrum can change shape, and its highest point may move even though the cavity’s temperature is constant.

This gives you an important experimental test: if the peak changes, first ask whether the source changed or whether something along the optical path changed the spectral balance.

07Decide whether Wien’s law can be used

Before calculating, identify what kind of shift the question actually describes.

One more detail: Wien’s constant \(b=2.90\times10^{-3}\ \mathrm{m\,K}\) applies to the maximum of a spectrum plotted per unit wavelength. A spectrum plotted per unit frequency has a differently positioned maximum when converted to a wavelength scale. Simply taking \(c/f_{\mathrm{peak}}\), where \(c\) is the speed of light and \(f_{\mathrm{peak}}\) is the frequency-spectrum maximum, does not give the wavelength-spectrum peak.

The equation is short. The difficult part of a real measurement is knowing whether the number called a “peak wavelength” actually represents the source’s thermal emission.

08Questions and solutions

Question 1

An ideal black-body cavity is initially at \(1000\ \mathrm{K}\), with a peak wavelength of \(2.90\ \mu\mathrm{m}\).

After the heater power increases, the peak wavelength becomes \(2.32\ \mu\mathrm{m}\).

Calculate:

a) The final temperature in kelvin.

b) The temperature increase in kelvin.

c) The percentage temperature increase.

Solution 1

The final temperature is \(1250\ \mathrm{K}\), representing an increase of \(250\ \mathrm{K}\), or 25%.

a) Final temperature

Using the inverse relationship between temperature and peak wavelength:

\[
\frac{T_2}{T_1}=\frac{\lambda_1}{\lambda_2}
\]

Substitute the measurements:

\[
\begin{aligned}
T_2&=1000\left(\frac{2.90}{2.32}\right)\\
&=1250\ \mathrm{K}
\end{aligned}
\]

The wavelength units cancel because both measurements are in micrometres.

b) Temperature increase

\[
\begin{aligned}
\Delta T&=T_2-T_1\\
&=1250-1000\\
&=250\ \mathrm{K}
\end{aligned}
\]

c) Percentage increase

\[
\begin{aligned}
\%\Delta T&=\frac{250}{1000}\times100\%\\
&=25\%
\end{aligned}
\]

The tempting mistake is to calculate the 20% decrease in wavelength and report a 20% increase in temperature. The inverse relationship gives 25%, not 20%.

Question 2

A black-body cavity’s initial complete spectrum has a peak wavelength of \(4.00\ \mu\mathrm{m}\).

After the heater setting is changed, a second spectrum is measured. The calibrated spectrometer initially covers only \(2.00\) to \(3.20\ \mu\mathrm{m}\), and the spectral intensity is still increasing at \(3.20\ \mu\mathrm{m}\).

A separate measurement at \(3.80\ \mu\mathrm{m}\) establishes that the spectral intensity is decreasing there.

Assume the source remains an ideal black body and all wavelength measurements are reliable. Use \(b=2.90\times10^{-3}\ \mathrm{m\,K}\).

a) Explain why \(3.20\ \mu\mathrm{m}\) cannot be taken as the second peak wavelength.

b) Determine the possible range of the second temperature.

c) Determine whether the cavity definitely heated or cooled, and give bounds for the temperature change.

d) Explain what further measurement would improve the estimate.

Solution 2

The cavity definitely heated, but its final temperature cannot be calculated exactly from these observations. It must lie between approximately \(763\) and \(906\ \mathrm{K}\), giving a temperature rise between approximately \(38\) and \(181\ \mathrm{K}\).

a) Why the measured boundary is not the peak

The first part of the second scan ends while spectral intensity is still increasing. Its largest recorded value is therefore not necessarily the maximum of the complete spectrum.

Because an ideal black-body wavelength spectrum has one maximum, and its intensity is increasing at \(3.20\ \mu\mathrm{m}\) but decreasing at \(3.80\ \mu\mathrm{m}\), the peak must satisfy:

\[
3.20<\lambda_2<3.80
\]

Here, all wavelengths are in micrometres.

b) Bounds on the second temperature

Using Wien’s law:

\[
T_2=\frac{b}{\lambda_2}
\]

Since temperature decreases as peak wavelength increases, the bounds reverse:

\[
\frac{b}{3.80\times10^{-6}}
<T_2<
\frac{b}{3.20\times10^{-6}}
\]

Substituting \(b=2.90\times10^{-3}\ \mathrm{m\,K}\):

\[
763.2\ \mathrm{K}<T_2<906.3\ \mathrm{K}
\]

c) Determine the direction and size of the change

The initial temperature was:

\[
\begin{aligned}
T_1&=\frac{2.90\times10^{-3}}
{4.00\times10^{-6}}\\
&=725\ \mathrm{K}
\end{aligned}
\]

Even the lowest possible final temperature is greater than \(725\ \mathrm{K}\).

Therefore, the cavity must have heated.

The temperature increase is bounded by:

\[
\begin{aligned}
\Delta T_{\min}&>763.2-725\\
&=38.2\ \mathrm{K}
\end{aligned}
\]

and

\[
\begin{aligned}
\Delta T_{\max}&<906.3-725\\
&=181.3\ \mathrm{K}
\end{aligned}
\]

Thus:

\[
38.2\ \mathrm{K}<\Delta T<181.3\ \mathrm{K}
\]

d) Improve the measurement

Scan the wavelength region between \(3.20\) and \(3.80\ \mu\mathrm{m}\) at finer intervals to locate the maximum more accurately.

The important distinction is that incomplete data may still allow a definite conclusion about the direction of temperature change, even when the exact change cannot be calculated.

Question 3

Extension: Combining Wien’s law, Stefan-Boltzmann radiation, and the inverse-square relationship.

A small black-body cavity aperture emits radiation towards a distant detector.

In the first measurement, the source’s peak wavelength is \(3.20\ \mu\mathrm{m}\).

After heating, the peak wavelength becomes \(2.40\ \mu\mathrm{m}\), but the total radiation power collected by the detector is only 79.0% of its earlier reading.

The cavity’s emitting area remains unchanged. The detector has the same collecting area, orientation, and sensitivity in both measurements, but its distance from the source may have changed.

Assume the detector is sufficiently far from the aperture that collected power follows an inverse-square relationship.

a) Find the temperature ratio and percentage temperature increase.

b) Determine the ratio of total power emitted by the cavity through its fixed aperture.

c) Determine the ratio of the detector distances in the second and first measurements.

d) Explain why the detector recorded less power even though the source became hotter.

Solution 3

The cavity becomes 33.3% hotter and emits approximately 3.16 times as much total power. The detector must have moved to approximately twice its original distance, causing its collected power to fall to 79.0% of the original reading.

a) Temperature change

Wien’s law gives:

\[
\begin{aligned}
\frac{T_2}{T_1}&=\frac{3.20}{2.40}\\
&=1.333\ldots
\end{aligned}
\]

Therefore:

\[
\%\Delta T=33.3\%
\]

b) Ratio of emitted power

The emitting area is constant. For an ideal black body, the Stefan-Boltzmann law gives:

\[
\begin{aligned}
\frac{P_2}{P_1}
&=\left(\frac{T_2}{T_1}\right)^4\\
&=\left(\frac{3.20}{2.40}\right)^4\\
&\approx3.16
\end{aligned}
\]

The cavity emits 3.16 times its initial total power.

c) Detector distance

Let \(D\) represent the power collected by the detector in watts, \(P\) the power emitted by the aperture in watts, and \(r\) the distance between the source and detector in metres.

For the specified geometry:

\[
D\propto\frac{P}{r^2}
\]

Therefore:

\[
\frac{D_2}{D_1}
=\frac{P_2}{P_1}
\left(\frac{r_1}{r_2}\right)^2
\]

Substitute the measurements:

\[
0.790=3.16\left(\frac{r_1}{r_2}\right)^2
\]

Rearrange:

\[
\begin{aligned}
\frac{r_2}{r_1}
&=\sqrt{\frac{3.16}{0.790}}\\
&=2.00
\end{aligned}
\]

The detector was twice as far away in the second measurement.

d) Explain the apparent contradiction

Doubling the distance reduces the collected power by a factor of four, under the inverse-square assumption.

Heating increases the emitted power by a factor of approximately 3.16.

Combining these effects:

\[
\begin{aligned}
\frac{D_2}{D_1}
&=\frac{3.16}{2^2}\\
&\approx0.790
\end{aligned}
\]

The detector receives less radiation because the increased distance more than compensates for the increased emission.

The tempting conclusion, “the detector got dimmer, so the source cooled”, confuses detected power with emitted power. The calibrated peak wavelengths still indicate heating.

Question 4

A black-body cavity is initially measured using a correctly calibrated infrared spectrometer. Its peak wavelength is \(3.00\ \mu\mathrm{m}\).

After the cavity’s heater setting is changed, the spectrometer reports a new peak wavelength of \(3.45\ \mu\mathrm{m}\).

However, the wavelength calibration may have changed between measurements.

During the second scan, two reference signals with known wavelengths produce the following instrument readings:

Assume the second scan’s wavelength readout is related to actual wavelength by:

\[
\lambda_{\mathrm{read}}
=a\lambda_{\mathrm{true}}+d
\]

Here, \(a\) is dimensionless, \(d\) is an offset measured in micrometres, and all wavelengths in this equation are expressed in micrometres.

The instrument’s intensity response remains calibrated, and the cavity’s emitting area is unchanged.

An independent radiometer also finds that total emitted power per unit surface area has increased by a factor of approximately 1.77.

a) Calculate \(a\) and \(d\).

b) Determine the corrected second peak wavelength.

c) Calculate the actual temperature change in kelvin and as a percentage.

d) Using the Stefan-Boltzmann law as an extension, assess whether the independent power measurement supports your result.

Solution 4

The cavity actually heated by approximately \(149\ \mathrm{K}\), or 15.4%, despite the reported peak apparently moving towards longer wavelengths. The independent radiation-power measurement supports this corrected result.

a) Determine the calibration relationship

We are given:

\[
\lambda_{\mathrm{read}}
=a\lambda_{\mathrm{true}}+d
\]

Using the two references:

\[
\begin{aligned}
2.70&=2.00a+d\\
5.20&=4.00a+d
\end{aligned}
\]

Subtract the first equation from the second:

\[
\begin{aligned}
2.50&=2.00a\\
a&=1.25
\end{aligned}
\]

Now substitute into the first equation:

\[
\begin{aligned}
2.70&=2.00(1.25)+d\\
d&=0.20\ \mu\mathrm{m}
\end{aligned}
\]

The second scan’s calibration relationship is:

\[
\lambda_{\mathrm{read}}
=1.25\lambda_{\mathrm{true}}+0.20
\]

b) Correct the second peak

The instrument reported \(3.45\ \mu\mathrm{m}\).

\[
\begin{aligned}
3.45&=1.25\lambda_2+0.20\\
\lambda_2&=\frac{3.45-0.20}{1.25}\\
&=2.60\ \mu\mathrm{m}
\end{aligned}
\]

The actual peak moved from \(3.00\) to \(2.60\ \mu\mathrm{m}\), not from \(3.00\) to \(3.45\ \mu\mathrm{m}\).

That reverses the initial temperature prediction.

c) Calculate the temperature change

Using Wien’s law with wavelengths in metres:

\[
\begin{aligned}
T_1&=\frac{2.90\times10^{-3}}
{3.00\times10^{-6}}\\
&=966.7\ \mathrm{K}\\[4pt]
T_2&=\frac{2.90\times10^{-3}}
{2.60\times10^{-6}}\\
&=1115.4\ \mathrm{K}
\end{aligned}
\]

The change is:

\[
\begin{aligned}
\Delta T&=1115.4-966.7\\
&\approx149\ \mathrm{K}
\end{aligned}
\]

The percentage increase is:

\[
\begin{aligned}
\%\Delta T
&=\left(\frac{3.00}{2.60}-1\right)100\%\\
&\approx15.4\%
\end{aligned}
\]

d) Compare with the independent radiation measurement

For the same ideal black-body surface:

\[
\begin{aligned}
\frac{M_2}{M_1}
&=\left(\frac{T_2}{T_1}\right)^4\\
&=\left(\frac{3.00}{2.60}\right)^4\\
&\approx1.77
\end{aligned}
\]

Here, \(M\) is total emitted power per square metre of surface.

The predicted factor of 1.77 agrees with the independent radiometer measurement.

This is stronger evidence than the uncorrected peak position alone. The corrected spectral measurement and the independently measured emitted power are consistent with the same increase in temperature.

The tempting approach was to apply Wien’s law immediately to the instrument’s displayed wavelengths. But a calibrated intensity signal cannot rescue an incorrect wavelength axis. The calibration error had to be corrected first.

Question 5

Two ideal black-body cavities, A and B, send radiation towards the same infrared spectrometer.

• Cavity A is maintained at \(1500\ \mathrm{K}\).
• Cavity B is maintained at \(750\ \mathrm{K}\).

Independent temperature probes confirm that neither cavity changes temperature during the experiment.

Each cavity has its own adjustable shutter. The shutters change how much radiation reaches the spectrometer, but their transmission is independent of wavelength. The source apertures may have different areas.

The wavelength calibration, detector sensitivity, and background correction have all been verified.

In the first scan, both shutters transmit only 1% of their respective available radiation, and no filter is installed. The combined spectrum has a measured peak at \(2.60\ \mu\mathrm{m}\).

Before the second scan, the shutter openings may have been increased, and a removable wavelength-selective infrared filter may have been inserted.

The second scan produces two surprising observations:

• The combined spectrum now peaks at \(1.70\ \mu\mathrm{m}\).
• The total power reaching the detector is 1.30 times its value in the first scan.

Assume only the shutters and possible filter insertion could have changed the optical setup.

a) Calculate the individual peak wavelengths of cavities A and B.

b) Can a change in shutter openings alone produce the second peak at \(1.70\ \mu\mathrm{m}\)? Establish a bound on the peak position of any unfiltered mixture of the two black-body spectra, using the shapes of the curves rather than Planck’s full radiation equation.

c) Calculate the temperatures that would be inferred by incorrectly applying Wien’s law to each combined-spectrum peak. Explain why these are not the actual temperatures of the cavities.

d) Determine what must have happened to the optical setup. Does the increase in total detected power rule out the insertion of a passive filter? Explain how the observations could coexist.

e) Propose one experimental test that would help verify your explanation.

Solution 5

The cavities did not heat. Under the stated experimental constraints, a wavelength-selective filter must have been inserted or otherwise introduced into the optical path. Changing the shutters alone cannot move the combined spectrum’s peak to \(1.70\ \mu\mathrm{m}\), even though increasing the shutter openings can explain the greater total detected power.

The key is that the peak of a mixture is not the peak of either component, and even a mixture of ideal black-body spectra is not generally a single black-body spectrum.

a) Calculate the two individual peaks

For cavity A:

\[
\begin{aligned}
\lambda_A&=\frac{b}{T_A}\\
&=\frac{2.90\times10^{-3}}{1500}\\
&=1.93\times10^{-6}\ \mathrm{m}\\
&=1.93\ \mu\mathrm{m}
\end{aligned}
\]

For cavity B:

\[
\begin{aligned}
\lambda_B&=\frac{b}{T_B}\\
&=\frac{2.90\times10^{-3}}{750}\\
&=3.87\times10^{-6}\ \mathrm{m}\\
&=3.87\ \mu\mathrm{m}
\end{aligned}
\]

Cavity A has the shorter peak because it is hotter.

b) Establish the bound for the combined spectrum

Consider wavelengths shorter than \(1.93\ \mu\mathrm{m}\).

At these wavelengths, both black-body spectral-intensity curves are increasing as wavelength increases towards their respective peaks.

The intensity from cavity A is increasing, and the intensity from cavity B is also increasing.

Adding two increasing curves, multiplied by positive constant factors, still gives an increasing curve.

Therefore, the combined unfiltered spectrum cannot reach its maximum below \(1.93\ \mu\mathrm{m}\).

Now consider wavelengths longer than \(3.87\ \mu\mathrm{m}\).

Both individual spectra are decreasing there. Their sum must also decrease.

Therefore, the combined spectrum cannot peak beyond \(3.87\ \mu\mathrm{m}\).

The maximum of any unfiltered mixture must consequently satisfy:

\[
1.93\leq\lambda_{\mathrm{peak}}
\leq3.87
\]

Here, wavelengths are measured in micrometres.

Changing the shutter openings changes the positive weighting of the two curves. It can move the combined peak within this interval, but not beyond either end.

The first observed peak, \(2.60\ \mu\mathrm{m}\), lies inside the interval and is physically possible.

The second observed peak, \(1.70\ \mu\mathrm{m}\), lies outside it.

So the second scan cannot result from shutter adjustments alone.

This conclusion doesn’t require the full mathematical form of Planck’s radiation law. It follows from the fact that each black-body wavelength spectrum increases before its peak and decreases after it.

c) Calculate the misleading apparent temperatures

If Wien’s law were incorrectly applied to the first combined peak:

\[
\begin{aligned}
T_{\mathrm{app},1}
&=\frac{2.90\times10^{-3}}
{2.60\times10^{-6}}\\
&\approx1115\ \mathrm{K}
\end{aligned}
\]

For the second combined peak:

\[
\begin{aligned}
T_{\mathrm{app},2}
&=\frac{2.90\times10^{-3}}
{1.70\times10^{-6}}\\
&\approx1706\ \mathrm{K}
\end{aligned}
\]

Here, \(T_{\mathrm{app}}\) means an apparent temperature produced by incorrectly treating the mixture as a single black body.

The first apparent temperature, \(1115\ \mathrm{K}\), lies between the actual cavity temperatures. That might look convincing, but neither cavity is at that temperature.

The second apparent temperature, \(1706\ \mathrm{K}\), is even higher than the hotter cavity’s actual temperature of \(1500\ \mathrm{K}\).

This is the deeper trap. An apparently reasonable temperature estimate from a combined peak can be wrong even when the numerical result lies between the actual temperatures.

d) Explain the filter and the increased detected power

A wavelength-selective filter changes the spectrum differently at different wavelengths.

Let \(F(\lambda)\) represent the filter’s transmission fraction at wavelength \(\lambda\). It is a dimensionless number between zero and one.

If \(I_{\mathrm{mix}}(\lambda)\) is the combined spectral intensity before filtering, the transmitted spectrum is:

\[
I_{\mathrm{out}}(\lambda)
=F(\lambda)I_{\mathrm{mix}}(\lambda)
\]

Unlike a neutral shutter, \(F(\lambda)\) need not be constant.

A filter that transmits radiation strongly around \(1.70\ \mu\mathrm{m}\) while suppressing longer wavelengths can make the measured output peak near \(1.70\ \mu\mathrm{m}\).

That does not mean either source’s intrinsic spectrum now peaks there.

But what about the increased total detected power? Doesn’t a passive filter reduce power?

For fixed shutter openings, a passive filter cannot increase the total power reaching the detector. However, the shutters were also allowed to change.

Initially, each shutter transmitted only 1% of its available radiation. Opening both shutters fully would increase the total radiation entering the common optical path by a factor of 100 before filtering.

A filter could then transmit only a small fraction of that larger input and still deliver more power than the first scan.

For example, if the fully opened shutters admit 100 times the first scan’s power, an overall filter transmission of 1.3% would give:

\[
\begin{aligned}
\frac{D_2}{D_1}
&=100\times0.013\\
&=1.30
\end{aligned}
\]

Here, \(D_1\) and \(D_2\) are the total powers reaching the detector before and after the change, in watts. The 1.3% is an illustrative spectrum-weighted average transmission, not a constant transmission at every wavelength.

A suitably selective passband near \(1.70\ \mu\mathrm{m}\) could also produce the observed spectral maximum.

Both observations are therefore compatible with unchanged source temperatures, larger shutter openings, and a wavelength-selective filter.

The misleading intuition was that greater total detected power must mean greater temperature. In this experiment, detected power is affected by both the radiation sources and the optical system.

e) Test the explanation

Keep both shutter settings fixed and repeat the scan with the filter removed.

If the source temperatures remain constant and the instrument is correctly calibrated, the unfiltered combined spectrum must have its peak somewhere between approximately \(1.93\) and \(3.87\ \mu\mathrm{m}\).

The recorded peak should no longer remain at \(1.70\ \mu\mathrm{m}\).

A second useful test would be to block each cavity in turn and measure its individual unfiltered spectrum. Those peaks should occur near \(1.93\) and \(3.87\ \mu\mathrm{m}\), respectively.

This leads to the next important decision in radiation measurements: when a source contains multiple temperatures, or when its light has been filtered, a single measured peak is no longer a reliable thermometer. You need to examine the full spectral shape and model the contributing sources before assigning a physical temperature.