301. A plane electromagnetic wave of frequency \(6.0\times10^8\,\text{Hz}\) travels along positive \(x\) through vacuum. At a certain event its electric field is at a positive maximum along \(+\hat j\), and its magnetic field is at a positive maximum along \(+\hat k\). A second observation is made at a position \(0.50\,\text{m}\) farther along positive \(x\) and at a time \(1.0\,\text{ns}\) later. Assume a sinusoidal travelling wave and take \(c=3.0\times10^8\,\text{m s}^{-1}\). Which conclusion correctly gives the signed electric and magnetic fields at the second event relative to the first?
ⓐ. Both fields reverse sign because the detector displacement equals one full wavelength.
ⓑ. Both fields are reduced to zero because the spatial and temporal phase changes add to an odd quarter-cycle.
ⓒ. The electric field reverses sign but the magnetic field does not, because they oscillate in different phases.
ⓓ. Both fields retain their original signs because the elapsed time exactly compensates the spatial phase change.
Correct Answer: Both fields retain their original signs because the elapsed time exactly compensates the spatial phase change.
Explanation: \(\textbf{Calculate the wavelength and period:}\)
The frequency is
\[
f=6.0\times10^8\,\text{Hz}
\]
Therefore,
\[
\lambda=\frac cf
\]
\[
\lambda=\frac{3.0\times10^8}{6.0\times10^8}
\]
\[
\lambda=0.50\,\text{m}
\]
The period is
\[
T=\frac1f
\]
\[
T=1.67\times10^{-9}\,\text{s}
\]
\[
T\approx1.67\,\text{ns}
\]
\(\textbf{Determine the spatial phase change:}\)
The observation position changes by
\[
\Delta x=0.50\,\text{m}
\]
For a wave travelling along positive \(x\),
\[
\phi=kx-\omega t
\]
Therefore,
\[
\Delta\phi_x=k\Delta x
\]
\[
\Delta\phi_x=\frac{2\pi}{\lambda}(0.50)
\]
\[
\Delta\phi_x=2\pi
\]
\(\textbf{Determine the temporal phase change:}\)
The elapsed time is
\[
\Delta t=1.0\,\text{ns}
\]
Thus,
\[
\Delta\phi_t=-\omega\Delta t
\]
\[
\Delta\phi_t=-2\pi f\Delta t
\]
\[
\Delta\phi_t=-2\pi(6.0\times10^8)(1.0\times10^{-9})
\]
\[
\Delta\phi_t=-1.2\pi
\]
\(\textbf{Combine the phase changes:}\)
\[
\Delta\phi=\Delta\phi_x+\Delta\phi_t
\]
\[
\Delta\phi=2\pi-1.2\pi
\]
\[
\Delta\phi=0.8\pi
\]
\(\textbf{Evaluate the second fields:}\)
At the first event, the phase corresponds to a positive maximum.
At the second event, the fields are multiplied by
\[
\cos(0.8\pi)\approx-0.809
\]
Therefore, both the electric and magnetic fields become negative relative to their original positive directions.
\(\textbf{Physical interpretation:}\)
The spatial displacement alone contributes one complete cycle.
However, the elapsed time introduces an additional nonzero phase change.
The two effects do not cancel exactly.
The electric and magnetic fields remain in phase with each other, so they change sign together.
The correct field ratio remains \(E/B=c\) whenever the fields are nonzero.
\(\textbf{Final answer:}\)
\[
\boxed{\vec E_2\approx-0.809E_0\hat j,\qquad\vec B_2\approx-0.809B_0\hat k}
\]
302. Two freely propagating sinusoidal plane electromagnetic waves, P and Q, travel through separate non-magnetic transparent media. Their frequencies are equal, and the electric-field amplitudes at their respective detectors are \(E_{0P}=120\,\text{V m}^{-1}\) and \(E_{0Q}=80\,\text{V m}^{-1}\). The relative permittivities are \(\varepsilon_{rP}=4\) and \(\varepsilon_{rQ}=9\). Assume \(\mu_r=1\) for both media, negligible absorption, and use
\[
v=\frac{c}{\sqrt{\varepsilon_r}},\qquad
\langle u\rangle=\frac12\varepsilon E_0^2,\qquad
I=v\langle u\rangle.
\]
Which conclusion correctly compares the average electromagnetic intensities at the detectors?
ⓐ. P has twice Q's intensity because its electric-field amplitude is \(3/2\) times as large.
ⓑ. Q has the greater intensity because its permittivity is greater.
ⓒ. The two average intensities are equal despite the different electric-field amplitudes and medium properties.
ⓓ. P has \(3/2\) times Q's intensity because its propagation speed is greater.
Correct Answer: The two average intensities are equal despite the different electric-field amplitudes and medium properties.
Explanation: \(\textbf{Determine the wave speeds:}\)
For the specified non-magnetic transparent media,
\[
v=\frac c{\sqrt{\varepsilon_r}}
\]
Therefore,
\[
v_P=\frac c{\sqrt4}
\]
\[
v_P=\frac c2
\]
For Q,
\[
v_Q=\frac c{\sqrt9}
\]
\[
v_Q=\frac c3
\]
\(\textbf{Express the electromagnetic intensities:}\)
The stated average total energy density is
\[
\langle u\rangle=\frac12\varepsilon E_0^2
\]
The average intensity is
\[
I=v\langle u\rangle
\]
Therefore,
\[
I=\frac12v\varepsilon E_0^2
\]
Using
\[
\varepsilon=\varepsilon_r\varepsilon_0,
\]
we obtain
\[
I=\frac12v\varepsilon_r\varepsilon_0E_0^2
\]
\(\textbf{Calculate the ratio:}\)
\[
\frac{I_P}{I_Q}
=\frac{v_P\varepsilon_{rP}E_{0P}^2}{v_Q\varepsilon_{rQ}E_{0Q}^2}
\]
Substituting,
\[
\frac{I_P}{I_Q}
=\frac{(c/2)(4)(120)^2}{(c/3)(9)(80)^2}
\]
Canceling the common factor \(c\),
\[
\frac{I_P}{I_Q}
=\frac{2(14400)}{3(6400)}
\]
\[
\frac{I_P}{I_Q}=\frac{28800}{19200}
\]
\[
\frac{I_P}{I_Q}=\frac32
\]
\(\textbf{Physical interpretation:}\)
P has the larger electric-field amplitude and greater wave speed.
Q has the larger permittivity.
These quantities enter the intensity relation through different factors.
The larger permittivity of Q partly compensates for its smaller electric-field amplitude and lower speed.
However, the compensation is not complete for the supplied values.
Therefore, P has \(3/2\) times Q's average intensity.
The calculation demonstrates why comparing electric-field amplitudes alone is insufficient when waves propagate through different media.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{I_P}{I_Q}=\frac32}
\]
303. A monochromatic plane electromagnetic wave propagates through vacuum along the positive \(z\)-axis. At one observation point, two perpendicular electric-field probes record the following time dependences:
\[
E_x(t)=30\cos(\omega t)\,\text{V m}^{-1},
\]
\[
E_y(t)=40\cos(\omega t)\,\text{V m}^{-1}.
\]
A student claims that the field direction rotates continuously within the transverse plane because both \(x\)- and \(y\)-components are present. Which conclusion is correct?
ⓐ. The electric-field direction rotates continuously, and the wave must be longitudinal.
ⓑ. The electric field oscillates along the fixed direction \(3\hat i+4\hat j\); its magnetic-field amplitude is \(50/c\).
ⓒ. The electric field has no fixed oscillation direction because its two components have different amplitudes.
ⓓ. The electric field oscillates along positive \(z\), while its magnetic field lies in the \(xy\)-plane.
Correct Answer: The electric field oscillates along the fixed direction \(3\hat i+4\hat j\); its magnetic-field amplitude is \(50/c\).
Explanation: The two electric-field components have the same angular frequency.
They also have the same phase factor \(\cos(\omega t)\).
Therefore, the complete electric-field vector can be written as
\[
\vec E(t)=(30\hat i+40\hat j)\cos(\omega t).
\]
The quantity in brackets has a fixed direction.
Only the common scalar factor varies with time.
Consequently, the field oscillates forwards and backwards along one fixed line in the transverse \(xy\)-plane.
The electric-field direction does not continuously rotate merely because two perpendicular components are present.
The fixed amplitude vector is
\[
\vec E_0=30\hat i+40\hat j.
\]
Its magnitude is
\[
E_0=\sqrt{30^2+40^2}
\]
\[
E_0=50\,\text{V m}^{-1}.
\]
The oscillation-axis direction is proportional to \(3\hat i+4\hat j\).
Because the wave travels along positive \(z\), its electric field is transverse to propagation.
For the freely propagating plane wave in vacuum,
\[
B_0=\frac{E_0}{c}.
\]
Therefore,
\[
B_0=\frac{50}{c}.
\]
The corresponding magnetic-field direction remains perpendicular to both the electric field and positive-\(z\) propagation.
If the electric-field components had different phase relations, the field direction could vary differently.
However, no such relative phase difference is present in the measured expressions.
The conclusion follows from factoring out the common time dependence rather than from comparing the component amplitudes alone.
304. A plane electromagnetic wave is described by
\[
\vec E(x,t)=E_0\hat j\cos(kx-\omega t)
\]
and
\[
\vec B(x,t)=B_0\hat k\cos(kx-\omega t+\delta).
\]
Independent measurements confirm \(E_0/B_0=\omega/k=c\), where \(c\) is the vacuum speed of light. The phase difference \(\delta\) is unknown. A detector at \(x=0\) records both electric and magnetic fields at \(t=0\). It measures \(E_y=+E_0\) and \(B_z=0\). Which conclusion is justified?
ⓐ. The proposed field pair is a valid freely propagating vacuum plane wave because its amplitudes and phase speed are correct.
ⓑ. The pair fails the phase requirement for a freely propagating vacuum plane wave, despite correct amplitude and speed ratios.
ⓒ. The fields are valid only if the propagation direction is reversed from positive \(x\) to negative \(x\).
ⓓ. The fields are valid because electric-field and magnetic-field maxima must occur a quarter-cycle apart.
Correct Answer: The pair fails the phase requirement for a freely propagating vacuum plane wave, despite correct amplitude and speed ratios.
Explanation: The electric-field expression has phase \(kx-\omega t\).
The magnetic-field expression has phase \(kx-\omega t+\delta\).
The wave's amplitude ratio is correctly specified as \(E_0/B_0=c\).
Its phase-speed ratio is also correctly specified as \(\omega/k=c\).
However, these conditions are not sufficient by themselves to establish a valid freely propagating plane electromagnetic wave.
The electric and magnetic fields of the standard vacuum plane-wave solution oscillate in phase.
At \(x=0,t=0\), the electric field is at its positive maximum:
\[
E_y(0,0)=E_0.
\]
If the fields were in phase, the magnetic field would simultaneously have the corresponding positive maximum:
\[
B_z(0,0)=B_0.
\]
Instead, the magnetic probe records zero.
The magnetic-field expression therefore requires
\[
\cos\delta=0
\]
at the specified event.
This corresponds to an inappropriate quarter-cycle phase difference between the two field oscillations.
Reversing the claimed propagation direction would not remove that relative temporal phase mismatch.
Nor does a correct amplitude ratio compensate for inconsistent phase behaviour.
The field axes are already perpendicular to each other and to the stated positive-\(x\) propagation direction.
Thus, the failure lies specifically in the relative phase, not in transverse geometry or the numerical propagation-speed ratio.
A physically consistent field pair must satisfy direction, amplitude and phase requirements simultaneously.
The measurements therefore expose a defect that could be missed by checking only \(E_0/B_0\) and \(\omega/k\).
305. A short electromagnetic pulse travels through vacuum and reaches a detector after passing two separate absorbing sheets. The pulse contains three distinguishable spectral components: visible, ultraviolet and X-ray radiation. The incident pulse energies and transmission fractions of the sheets are:
| Spectral component | Incident energy | Sheet P transmission | Sheet Q transmission |
|---|
| Visible | \(40\,\text{mJ}\) | \(80\%\) | \(50\%\) |
| Ultraviolet | \(20\,\text{mJ}\) | \(20\%\) | \(25\%\) |
| X-rays | \(10\,\text{mJ}\) | \(50\%\) | \(40\%\) |
Assume no frequency conversion, negligible reflection back into the sheets and no additional propagation losses. Which statement correctly identifies the largest and smallest spectral contributions reaching the detector and their energy ratio?
ⓐ. X-rays are largest, ultraviolet is smallest, with ratio \(2:1\).
ⓑ. Visible is largest, X-rays are smallest, with ratio \(8:1\).
ⓒ. Ultraviolet is largest, X-rays are smallest, with ratio \(2:1\).
ⓓ. Visible is largest, ultraviolet is smallest, with ratio \(16:1\).
Correct Answer: Visible is largest, ultraviolet is smallest, with ratio \(16:1\).
Explanation: \(\textbf{Apply both transmission fractions separately:}\)
Each spectral component passes through Sheet P and then Sheet Q.
For any one component,
\[
U_{\text{out}}=U_{\text{in}}T_PT_Q
\]
The transmission fractions differ by spectral region.
Therefore, the detected energy distribution must be calculated separately for each component.
\(\textbf{Calculate transmitted visible energy:}\)
The incident visible energy is
\[
U_{V0}=40\,\text{mJ}
\]
Its overall transmission fraction is
\[
T_V=(0.80)(0.50)
\]
\[
T_V=0.40
\]
Therefore,
\[
U_V=40(0.40)
\]
\[
U_V=16\,\text{mJ}
\]
\(\textbf{Calculate transmitted ultraviolet energy:}\)
The incident ultraviolet energy is
\[
U_{\text{UV},0}=20\,\text{mJ}
\]
The overall transmission is
\[
T_{\text{UV}}=(0.20)(0.25)
\]
\[
T_{\text{UV}}=0.05
\]
Thus,
\[
U_{\text{UV}}=20(0.05)
\]
\[
U_{\text{UV}}=1.0\,\text{mJ}
\]
\(\textbf{Calculate transmitted X-ray energy:}\)
The incident X-ray energy is
\[
U_{X0}=10\,\text{mJ}
\]
The overall transmission is
\[
T_X=(0.50)(0.40)
\]
\[
T_X=0.20
\]
Therefore,
\[
U_X=10(0.20)
\]
\[
U_X=2.0\,\text{mJ}
\]
\(\textbf{Compare the detected spectral components:}\)
The detected energies are
\[
U_V=16\,\text{mJ}
\]
\[
U_{\text{UV}}=1.0\,\text{mJ}
\]
\[
U_X=2.0\,\text{mJ}
\]
Therefore, visible radiation is the largest contribution and ultraviolet the smallest.
Their energy ratio is
\[
\frac{U_V}{U_{\text{UV}}}=\frac{16}{1}
\]
\[
\frac{U_V}{U_{\text{UV}}}=16
\]
\(\textbf{Physical interpretation:}\)
The highest-frequency component need not deliver the greatest total energy to the detector.
The detected energy depends on the source energy and the cumulative transmission through the intervening materials.
All three components remain electromagnetic radiation and share the same vacuum propagation speed.
Their different transmission fractions result from differing interactions with the sheets rather than differing fundamental propagation laws.
\(\textbf{Final answer:}\)
\[
\boxed{U_V:U_{\text{UV}}=16:1}
\]
306. A laboratory source produces a mixture of X-rays and gamma rays with overlapping wavelength ranges. The emitted radiation is separated into two beams using their sources rather than their wavelengths. Beam P is produced by energetic electrons striking a metal target. Beam Q is emitted by excited atomic nuclei. Both beams are observed to produce ionisation in suitable detector matter. Which statement correctly evaluates the following claims?
Statement I: The common ability to produce ionisation does not establish that the beams have the same production mechanism.
Statement II: The wavelengths of X-rays and gamma rays may overlap even though the conventional source descriptions differ.
Statement III: Beam P must consist of the target-striking electrons because it can cause ionisation.
Statement IV: Beam Q is electromagnetic radiation even though its emission originates in nuclear transitions.
ⓐ. Statements I, II and IV only
ⓑ. Statements I and III only
ⓒ. Statements II, III and IV only
ⓓ. Statements I, II and III only
Correct Answer: Statements I, II and IV only
Explanation: Statement I is correct because radiation detection and radiation production describe different physical stages.
Both X-rays and gamma rays can interact with suitable matter and produce ionisation.
An ionisation detector responds to charged particles produced or affected through such interactions.
This common detector response does not establish identical source mechanisms.
Statement II is correct because the conventional X-ray and gamma-ray spectrum regions are not separated by an absolutely sharp, non-overlapping wavelength boundary.
Their characteristic wavelength ranges can overlap.
In such cases, the production mechanism remains an important distinction.
Energetic electrons striking a metal target characteristically produce X-rays.
Transitions in excited atomic nuclei are characteristic sources of gamma radiation.
Statement III is incorrect because the X-rays emitted by the metal target are electromagnetic radiation.
They are distinct from the material electrons that struck the target.
The ability of radiation to produce ionisation does not require the incoming beam to consist of those original electrons.
Statement IV is correct because nuclear emission can produce electromagnetic gamma radiation.
A gamma ray is not itself an emitted massive nucleus merely because it originates from a nuclear transition.
Both emitted beams can propagate through vacuum.
They travel at the common electromagnetic-wave speed \(c\).
Thus, the distinction between them can involve their sources even when their wavelengths or detector interactions are similar.
The correct combination is Statements I, II and IV.
307. An infrared observation instrument contains two independent detector channels. Channel P measures radiation in a wavelength band centred near \(2\,\mu\text{m}\), while Channel Q measures radiation near \(10\,\mu\text{m}\). Both channels view the same object through the same region of atmosphere. The emitted powers and atmospheric transmission fractions are:
| Band | Emitted power | Atmospheric transmission |
|---|
| P | \(24\,\text{W}\) | \(25\%\) |
| Q | \(15\,\text{W}\) | \(60\%\) |
The instrument is then relocated so that the atmospheric transmission in P's band doubles, while the transmission in Q's band is reduced to half its previous value. The source emission and detector geometry remain unchanged. Which statement correctly compares the original and final detected power ratios \(P:Q\)?
ⓐ. Originally \(2:3\), finally \(8:3\).
ⓑ. Originally \(3:2\), finally \(4:3\).
ⓒ. Originally \(2:3\), finally \(3:8\).
ⓓ. Originally \(3:2\), finally \(3:4\).
Correct Answer: Originally \(2:3\), finally \(8:3\).
Explanation: \(\textbf{Calculate the original detected powers:}\)
The detected power in a spectral band is
\[
P_{\text{det}}=TP_{\text{emitted}}
\]
For the shorter-wavelength infrared band P,
\[
P_{P1}=0.25(24)
\]
\[
P_{P1}=6.0\,\text{W}
\]
For the longer-wavelength infrared band Q,
\[
P_{Q1}=0.60(15)
\]
\[
P_{Q1}=9.0\,\text{W}
\]
Therefore, the original detected power ratio is
\[
P_{P1}:P_{Q1}=6:9
\]
\[
P_{P1}:P_{Q1}=2:3
\]
\(\textbf{Apply the changed atmospheric transmission in P:}\)
The transmission fraction in P's band doubles:
\[
T_{P2}=2(0.25)
\]
\[
T_{P2}=0.50
\]
The new detected power is
\[
P_{P2}=0.50(24)
\]
\[
P_{P2}=12\,\text{W}
\]
\(\textbf{Apply the changed transmission in Q:}\)
The transmission fraction in Q's band is halved:
\[
T_{Q2}=\frac{0.60}{2}
\]
\[
T_{Q2}=0.30
\]
Therefore,
\[
P_{Q2}=0.30(15)
\]
\[
P_{Q2}=4.5\,\text{W}
\]
\(\textbf{Compare the final powers:}\)
\[
P_{P2}:P_{Q2}=12:4.5
\]
Multiplying both terms by two,
\[
P_{P2}:P_{Q2}=24:9
\]
\[
P_{P2}:P_{Q2}=8:3
\]
\(\textbf{Interpret the reversal:}\)
Originally, Q produces the greater detector signal despite having the smaller emitted power.
After the atmospheric change, P produces the greater detected signal.
The object itself has not changed its emitted powers.
The reversal arises from the wavelength-dependent transmission of the atmosphere.
This illustrates why infrared measurements must distinguish the radiation emitted by an object from the radiation that reaches the detector.
Different spectral bands may be attenuated differently even along the same observation path.
\(\textbf{Final answer:}\)
\[
\boxed{(P:Q)_{\text{original}}=2:3,\qquad(P:Q)_{\text{final}}=8:3}
\]
308. A medical-imaging detector is illuminated by an X-ray beam that has passed through an object containing two regions. The transmitted powers reaching the detector are \(P_A=12\,\text{mW}\) from Region A and \(P_B=8\,\text{mW}\) from Region B. A uniform protective filter is then inserted before the detector. It transmits \(50\%\) of the X-rays from both regions. The detector's response is linear in incident X-ray power, and its background response is negligible. Which statement is correct?
ⓐ. The absolute difference in detector powers is unchanged, but their ratio is halved.
ⓑ. Both the absolute difference and the ratio of detector powers are halved.
ⓒ. The absolute difference in detector powers is halved, but the ratio of detector powers remains unchanged.
ⓓ. The detector powers become equal because the same filter is placed in both radiation paths.
Correct Answer: The absolute difference in detector powers is halved, but the ratio of detector powers remains unchanged.
Explanation: \(\textbf{Identify the initial detector signals:}\)
The X-ray power reaching the detector from Region A is
\[
P_A=12\,\text{mW}
\]
From Region B,
\[
P_B=8\,\text{mW}
\]
The initial ratio is
\[
\frac{P_A}{P_B}=\frac{12}{8}
\]
\[
\frac{P_A}{P_B}=\frac32
\]
The initial absolute difference is
\[
\Delta P_1=P_A-P_B
\]
\[
\Delta P_1=12-8
\]
\[
\Delta P_1=4.0\,\text{mW}
\]
\(\textbf{Apply the common filter:}\)
The filter transmits the same fraction of either beam:
\[
T=0.50
\]
Therefore,
\[
P'_A=TP_A
\]
\[
P'_A=0.50(12)
\]
\[
P'_A=6.0\,\text{mW}
\]
For Region B,
\[
P'_B=TP_B
\]
\[
P'_B=0.50(8)
\]
\[
P'_B=4.0\,\text{mW}
\]
\(\textbf{Compare the new signals:}\)
The new power ratio is
\[
\frac{P'_A}{P'_B}=\frac6{4}
\]
\[
\frac{P'_A}{P'_B}=\frac32
\]
Thus, the ratio is unchanged.
However, the new absolute difference is
\[
\Delta P_2=6.0-4.0
\]
\[
\Delta P_2=2.0\,\text{mW}
\]
Therefore,
\[
\frac{\Delta P_2}{\Delta P_1}=\frac{2.0}{4.0}
\]
\[
\frac{\Delta P_2}{\Delta P_1}=\frac12
\]
\(\textbf{Physical interpretation:}\)
The filter reduces the transmitted X-ray powers from both regions by the same fraction.
Their relative ratio remains unchanged because the common factor cancels.
However, the absolute difference between the detector readings becomes smaller.
This distinction is important when interpreting measurements based on transmitted electromagnetic radiation.
A common attenuation can preserve relative transmission contrast while reducing the available signal magnitude.
The reduction in detector exposure does not automatically imply that every aspect of image quality remains unchanged.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{P'_A}{P'_B}=\frac32,\qquad\frac{\Delta P_2}{\Delta P_1}=\frac12}
\]
309. A laboratory performs three experiments involving electromagnetic phenomena.
Experiment P: A capacitor is being charged, and a magnetic-field probe near its insulating gap detects a field associated with changing electric flux. Experiment Q: A transmitting aerial emits a radio pulse that continues travelling through vacuum after the source is switched off. Experiment R: A suitable detector records gamma radiation emitted by excited nuclei, although no charged nuclear particles are observed travelling from the source to the detector.
Consider the following statements.
Statement I: The magnetic field in P can be explained using Maxwell's displacement-current contribution without requiring conduction electrons to cross the capacitor gap.
Statement II: The continued travel of the pulse in Q is consistent with electromagnetic energy propagation without a material medium.
Statement III: The detector observation in R proves that the gamma radiation is a beam of charged nuclei.
Statement IV: The observations in P and Q are connected conceptually because time-varying electric and magnetic fields are central to electromagnetic-wave theory.
Which combination is correct?
ⓐ. Statements I, II and III only
ⓑ. Statements I, II and IV only
ⓒ. Statements II, III and IV only
ⓓ. Statements I, III and IV only
Correct Answer: Statements I, II and IV only
Explanation: Statement I is correct because the magnetic field near a charging capacitor gap is associated with a changing electric flux.
Maxwell introduced the displacement-current contribution
\[
I_d=\varepsilon_0\frac{d\Phi_E}{dt}
\]
to account for this changing-field effect.
The displacement current does not require electrons to travel across the insulating gap.
The conduction current in the connecting wires and displacement current in the gap represent different physical mechanisms.
Statement II is correct because electromagnetic radiation can propagate through vacuum.
Once a radio pulse has been emitted, it carries electromagnetic energy away from the source.
Switching off the transmitting aerial prevents subsequent emission but does not immediately eliminate a pulse already travelling through space.
The electromagnetic disturbance does not require the original source charges to accompany it to a distant receiver.
Statement III is incorrect because gamma radiation is electromagnetic radiation associated with nuclear processes.
Its origin in a nuclear transition does not mean that the emitted gamma ray is a charged nucleus.
A suitable detector can register gamma radiation through interactions with matter, including ionisation.
The charged particles collected inside a detector need not be the original incident radiation.
Statement IV is correct because Maxwell's displacement-current contribution connects changing electric fields with magnetic fields.
Faraday's electromagnetic-induction principle connects changing magnetic fields with electric fields.
These linked field relationships provide the qualitative physical basis for electromagnetic-wave propagation.
Thus, Experiment P illustrates an important changing-field mechanism, while Experiment Q illustrates the propagation of an electromagnetic disturbance.
The three experiments also demonstrate the distinction between radiation sources, propagating electromagnetic fields and detector responses.
The correct combination is Statements I, II and IV.
310. In a simplified recreation of Hertz's electromagnetic-wave experiment, a high-voltage arrangement produces repeated sparks across a small gap in a transmitting circuit. A separate receiving loop, having a narrow spark gap, is placed several metres away without any conducting connection to the transmitter. Small sparks appear in the receiver when the transmitter operates. An opaque insulating screen is then placed between the two circuits, and the receiver sparks continue to appear. Which interpretation is most appropriate?
ⓐ. Electrons from the transmitter must travel through the air and cross the receiving gap to produce the observed sparks.
ⓑ. The receiving sparks establish that the transmitted disturbance is a longitudinal mechanical wave.
ⓒ. The observation proves that the transmitting circuit produces only a static electric field because the screen is insulating.
ⓓ. Time-varying transmitter charges and currents can radiate electromagnetic waves that induce effects in the disconnected receiver.
Correct Answer: Time-varying transmitter charges and currents can radiate electromagnetic waves that induce effects in the disconnected receiver.
Explanation: The transmitting arrangement contains a rapidly varying electrical disturbance associated with repeated sparks.
Accelerated charges can generate electromagnetic radiation.
The resulting disturbance contains time-varying electric and magnetic fields.
These fields can propagate through space without requiring a conducting connection between the source and receiver.
A suitably arranged receiving loop can respond to the incident electromagnetic disturbance.
The electrical response may become sufficiently large to produce small sparks across its own gap.
The receiver sparks do not require electrons from the transmitting circuit to travel all the way to the receiving circuit.
Charges already present in the receiver respond to the arriving electromagnetic fields.
The insulating screen is not specified to block the electromagnetic radiation effectively.
Its failure to prevent reception is therefore compatible with electromagnetic-wave propagation.
The screen's electrical insulating property alone does not establish its behaviour towards every electromagnetic wavelength.
A longitudinal mechanical disturbance is not required to explain the receiving-loop response.
Hertz's experiments provided important experimental confirmation of electromagnetic waves predicted by Maxwell's theory.
The key observation is that an electrical disturbance can produce a detectable effect at a separated receiver.
The experiment also distinguishes radiation propagation from conduction through a continuous material circuit.
311. A laboratory compares two electromagnetic-wave experiments. Experiment P uses a transmitting aerial operating at \(150\,\text{MHz}\). Experiment Q uses a short-wavelength source producing radiation of wavelength \(5.0\,\text{mm}\), similar in scale to the millimetre-wave investigations associated with Jagadish Chandra Bose. Both radiations propagate through vacuum. A detector in Experiment Q responds only to frequencies between \(50\) and \(70\,\text{GHz}\). Taking \(c=3.0\times10^8\,\text{m s}^{-1}\), which statement correctly compares the radiations and the detector compatibility?
ⓐ. Experiment Q has \(400\) times the frequency of P and is compatible with the specified detector.
ⓑ. Experiment P has \(400\) times the frequency of Q, while Q lies in the infrared region.
ⓒ. Experiment Q has \(40\) times the frequency of P, but its wavelength is too short for the detector.
ⓓ. Both experiments have equal frequencies because they involve electromagnetic waves in vacuum.
Correct Answer: Experiment Q has \(400\) times the frequency of P and is compatible with the specified detector.
Explanation: \(\textbf{Determine the frequency of Experiment P:}\)
The stated radio frequency is
\[
f_P=150\,\text{MHz}
\]
Converting,
\[
f_P=1.50\times10^8\,\text{Hz}
\]
Its corresponding vacuum wavelength is
\[
\lambda_P=\frac{c}{f_P}
\]
\[
\lambda_P=\frac{3.0\times10^8}{1.50\times10^8}
\]
\[
\lambda_P=2.0\,\text{m}
\]
\(\textbf{Calculate the frequency of the millimetre wave:}\)
For Experiment Q,
\[
\lambda_Q=5.0\,\text{mm}
\]
\[
\lambda_Q=5.0\times10^{-3}\,\text{m}
\]
Using
\[
f_Q=\frac{c}{\lambda_Q},
\]
we obtain
\[
f_Q=\frac{3.0\times10^8}{5.0\times10^{-3}}
\]
\[
f_Q=6.0\times10^{10}\,\text{Hz}
\]
\[
f_Q=60\,\text{GHz}
\]
\(\textbf{Compare the source frequencies:}\)
\[
\frac{f_Q}{f_P}=\frac{6.0\times10^{10}}{1.50\times10^8}
\]
\[
\frac{f_Q}{f_P}=400
\]
\(\textbf{Check detector compatibility:}\)
The Q detector accepts frequencies from \(50\) to \(70\,\text{GHz}\).
The calculated source frequency is \(60\,\text{GHz}\).
Therefore, Q lies within the detector's stated operating range.
Its wavelength belongs to the millimetre-wave portion of the microwave region.
\(\textbf{Physical interpretation:}\)
The two experiments operate at very different wavelengths.
Nevertheless, both investigate electromagnetic radiation.
Jagadish Chandra Bose's work on short electromagnetic waves helped demonstrate their properties and experimental applications.
Different wavelengths require appropriately designed sources and receivers.
Their differing frequencies do not imply differing speeds through vacuum.
The comparison therefore connects historical experimental work with the frequency–wavelength relation and practical receiver compatibility.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{f_Q}{f_P}=400,\qquad f_Q=60\,\text{GHz}}
\]
312. Three charged particles undergo different motions:
Particle P moves at constant velocity along a straight line in vacuum. Particle Q moves at constant speed along a circular path under a suitable external force. Particle R performs simple harmonic motion along a straight line and is observed at an extreme position, where its instantaneous velocity is zero. Assume that all three particles are electrically charged and that radiation associated with acceleration can be considered.
Which conclusion about their potential electromagnetic-radiation production is correct?
ⓐ. Only P can radiate because its velocity remains constant.
ⓑ. Only Q can radiate because an accelerating charged particle must have nonzero instantaneous speed.
ⓒ. Q and R can radiate because both undergo acceleration, even though Q's speed is constant and R is instantaneously at rest.
ⓓ. All three must radiate equally because they are electrically charged.
Correct Answer: Q and R can radiate because both undergo acceleration, even though Q's speed is constant and R is instantaneously at rest.
Explanation: Accelerated electric charges can produce electromagnetic radiation.
The relevant physical property is acceleration rather than merely the existence of electric charge.
Particle P moves with constant velocity along a straight line.
Its velocity vector therefore does not change with time.
Consequently, its acceleration is zero under the stated conditions.
Uniform straight-line motion alone does not provide the acceleration responsible for electromagnetic radiation.
Particle Q moves with constant speed along a circular path.
Although its speed remains unchanged, its velocity direction continuously changes.
Its acceleration is directed towards the centre of the circular path.
Therefore, Q is an accelerated charged particle and can radiate.
Particle R performs simple harmonic motion.
At an extreme position, its instantaneous velocity is zero.
However, the restoring acceleration has maximum magnitude at an extreme position.
Thus, R can also produce electromagnetic radiation associated with its acceleration.
The absence of instantaneous motion at the turning point does not imply the absence of acceleration.
Likewise, constant speed does not imply zero acceleration when the direction of motion changes.
The three cases distinguish velocity magnitude from acceleration of the full velocity vector.
The conclusion does not require the radiation powers of Q and R to be equal.
Such a comparison would require additional information about their motions and charges.
Therefore, Q and R are the appropriate accelerating-charge cases.
313. An ideal capacitor has constant capacitance and initially stores a positive plate charge of \(5.0\,\mu\text{C}\). Its signed charging current follows the graph described below.
The horizontal axis is time \(t\), measured in milliseconds, from \(0\) to \(6\,\text{ms}\). The vertical axis is signed capacitor current \(i\), measured in milliamperes. From \(0\) to \(2\,\text{ms}\), the current is constant at \(+4.0\,\text{mA}\). From \(2\) to \(4\,\text{ms}\), it is constant at \(-3.0\,\text{mA}\). From \(4\) to \(6\,\text{ms}\), it is zero. The current changes instantaneously at the stated boundaries in the idealised model. Positive current increases the stated positive plate charge.
A magnetic-field probe surrounds the capacitor gap with a fixed orientation, and its field direction follows the sign of the enclosed displacement current. Which conclusion correctly gives the final plate charge and the magnetic-field behaviour during the final interval?
ⓐ. Final charge \(7.0\,\mu\text{C}\); the displacement-current contribution and its associated magnetic-field circulation vanish during \(4\)–\(6\,\text{ms}\).
ⓑ. Final charge \(13\,\mu\text{C}\); the magnetic field remains in its original direction throughout the final interval.
ⓒ. Final charge \(-1.0\,\mu\text{C}\); the displacement current remains negative during the final interval.
ⓓ. Final charge \(7.0\,\mu\text{C}\); the magnetic field remains nonzero because the capacitor retains charge.
Correct Answer: Final charge \(7.0\,\mu\text{C}\); the displacement-current contribution and its associated magnetic-field circulation vanish during \(4\)–\(6\,\text{ms}\).
Explanation: \(\textbf{Relate signed current to change in capacitor charge:}\)
The signed current is
\[
i=\frac{dQ}{dt}
\]
Therefore, the charge change over an interval is the signed area beneath the current–time graph:
\[
\Delta Q=\int i\,dt
\]
The graph contains three constant-current intervals.
\(\textbf{Calculate charge acquired during the first interval:}\)
The current is
\[
i_1=+4.0\,\text{mA}
\]
The duration is
\[
\Delta t_1=2.0\,\text{ms}
\]
Therefore,
\[
\Delta Q_1=i_1\Delta t_1
\]
\[
\Delta Q_1=(4.0\times10^{-3})(2.0\times10^{-3})
\]
\[
\Delta Q_1=+8.0\,\mu\text{C}
\]
\(\textbf{Account for discharge during the second interval:}\)
The second current is
\[
i_2=-3.0\,\text{mA}
\]
The duration is again
\[
\Delta t_2=2.0\,\text{ms}
\]
Thus,
\[
\Delta Q_2=(-3.0\times10^{-3})(2.0\times10^{-3})
\]
\[
\Delta Q_2=-6.0\,\mu\text{C}
\]
\(\textbf{Determine the final charge:}\)
During the third interval,
\[
i_3=0
\]
so
\[
\Delta Q_3=0
\]
The initial charge is
\[
Q_0=5.0\,\mu\text{C}
\]
Therefore,
\[
Q_f=Q_0+\Delta Q_1+\Delta Q_2+\Delta Q_3
\]
\[
Q_f=5.0+8.0-6.0+0
\]
\[
Q_f=7.0\,\mu\text{C}
\]
\(\textbf{Interpret the displacement-current measurement:}\)
For the ideal capacitor,
\[
I_d=\varepsilon_0\frac{d\Phi_E}{dt}=\frac{dQ}{dt}
\]
During the final interval, \(dQ/dt=0\).
The electric flux is then constant, even though the capacitor retains nonzero charge.
The displacement-current contribution vanishes.
Accordingly, the associated magnetic-field circulation about the stated path also vanishes in the ideal model.
\(\textbf{Physical interpretation:}\)
Stored charge and displacement current are different quantities.
A charged capacitor can possess an electrostatic field and stored energy without a continuing displacement current.
The magnetic-field contribution is associated with changing electric flux, not simply with the existence of capacitor charge.
\(\textbf{Final answer:}\)
\[
\boxed{Q_f=7.0\,\mu\text{C},\qquad I_d=0\text{ during }4\text{–}6\,\text{ms}}
\]
314. The circular gap of an ideal capacitor is divided conceptually into two equal semicircular regions. During a short interval, the electric field in one half increases uniformly along the chosen positive normal at rate \(+\alpha\), while the electric field in the other half decreases at rate \(-\alpha\). The electric field remains perpendicular to the plates, and no conduction current crosses the gap. A circular Amperian path encloses the entire plate projection. Which conclusion is justified?
ⓐ. The magnetic field must vanish at every point on the path because the electric field is nonzero in both regions.
ⓑ. The magnetic-field circulation must be positive because displacement currents cannot have negative signs.
ⓒ. The magnetic-field circulation is zero, so the electric field must be static throughout the gap.
ⓓ. The net magnetic-field circulation is zero, but this alone does not establish that the local magnetic field is zero everywhere on the path.
Correct Answer: The net magnetic-field circulation is zero, but this alone does not establish that the local magnetic field is zero everywhere on the path.
Explanation: The displacement-current contribution depends on the rate of change of electric flux through a chosen surface.
For a stationary surface,
\[
I_d=\varepsilon_0\frac{d\Phi_E}{dt}.
\]
The two semicircular regions have equal areas.
In the first half, the normal component of electric field increases at rate \(+\alpha\).
In the second half, it changes at rate \(-\alpha\).
Therefore, their signed electric-flux change rates are equal in magnitude and opposite in sign.
When the complete plate projection is enclosed, the contributions cancel.
The net rate of change of electric flux through that surface is consequently zero.
Since no conduction current crosses the gap, the Ampere–Maxwell law gives zero magnetic-field circulation around the stated closed path.
However, circulation is the line integral of the tangential magnetic-field component around the path.
A zero value of this integral does not generally require the magnetic field to vanish at every point.
Positive and negative contributions to a closed-path integral can cancel.
Furthermore, the semicircular distribution does not possess the cylindrical symmetry of a uniformly charging circular capacitor.
It would therefore be unjustified to treat the magnetic field as having constant magnitude and identical tangential orientation everywhere around the chosen circle.
The changing electric fields remain present locally in both semicircular regions.
The cancellation concerns the net enclosed changing flux, not the disappearance of all local field variations.
The experiment distinguishes zero magnetic circulation from an identically zero magnetic field.
It also illustrates why symmetry must be established before replacing a circulation integral with \(B(2\pi r)\).
315. Two identical sinusoidal plane electromagnetic waves travel through vacuum along opposite directions of the \(x\)-axis. Their electric fields both oscillate along the \(y\)-axis and are given by
\[
\vec E_1=E_0\hat j\cos(kx-\omega t),
\qquad
\vec E_2=E_0\hat j\cos(kx+\omega t).
\]
Their corresponding magnetic fields have the directions required by their respective propagation directions. Let \(\lambda=2\pi/k\) and \(T=2\pi/\omega\). Which statement correctly describes the resultant fields at \(x=0,t=0\) and at \(x=\lambda/4,t=T/4\)?
ⓐ. At both events, the electric field is \(2E_0\hat j\) and the magnetic field vanishes.
ⓑ. At the first event, the magnetic field is \(2E_0\hat k/c\); at the second, both resultant fields vanish.
ⓒ. At the first event, \(\vec E=2E_0\hat j\) and \(\vec B=0\); at the second, \(\vec E=0\) and \(\vec B=2E_0\hat k/c\).
ⓓ. At both events, the electric and magnetic fields have equal nonzero magnitudes because both waves propagate at the same speed.
Correct Answer: At the first event, \(\vec E=2E_0\hat j\) and \(\vec B=0\); at the second, \(\vec E=0\) and \(\vec B=2E_0\hat k/c\).
Explanation: \(\textbf{Construct the magnetic fields of the opposing waves:}\)
Wave 1 travels along positive \(x\).
For its electric field along \(y\), its magnetic field points along \(z\) at a positive electric maximum:
\[
\vec B_1=\frac{E_0}{c}\hat k\cos(kx-\omega t)
\]
Wave 2 travels along negative \(x\).
For the same positive-\(y\) electric-field direction, its magnetic field must point along negative \(z\):
\[
\vec B_2=-\frac{E_0}{c}\hat k\cos(kx+\omega t)
\]
\(\textbf{Add the electric fields:}\)
The resultant is
\[
\vec E=\vec E_1+\vec E_2
\]
Therefore,
\[
\vec E=E_0\hat j[\cos(kx-\omega t)+\cos(kx+\omega t)]
\]
Using the cosine addition identity,
\[
\vec E=2E_0\hat j\cos(kx)\cos(\omega t)
\]
\(\textbf{Add the magnetic fields:}\)
\[
\vec B=\frac{E_0}{c}\hat k[\cos(kx-\omega t)-\cos(kx+\omega t)]
\]
Hence,
\[
\vec B=\frac{2E_0}{c}\hat k\sin(kx)\sin(\omega t)
\]
\(\textbf{Evaluate the first event:}\)
At
\[
x=0,\qquad t=0,
\]
we have
\[
\cos(kx)=1,\qquad\cos(\omega t)=1
\]
Therefore,
\[
\vec E(0,0)=2E_0\hat j
\]
Also,
\[
\sin(kx)=0
\]
so
\[
\vec B(0,0)=0
\]
\(\textbf{Evaluate the second event:}\)
At
\[
x=\frac{\lambda}{4},
\qquad
t=\frac T4,
\]
the phases are
\[
kx=\frac{\pi}{2},\qquad\omega t=\frac{\pi}{2}
\]
Therefore,
\[
\vec E=2E_0\hat j\cos\frac{\pi}{2}\cos\frac{\pi}{2}
\]
\[
\vec E=0
\]
For the magnetic field,
\[
\vec B=\frac{2E_0}{c}\hat k\sin\frac{\pi}{2}\sin\frac{\pi}{2}
\]
\[
\vec B=\frac{2E_0}{c}\hat k
\]
\(\textbf{Physical interpretation:}\)
The fields of the two waves superpose as vectors.
At the first event, their electric fields reinforce while their magnetic fields cancel.
At the second event, their electric fields vanish while their magnetic fields reinforce.
The resultant field need not satisfy the local travelling-wave ratio \(E=cB\) because it is a superposition of waves propagating in opposite directions.
Each individual wave nevertheless satisfies the vacuum plane-wave relation.
\(\textbf{Final answer:}\)
\[
\boxed{\begin{aligned}
(0,0):\;&\vec E=2E_0\hat j,\quad\vec B=0\\
(\lambda/4,T/4):\;&\vec E=0,\quad\vec B=\frac{2E_0}{c}\hat k
\end{aligned}}
\]
316. An X-ray imaging system produces two monochromatic beams of different wavelengths. Each beam is used separately to image an object containing regions P and Q. The transmission data are:
| Region | Fraction transmitted at wavelength \(\lambda_1\) | Fraction transmitted at wavelength \(\lambda_2\) |
|---|
| P | \(20\%\) | \(70\%\) |
| Q | \(60\%\) | \(30\%\) |
The incident intensity is uniform and the same for both wavelengths. A photographic film placed immediately behind the object becomes darker where more X-rays reach it. Which conclusion is supported by the data?
ⓐ. P produces the darker film region at both wavelengths because it is always more transparent.
ⓑ. Q produces the darker film region at \(\lambda_1\), but P produces the darker region at \(\lambda_2\).
ⓒ. Q produces the darker region at both wavelengths because X-ray contrast does not depend on wavelength.
ⓓ. Both regions produce equal darkening at \(\lambda_2\) because the X-rays have the same vacuum propagation speed.
Correct Answer: Q produces the darker film region at \(\lambda_1\), but P produces the darker region at \(\lambda_2\).
Explanation: The photographic film responds to X-rays reaching it after passage through the object.
The film-darkening convention is explicitly specified.
Greater transmitted X-ray intensity therefore produces greater darkening.
At wavelength \(\lambda_1\), Region P transmits only \(20\%\) of the incident radiation.
Region Q transmits \(60\%\).
Consequently, more X-rays reach the film behind Q.
Thus, Q appears darker than P on the developed film in the first exposure.
At wavelength \(\lambda_2\), the transmission fractions change.
Region P now transmits \(70\%\), whereas Q transmits \(30\%\).
Therefore, more X-rays reach the film behind P.
The relative darkening pattern reverses.
This reversal does not require the X-rays to change their vacuum propagation speeds.
Both beams are electromagnetic radiation and travel at speed \(c\) in vacuum.
The different contrast arises because the two regions interact differently with radiation at the selected wavelengths.
The ordering of X-ray attenuation is therefore not determined solely by region identity.
It may also depend on the wavelength used for observation.
The incident intensity is the same, so the reversal cannot be attributed merely to unequal source strengths.
The experiment demonstrates why a material's transmission properties must be understood in relation to the radiation band used.
It also separates attenuation within the object from darkening of the detector placed behind it.
317. A security marker is illuminated by a lamp that emits ultraviolet radiation but negligible visible light. The marker contains a fluorescent substance that absorbs part of the incident ultraviolet radiation and emits visible green light. Two instruments are used: Detector P responds only to the original ultraviolet band, while Detector Q responds only to the emitted visible green band. Which statement correctly interprets their possible responses?
ⓐ. Only P can respond because radiation emitted by a material must retain the wavelength of the radiation it absorbed.
ⓑ. Q can respond only if the ultraviolet radiation becomes visible spontaneously while propagating through empty space.
ⓒ. Both detectors must report identical powers because fluorescence cannot change the spectral region of emitted radiation.
ⓓ. P may detect remaining ultraviolet radiation, while Q may detect visible light newly emitted by the marker after ultraviolet absorption.
Correct Answer: P may detect remaining ultraviolet radiation, while Q may detect visible light newly emitted by the marker after ultraviolet absorption.
Explanation: The ultraviolet lamp supplies electromagnetic radiation in a wavelength region outside ordinary human vision.
The marker contains a material capable of absorbing part of that ultraviolet radiation.
Absorption transfers electromagnetic energy to the material.
The fluorescent substance can subsequently emit electromagnetic radiation in the visible region.
The question specifically identifies the emitted light as green.
Therefore, the marker can appear visibly luminous even when the lamp itself emits negligible visible light.
Detector P is sensitive to the original ultraviolet band.
It can respond to ultraviolet radiation that reaches it without being absorbed by the marker.
Detector Q is sensitive to the green visible band.
It can respond to visible radiation emitted by the marker.
The two signals need not have the same power because material absorption and re-emission are distinct processes.
The visible light does not have to originate directly from the ultraviolet lamp.
Nor does the ultraviolet radiation spontaneously change wavelength merely by travelling through empty space.
The spectral change occurs through interaction with the fluorescent material.
This distinguishes propagation of incident radiation from absorption and subsequent emission by matter.
The experiment also illustrates why the visible appearance of an ultraviolet-illuminated material need not reveal the wavelength of the original illuminating radiation.
318. A microwave transmitter operates at \(3.0\,\text{GHz}\). Its radiation enters a uniform transparent medium in which the electromagnetic-wave speed is \(2.0\times10^8\,\text{m s}^{-1}\). A spatial probe measures the transmitted wavelength. An engineer mistakenly calculates the source frequency from this measured wavelength using the vacuum speed \(3.0\times10^8\,\text{m s}^{-1}\) rather than the medium speed. Assume the actual frequency remains unchanged at the interface. What frequency does the engineer incorrectly infer, and which frequency should a correctly tuned receiver accept?
ⓐ. Incorrectly inferred \(2.0\,\text{GHz}\); receiver tuned to \(3.0\,\text{GHz}\).
ⓑ. Incorrectly inferred \(4.5\,\text{GHz}\); receiver tuned to \(3.0\,\text{GHz}\).
ⓒ. Incorrectly inferred \(4.5\,\text{GHz}\); receiver tuned to \(4.5\,\text{GHz}\).
ⓓ. Incorrectly inferred \(3.0\,\text{GHz}\); receiver tuned to \(2.0\,\text{GHz}\).
Correct Answer: Incorrectly inferred \(4.5\,\text{GHz}\); receiver tuned to \(3.0\,\text{GHz}\).
Explanation: \(\textbf{Identify the source frequency:}\)
The transmitter operates at
\[
f=3.0\,\text{GHz}
\]
\[
f=3.0\times10^9\,\text{Hz}
\]
The frequency remains unchanged when the wave enters the stationary transparent medium.
However, its propagation speed and wavelength change.
\(\textbf{Calculate the wavelength in the medium:}\)
The medium speed is
\[
v_m=2.0\times10^8\,\text{m s}^{-1}
\]
Using
\[
v_m=f\lambda_m,
\]
we obtain
\[
\lambda_m=\frac{v_m}{f}
\]
Substituting,
\[
\lambda_m=\frac{2.0\times10^8}{3.0\times10^9}
\]
\[
\lambda_m=6.67\times10^{-2}\,\text{m}
\]
\[
\lambda_m\approx6.67\,\text{cm}
\]
\(\textbf{Reproduce the engineer's incorrect calculation:}\)
The engineer uses the vacuum speed with the wavelength measured inside the material:
\[
f_{\text{wrong}}=\frac{c}{\lambda_m}
\]
Substituting,
\[
f_{\text{wrong}}=\frac{3.0\times10^8}{6.67\times10^{-2}}
\]
\[
f_{\text{wrong}}\approx4.5\times10^9\,\text{Hz}
\]
\[
f_{\text{wrong}}\approx4.5\,\text{GHz}
\]
\(\textbf{Determine the actual receiver frequency:}\)
The transmitted frequency remains
\[
f_{\text{actual}}=3.0\,\text{GHz}
\]
Therefore, a receiver designed to accept the transmitted signal must be compatible with \(3.0\,\text{GHz}\), not \(4.5\,\text{GHz}\).
\(\textbf{Physical interpretation:}\)
The engineer has mixed a wavelength measured in one medium with a propagation speed appropriate to another medium.
The relation \(v=f\lambda\) must use speed and wavelength belonging to the same propagation conditions.
The source frequency remains unchanged at the stationary interface.
Therefore, the material does not shift the microwave source from \(3.0\) to \(4.5\,\text{GHz}\).
The error is a measurement-interpretation error rather than a physical frequency conversion.
\(\textbf{Final answer:}\)
\[
\boxed{f_{\text{wrong}}=4.5\,\text{GHz},\qquad f_{\text{receiver}}=3.0\,\text{GHz}}
\]
319. A remote-control receiver is designed to recognise a particular coded pattern of infrared pulses. Two infrared sources operate near the receiver's sensitive wavelength. Source P is an infrared LED transmitting the correct timed pulse pattern. Source Q is a warm object emitting infrared radiation continuously in the same wavelength band. At the receiver, Q produces a larger average infrared detector signal than P, but it does not reproduce P's coded sequence. Which conclusion is most appropriate?
ⓐ. Q must activate the remote-controlled device because its average received infrared power is greater.
ⓑ. P cannot activate the device because coded information cannot be carried by electromagnetic radiation.
ⓒ. P can activate the receiver but Q cannot, because reception requires the coded pulse pattern as well as the correct wavelength.
ⓓ. Q must be emitting visible radiation because continuous infrared emission cannot be detected electronically.
Correct Answer: P can activate the receiver but Q cannot, because reception requires the coded pulse pattern as well as the correct wavelength.
Explanation: Infrared radiation is electromagnetic radiation lying beyond the red end of the visible spectrum.
Suitable infrared LEDs are used in remote-control systems.
Their emitted radiation can be varied in time to carry coded information.
The receiver must first be sensitive to the emitted wavelength region.
However, spectral compatibility alone does not guarantee that a particular command has been received.
The electronic system is specified to recognise a particular pattern of pulses.
Source P emits that required pattern.
Therefore, it can provide the information needed to activate the corresponding command.
Source Q is a warm object emitting infrared radiation continuously.
Its radiation can reach the same infrared detector and may produce a substantial detector response.
However, it does not reproduce the required coded sequence.
Consequently, its larger average received power does not establish that the correct command has been transmitted.
The distinction is between detecting infrared electromagnetic energy and identifying information carried by a deliberately varied signal.
Both sources can emit radiation in the same spectral region while serving different physical purposes.
The useful remote-control operation depends on the source, detector sensitivity and the temporal structure of the received signal.
The observation also explains why an infrared thermal source is not automatically interchangeable with an infrared information transmitter.
The correct interpretation therefore considers both spectral matching and the specified signalling requirement.
320. A suitable microwave source directs radiation towards a water-containing sample inside an evacuated chamber. The sample absorbs part of the microwave energy and becomes warmer. The microwave source is then switched off, but an infrared-sensitive detector continues registering radiation emitted by the warmed sample. A student proposes that the detected infrared radiation is simply the original microwave beam whose frequency increased spontaneously during travel through vacuum. Which explanation is more appropriate?
ⓐ. The sample absorbed microwave energy, became warmer and subsequently emitted infrared radiation through a separate radiation–matter process.
ⓑ. The original microwaves necessarily changed into infrared waves simply because the microwave source was switched off.
ⓒ. The infrared detector must be responding to electrons travelling from the microwave transmitter through the vacuum.
ⓓ. The observation is impossible because an object cannot emit infrared radiation unless it glows visibly.
Correct Answer: The sample absorbed microwave energy, became warmer and subsequently emitted infrared radiation through a separate radiation–matter process.
Explanation: Microwaves and infrared radiation are both electromagnetic waves.
They occupy different characteristic wavelength and frequency regions of the electromagnetic spectrum.
Suitable water-containing materials can absorb microwave energy.
The absorbed electromagnetic energy can increase the internal energy of the sample.
The sample consequently becomes warmer under the specified conditions.
Warm bodies can emit infrared radiation.
This infrared emission can continue after the microwave transmitter is switched off.
The emitted infrared radiation originates from the warmed material rather than from the original microwaves simply changing frequency in empty space.
The vacuum between the sample and infrared detector allows the emitted electromagnetic radiation to propagate without a material carrier.
The infrared detector can register suitable radiation arriving from the sample.
Its response does not require conduction electrons from the microwave transmitter to reach the detector.
Nor does the sample need to glow visibly to emit infrared radiation.
The experiment involves a sequence of distinct processes: microwave propagation, absorption by matter, heating and subsequent infrared emission.
Energy supplied by one electromagnetic-spectrum region can therefore contribute to emission in another region through an intervening material process.
This does not violate the fact that the frequency of an electromagnetic wave remains unchanged during ordinary propagation through stationary vacuum.
The student's explanation incorrectly treats a source–material–detector energy-transfer sequence as spontaneous spectral conversion during free propagation.
