Physics 2 Quiz: Em Wave Relationships
18 questions · exam conditions
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Em Wave RelationshipsQuestion 1 of 18

A Wi-Fi router operates at 5.0 GHz. What is its wavelength in centimeters?

0.60 cm
60 cm
6.0 cm
600 cm
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Physics 2 Quiz

Physics 2 Quiz: Em Wave Relationships

Practice Em Wave Relationships in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Em Wave Relationships, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A Wi-Fi router operates at 5.0 GHz. What is its wavelength in centimeters?

  1. 0.60 cm
  2. 60 cm
  3. 6.0 cm (correct answer)
  4. 600 cm
Explanation: Use wavelength = speed divided by frequency: 300,000,000 m/s divided by 5,000,000,000 Hz gives 0.06 m, and 0.06 m is 6 cm. The tempting 60 cm appears if you treat 5 GHz as 500 million hertz instead of 5 billion, which moves the decimal one place too far.

Question 2

An EM wave in vacuum has a period of 8.0 ns. What is its wavelength?

  1. 0.24 m
  2. 1.2 m
  3. 24 m
  4. 2.4 m (correct answer)
Explanation: Since wavelength = speed x period, multiply 3 x 10^8 m/s by 8 x 10^-9 s. The powers of 10 give 10^-1, so 3 x 8 x 10^-1 = 2.4 m. A tempting error is 24 m, which drops that 10^-1 factor from the nanosecond conversion.

Question 3

How many times longer is the wavelength of a 100 MHz wave than a 2.45 GHz wave?

  1. 0.041
  2. 24.5 (correct answer)
  3. 2.45
  4. 245
Explanation: Wavelength is inversely proportional to frequency, so compare the frequencies: 2.45 GHz is 2450 MHz, and 2450 / 100 = 24.5. Therefore the 100 MHz wave has a wavelength 24.5 times longer than the 2.45 GHz wave. The tempting wrong answer is 0.041, which is the inverse ratio and tells you how much shorter the higher-frequency wave is, not how many times longer the lower-frequency wave is.

Question 4

A photon has a wavelength of 250 nm in vacuum. What is its frequency?

  1. 1.2 × 10^18 Hz
  2. 1.2 × 10^12 Hz
  3. 1.2 × 10^16 Hz
  4. 1.2 × 10^15 Hz (correct answer)
Explanation: Convert 250 nm to 2.5 x 10^-7 m. Frequency is speed of light divided by wavelength: f = (3.0 x 10810^8) / (2.5 x 10^-7) = 1.2 x 10^15 Hz. The tempting error is reporting 1.2 x 10^18 Hz by treating 250 nm as 2.5 x 10^-10 m instead of 2.5 x 10^-7 m.

Question 5

A 2.4 GHz signal crosses a 30 m room. How many wavelengths span the room?

  1. 240 (correct answer)
  2. 24
  3. 2.4
  4. 0.24
Explanation: A 2.4 GHz wave has wavelength (3 x 10810^8) / (2.4 x 10910^9) = 0.125 m, or 12.5 cm. Dividing the room length by that gives 30 / 0.125 = 240 wavelengths. The common slip is treating the wavelength as 1.25 m, which would give 24, but the frequency is 2.4 billion, not 2.4 million.

Question 6

An EM wave traveling through a non-dispersive dielectric medium has a measured wavelength of 400400 nm and a phase velocity of 2.0×1082.0 \times 10^8 m/s. If this same wave then enters vacuum, which of the following describes its new wavelength and frequency?

  1. Both the wavelength and the frequency increase, because in vacuum the wave speed is greater and the wave must compensate by raising both parameters proportionally.
  2. The frequency remains unchanged at 5.0×10145.0 \times 10^{14} Hz, and the wavelength increases to 600600 nm, because only the phase velocity and wavelength change at a medium boundary while frequency is conserved. (correct answer)
  3. The frequency increases to 7.5×10147.5 \times 10^{14} Hz, and the wavelength remains at 400400 nm, because the shorter wavelength is locked to the atomic spacing of the medium that originally emitted the wave.
  4. The frequency remains unchanged at 5.0×10145.0 \times 10^{14} Hz, and the wavelength decreases to 267267 nm, because the wave must conserve energy as it enters a less optically dense medium.
Explanation: Whenever an EM wave crosses a boundary between two media, you need to track which quantities change and which stay fixed. The key principle: frequency is conserved across boundaries, because it's determined by the source and must match at the interface to satisfy boundary conditions. Only the wave speed and wavelength adjust. Start by finding the frequency in the dielectric using v=fλv = f\lambda: f=vλ=2.0×108 m/s400×10−9 m=5.0×1014 Hzf = \frac{v}{\lambda} = \frac{2.0 \times 10^8 \text{ m/s}}{400 \times 10^{-9} \text{ m}} = 5.0 \times 10^{14} \text{ Hz} This frequency is locked — it doesn't change when the wave enters vacuum. In vacuum, the speed becomes c=3.0×108c = 3.0 \times 10^8 m/s, so the new wavelength is: λvac=cf=3.0×1085.0×1014=600 nm\lambda_{\text{vac}} = \frac{c}{f} = \frac{3.0 \times 10^8}{5.0 \times 10^{14}} = 600 \text{ nm} This confirms answer B: frequency stays at 5.0×10145.0 \times 10^{14} Hz and wavelength stretches to 600 nm. A is wrong because frequency never changes at a boundary — it's the wavelength alone (not both quantities) that compensates for the speed change. C is wrong on two counts: frequency doesn't increase, and wavelength is not "locked" to atomic spacing — that's a fabricated concept with no physical basis. D is wrong because energy conservation doesn't compress the wavelength; in fact, moving into vacuum (faster speed) means the wavelength increases, not decreases. A useful memory rule: frequency is the traveler's identity — it stays the same no matter what medium the wave passes through. Speed and wavelength are what adapt.

Question 7

A communications engineer needs a channel with a free-space wavelength of exactly λ=1550\lambda = 1550 nm (the standard telecom C-band). She tunes an oscillator to produce this wavelength in vacuum. A colleague argues that because glass fiber has a refractive index of approximately n=1.46n = 1.46 at this wavelength, the oscillator must be retuned to a longer free-space wavelength of n×1550≈2263n \times 1550 \approx 2263 nm so that the fiber compresses it to 1550 nm inside the glass, preserving the channel specification. Which of the following best evaluates the colleague's argument?

  1. The colleague is correct: the free-space wavelength must be set to approximately 2263 nm so that the fiber compresses it to 1550 nm, because the in-fiber wavelength is what determines the channel's propagation characteristics and must match the specification.
  2. The colleague is incorrect: the oscillator frequency is invariant across the boundary, but because silica glass is optically denser than air (n>1n > 1), the wavelength inside the fiber actually decreases relative to the free-space value, meaning the in-fiber wavelength will be shorter than 1550 nm — the opposite of what the colleague claims — and no retuning is needed to meet the free-space channel specification.
  3. The colleague is incorrect: the wavelength, not the frequency, is conserved across the air-fiber interface, so the 1550 nm oscillator setting already guarantees a 1550 nm wavelength inside the fiber without any adjustment to the oscillator.
  4. The colleague is incorrect: the oscillator frequency, not wavelength, is the invariant quantity across a boundary. The free-space frequency corresponding to 1550 nm is f=c/λ≈1.94×1014f = c/\lambda \approx 1.94 \times 10^{14} Hz, and this frequency is preserved in the fiber. The wavelength inside the fiber shortens to λfiber=λ/n≈1062\lambda_{fiber} = \lambda/n \approx 1062 nm, but the telecom channel is specified and selected by its free-space equivalent frequency, not its in-medium wavelength. (correct answer)
Explanation: Whenever a wave crosses a boundary between two media, ask yourself: what quantity is conserved? The key principle here is that frequency is invariant across an interface, while wavelength adjusts to match the new medium's wave speed. This is because the wave source sets a fixed oscillation rate — the boundary cannot create or destroy cycles. The free-space wavelength of 1550 nm corresponds to a frequency of f=c/λ=(3×108)/(1550×10−9)≈1.94×1014f = c/\lambda = (3 \times 10^8)/(1550 \times 10^{-9}) \approx 1.94 \times 10^{14} Hz. When this wave enters glass with n=1.46n = 1.46, the speed drops to v=c/nv = c/n, but the frequency stays the same. The in-fiber wavelength therefore shortens: λfiber=λfree/n≈1550/1.46≈1062\lambda_{\text{fiber}} = \lambda_{\text{free}}/n \approx 1550/1.46 \approx 1062 nm. Crucially, telecom channels are identified by their free-space frequency (or equivalently, their free-space wavelength), not the compressed in-medium wavelength. The 1550 nm oscillator already satisfies the C-band specification — no retuning needed. Answer D captures all of this correctly. Answer A is the colleague's flawed claim — it inverts the physics, imagining the fiber expands a longer wavelength down to 1550 nm, which contradicts how refraction works. Answer B correctly identifies that the fiber shortens the wavelength, which is good, but then wrongly concludes no retuning is needed on the grounds that the channel spec isn't affected by the in-fiber wavelength — the reasoning is muddled even if the conclusion happens to be right. Answer C commits the foundational error of treating wavelength (not frequency) as the conserved quantity across the boundary. Your study tip: always anchor on frequency as the invariant when light crosses media — it's the "identity tag" of the wave. Wavelength is a consequence of the medium, frequency is not.

Question 8

A gamma-ray photon has a wavelength of λ=1.0×10−13\lambda = 1.0 \times 10^{-13} m. Approximately how many times greater is the frequency of this gamma ray compared to that of a 100 MHz FM radio wave?

  1. 3.0×10133.0 \times 10^{13} times greater, computed by using the wavelength ratio λFM/λγ\lambda_{FM}/\lambda_{\gamma} directly, where λFM=c/fFM=3.0\lambda_{FM} = c/f_{FM} = 3.0 m, giving 3.0/(1.0×10−13)=3.0×10133.0/(1.0 \times 10^{-13}) = 3.0 \times 10^{13}.
  2. 3.0×10123.0 \times 10^{12} times greater, computed by correctly finding fγ=c/λγ=3.0×1021f_{\gamma} = c/\lambda_{\gamma} = 3.0 \times 10^{21} Hz but making a powers-of-ten subtraction error when dividing by fFM=108f_{FM} = 10^8 Hz, yielding an exponent of 12 instead of 13.
  3. 3.0×10133.0 \times 10^{13} times greater, computed by finding fγ=c/λγ=3.0×1021f_{\gamma} = c/\lambda_{\gamma} = 3.0 \times 10^{21} Hz and dividing by fFM=1.0×108f_{FM} = 1.0 \times 10^{8} Hz, giving a ratio of 3.0×10133.0 \times 10^{13}. (correct answer)
  4. 3.0×1063.0 \times 10^{6} times greater, computed by incorrectly using λFM=3.0×10−3\lambda_{FM} = 3.0 \times 10^{-3} m (confusing the FM wavelength with a millimeter-scale value) when forming the wavelength ratio λFM/λγ\lambda_{FM}/\lambda_{\gamma}.
Explanation: When comparing frequencies across the electromagnetic spectrum, your go-to relationship is f=c/λf = c/\lambda, where c=3.0×108c = 3.0 \times 10^8 m/s. The ratio of two frequencies equals the inverse ratio of their wavelengths — so a shorter wavelength means a higher frequency. For the gamma ray: fγ=c/λγ=(3.0×108)/(1.0×10−13)=3.0×1021f_{\gamma} = c/\lambda_{\gamma} = (3.0 \times 10^8)/(1.0 \times 10^{-13}) = 3.0 \times 10^{21} Hz. The FM radio wave is given directly as fFM=100 MHz=1.0×108f_{FM} = 100 \text{ MHz} = 1.0 \times 10^8 Hz. Dividing gives the ratio: (3.0×1021)/(1.0×108)=3.0×1013(3.0 \times 10^{21})/(1.0 \times 10^8) = 3.0 \times 10^{13}. That confirms C is correct. Choice A arrives at the same numerical answer but through a partially flawed setup — it treats the wavelength ratio as the answer without explicitly computing fγf_{\gamma}, which can mask errors in more complex problems. Coincidentally it gets the right number here because the wavelength ratio is equivalent, but the reasoning skips the FM frequency calculation entirely, which is sloppy and risky. Choice B makes a classic powers-of-ten arithmetic error: it correctly finds fγ=3.0×1021f_{\gamma} = 3.0 \times 10^{21} Hz but then miscalculates 21−8=1221 - 8 = 12 instead of 13, landing on 3.0×10123.0 \times 10^{12}. Choice D incorrectly assumes the FM wavelength is 3.0×10−33.0 \times 10^{-3} m — confusing it with a millimeter-scale value — which drastically underestimates the ratio. Study tip: Always handle powers-of-ten division by subtracting exponents explicitly and double-checking: 1021/108=1021−8=101310^{21}/10^{8} = 10^{21-8} = 10^{13}. Writing this step out prevents the error in B.

Question 9

Compared to a 100 THz infrared wave, how many times longer is a 100 MHz radio wave's wavelength?

  1. 10−610^{-6}
  2. 10610^6 (correct answer)
  3. 10310^3
  4. 10−310^{-3}
Explanation: Wavelength equals speed divided by frequency, so it scales inversely with frequency. 100 THz is 100,000,000,000,000 Hz (101410^14 Hz) and 100 MHz is 100,000,000 Hz (10810^8 Hz). Dividing 10^14 by 10^8 gives 10^6, so the radio wave is 10^6 times longer. The tempting 10^-6 reverses the ratio and would describe the infrared wave's wavelength compared to the radio wave.

Question 10

A 600 nm vacuum wave enters glass with n = 1.5. What is its frequency?

  1. 3.3×10143.3 \times 10^{14} Hz
  2. 7.5×10147.5 \times 10^{14} Hz
  3. 5.0×10145.0 \times 10^{14} Hz (correct answer)
  4. 2.2×10142.2 \times 10^{14} Hz
Explanation: Frequency does not change when light enters a new medium; only wavelength and speed change. Use the vacuum values: f = c / lambda = (3.0 x 10810^8) / (600 x 10^-9) = 5.0 x 10^14 Hz. The tempting mistake is using the shorter glass wavelength (400 nm) with c, which gives 7.5 x 10^14 Hz and ignores that the wave's frequency is fixed.

Question 11

A 98.5 MHz radio wave: what is its wavelength in cm?

  1. 305 cm (correct answer)
  2. 30.5 cm
  3. 3050 cm
  4. 0.305 cm
Explanation: Use wavelength = speed / frequency. 98.5 MHz is 98,500,000 Hz, so 3.00 x 10^8 m/s divided by 98,500,000 Hz gives 3.05 m. Since 1 m = 100 cm, that is 305 cm. A tempting mistake is 30.5 cm, which comes from treating 98.5 MHz as 985 MHz or dropping one zero in the meter-to-centimeter conversion; keep 98.5 MHz at 98,500,000 Hz.

Question 12

What is the frequency, in THz, of 450 nm light in vacuum?

  1. 0.667 THz
  2. 6.67 THz
  3. 66.7 THz
  4. 667 THz (correct answer)
Explanation: Light speed divided by wavelength: (3 x 10810^8 m/s) / (450 x 10^-9 m) = 6.67 x 10^14 Hz = 667 THz. The tempting wrong answer is 66.7 THz, which comes from misplacing the power of ten when converting nanometers to meters. Keep the conversion as 450 x 10^-9 m and the result is 667 THz.

Question 13

An ultraviolet photon and an infrared photon are both traveling through vacuum. The UV photon has a wavelength of 150150 nm and the IR photon has a wavelength of 15001500 nm.

By what factor does the frequency of the UV photon exceed the frequency of the IR photon, and what is the approximate frequency of the UV photon?

  1. The UV frequency is 10 times higher than the IR frequency, and the UV photon frequency is approximately 2.0×10152.0 \times 10^{15} Hz. (correct answer)
  2. The UV frequency is 10 times higher than the IR frequency, and the UV photon frequency is approximately 2.0×10142.0 \times 10^{14} Hz.
  3. The UV frequency is 100 times higher than the IR frequency, and the UV photon frequency is approximately 2.0×10152.0 \times 10^{15} Hz.
  4. The UV frequency is 10 times higher than the IR frequency, and the UV photon frequency is approximately 2.0×10132.0 \times 10^{13} Hz.
Explanation: When a question gives you two wavelengths and asks you to compare frequencies, your first instinct should be to reach for the wave equation: c=fλc = f\lambda, which rearranges to f=cλf = \frac{c}{\lambda}. This tells you frequency and wavelength are inversely proportional — so the shorter the wavelength, the higher the frequency. The UV photon has wavelength 150 nm and the IR photon has 1500 nm. The ratio of wavelengths is 1500150=10\frac{1500}{150} = 10, meaning the IR wavelength is 10 times longer. Because frequency is inversely proportional to wavelength, the UV frequency must be 10 times higher than the IR frequency. Now calculate the UV frequency directly: f=cλ=3.0×108 m/s150×10−9 m=3.0×1081.5×10−7=2.0×1015 Hzf = \frac{c}{\lambda} = \frac{3.0 \times 10^8 \text{ m/s}}{150 \times 10^{-9} \text{ m}} = \frac{3.0 \times 10^8}{1.5 \times 10^{-7}} = 2.0 \times 10^{15} \text{ Hz}. This confirms answer A is correct. Answer B gets the factor of 10 right but lands on 2.0×10142.0 \times 10^{14} Hz — that's off by a factor of 10, likely from a powers-of-ten arithmetic error when converting nanometers to meters (forgetting that 1 nm = 10−910^{-9} m, not 10−810^{-8} m). Answer C claims the factor is 100, which would only be true if the wavelengths differed by a factor of 100 — they don't, they differ by a factor of 10. Answer D gets the factor right but makes an even larger exponent error, yielding 2.0×10132.0 \times 10^{13} Hz, two full powers of ten off. The big study tip here: always convert nanometers to meters carefully. Write out 150 nm=150×10−9 m=1.5×10−7 m150 \text{ nm} = 150 \times 10^{-9} \text{ m} = 1.5 \times 10^{-7} \text{ m} explicitly — this single step prevents most frequency calculation errors on this exam.

Question 14

A microwave oven operates at a frequency of 2.452.45 GHz. A student claims that if she doubles the power output of the oven while keeping the frequency fixed, the wavelength of the microwaves inside the cavity will decrease by a factor of two. Which of the following most precisely evaluates this claim?

  1. The claim is correct, because increasing power increases the energy per photon, which by the de Broglie relation shortens the effective wavelength proportionally.
  2. The claim is incorrect, because the wavelength is determined solely by λ=c/f\lambda = c/f, and since frequency is held constant and cc is invariant in vacuum, the wavelength cannot change regardless of power output. (correct answer)
  3. The claim is incorrect, because doubling power doubles the amplitude of the wave, and since wavelength scales as the square root of amplitude in a standing-wave cavity, the wavelength increases by 2\sqrt{2}.
  4. The claim is correct, because inside the cavity the superposition of incident and reflected waves creates a standing wave whose nodal spacing—and hence effective wavelength—depends on the wave intensity.
Explanation: Whenever you see a question mixing wave properties with power or intensity, your first instinct should be to separate what determines wavelength from what determines amplitude or energy. These are fundamentally independent quantities in classical wave physics. For electromagnetic waves in free space or a cavity, the wavelength is locked in by a single relationship: λ=c/f\lambda = c/f. The speed of light cc is a universal constant, and the frequency ff is set by the source — in this case, the magnetron oscillating at 2.45 GHz. Doubling the power output means the oven produces more photons per second and the electric field oscillates with greater amplitude, but the rate of oscillation — the frequency — doesn't change. Since λ=c/f\lambda = c/f and neither cc nor ff changes, the wavelength is completely unaffected. Answer B is correct. Answer A confuses photon energy with de Broglie wavelength. The de Broglie relation applies to matter particles, not photons, and even photon energy E=hfE = hf depends only on frequency — not power. Power increases photon count, not photon energy. Answer C invents a false rule: wavelength does not scale with the square root of amplitude. Amplitude and wavelength are independent wave properties. Answer D is similarly wrong; nodal spacing in a standing wave equals λ/2\lambda/2, which again depends only on frequency, not intensity. Study tip: On any wave question, mentally sort the properties into two buckets — those set by frequency (wavelength, photon energy) and those set by power (amplitude, intensity). Changing one bucket never affects the other.

Question 15

An FM station broadcasts at 98.5 MHz. How many wavelengths long is a 3.00 m antenna?

  1. 0.985 (correct answer)
  2. 1.02
  3. 3.05
  4. 9.85
Explanation: Find the wavelength: 3.00e8 m/s divided by 98.5e6 Hz gives about 3.05 m. The antenna is 3.00 m, so divide length by wavelength: 3.00 / 3.05 = 0.985 wavelengths. The tempting wrong answer is 1.02, which reverses the ratio and gives wavelength divided by antenna length instead.

Question 16

A spacecraft transmits a carrier signal at f0=8.4f_0 = 8.4 GHz from deep space. Due to the Doppler effect, ground stations receive the signal at a slightly shifted frequency. Setting aside the Doppler shift, mission engineers need to know the vacuum wavelength of the carrier.

What is the vacuum wavelength of the 8.4 GHz carrier signal, and to which region of the EM spectrum does it belong?

  1. λ≈3.6\lambda \approx 3.6 cm; the signal is in the microwave region, consistent with X-band deep-space communication frequencies. (correct answer)
  2. λ≈3.6\lambda \approx 3.6 mm; the signal is in the millimeter-wave (EHF) region, consistent with atmospheric absorption windows used by weather radar.
  3. λ≈3.6\lambda \approx 3.6 cm; the signal is in the radio (HF/VHF) region, because any frequency below 10 GHz is classified as a conventional radio wave rather than a microwave.
  4. λ≈2.8\lambda \approx 2.8 cm; the signal is in the microwave region, obtained by dividing cc by f0f_0 after converting GHz to MHz rather than Hz.
Explanation: Whenever you see a wavelength problem, your instinct should be to reach for λ=c/f\lambda = c/f, but the trap is almost always in the unit conversion — and this question is no exception. Starting with the correct approach: the speed of light is c=3.0×108 m/sc = 3.0 \times 10^8 \text{ m/s}, and the frequency must be in Hz. Converting 8.4 GHz gives f0=8.4×109 Hzf_0 = 8.4 \times 10^9 \text{ Hz}. Dividing: λ=3.0×1088.4×109≈0.036 m=3.6 cm\lambda = \frac{3.0 \times 10^8}{8.4 \times 10^9} \approx 0.036 \text{ m} = 3.6 \text{ cm} This places the signal squarely in the microwave region, and specifically within the X-band (roughly 8–12 GHz), which NASA and ESA routinely use for deep-space communication. That makes A the correct answer. B is wrong on both the value and the classification. 3.6 mm3.6 \text{ mm} would correspond to a frequency around 83 GHz — that's EHF, not 8.4 GHz. This error likely comes from misplacing the decimal by a factor of 10 during conversion. C gets the wavelength right (3.6 cm) but applies a false classification rule. There is no hard cutoff where "below 10 GHz = radio, not microwave." Microwaves span roughly 300 MHz to 300 GHz; 8.4 GHz falls well within that range. D uses the right formula but converts GHz to MHz instead of Hz, introducing a factor-of-1000 error and landing on a wavelength that doesn't match any real calculation cleanly. Study tip: Always convert frequency fully to Hz (not MHz or GHz) before plugging into λ=c/f\lambda = c/f. One missed prefix is enough to shift your answer by an order of magnitude and send you to the wrong spectral region entirely.

Question 17

A radio astronomy observatory detects two signals arriving simultaneously from a distant galaxy. Signal A has a wavelength of 2.5×10−22.5 \times 10^{-2} m, and Signal B has a frequency of 2.4×10102.4 \times 10^{10} Hz. Both signals travel through vacuum.

Which of the following correctly characterizes the relationship between Signal A and Signal B?

  1. Signal A has a higher frequency than Signal B, because its longer wavelength corresponds to more wave cycles per unit distance traveled at speed cc.
  2. Signal A and Signal B have the same frequency, because all electromagnetic waves detected by the same instrument must share the same oscillation rate in vacuum.
  3. Signal A has a lower frequency than Signal B, because its wavelength of 2.5×10−22.5 \times 10^{-2} m yields a frequency of 1.2×10101.2 \times 10^{10} Hz via f=c/λf = c/\lambda, which is less than 2.4×10102.4 \times 10^{10} Hz. (correct answer)
  4. Signal A has a higher frequency than Signal B, because its wavelength of 2.5×10−22.5 \times 10^{-2} m yields a frequency of 1.2×10111.2 \times 10^{11} Hz via f=c/λf = c/\lambda, which exceeds 2.4×10102.4 \times 10^{10} Hz.
Explanation: When a question gives you one signal's wavelength and another's frequency, your job is to convert to a common unit — frequency — and compare. The governing equation is f=c/λf = c/\lambda, where c=3.0×108c = 3.0 \times 10^8 m/s in vacuum. For Signal A: f=3.0×1082.5×10−2=1.2×1010f = \frac{3.0 \times 10^8}{2.5 \times 10^{-2}} = 1.2 \times 10^{10} Hz. Signal B is already given as 2.4×10102.4 \times 10^{10} Hz. Since 1.2×1010<2.4×10101.2 \times 10^{10} < 2.4 \times 10^{10}, Signal A has the lower frequency — confirming C is correct. Choice A gets both the conclusion and the reasoning backwards. A longer wavelength actually means fewer wave cycles per unit distance, which corresponds to lower frequency, not higher. The inverse relationship f=c/λf = c/\lambda means wavelength and frequency move in opposite directions. Choice B reflects a fundamental misconception: electromagnetic waves detected by the same instrument share no obligation to have the same frequency. Radio telescopes routinely detect signals across a wide frequency range simultaneously — the detector doesn't impose a common oscillation rate. Choice D uses the right formula but makes an arithmetic error, misplacing a power of ten. It reports Signal A's frequency as 1.2×10111.2 \times 10^{11} Hz instead of 1.2×10101.2 \times 10^{10} Hz — a factor-of-10 mistake that flips the comparison entirely. Study tip: On any EM wave comparison question, always convert everything to the same quantity before comparing. Watch your powers of ten carefully — a single exponent error is enough to reverse your conclusion, exactly as trap D demonstrates.

Question 18

The period of oscillation of a 500 nm visible-light wave is most nearly:

  1. 1.67×10−121.67 \times 10^{-12} s, obtained by computing T=λ/cT = \lambda/c but misplacing the decimal point by treating 500 nm as 500×10−6500 \times 10^{-6} m rather than 500×10−9500 \times 10^{-9} m.
  2. 1.67×10−61.67 \times 10^{-6} s, obtained by computing T=1/fT = 1/f after incorrectly expressing the frequency in MHz rather than Hz, leading to a microsecond-scale period.
  3. 1.67×10−91.67 \times 10^{-9} s, obtained by computing T=λ/cT = \lambda/c but treating 500 nm as 500×10−6500 \times 10^{-6} m (confusing nanometers with micrometers) and then incorrectly applying an additional factor of 10310^3.
  4. 1.67×10−151.67 \times 10^{-15} s, obtained by computing T=λ/c=(500×10−9)/(3.0×108)T = \lambda/c = (500 \times 10^{-9})/(3.0 \times 10^8), which equals the time for one wavefront to advance one wavelength. (correct answer)
Explanation: When you see a question about the period of a light wave, your instinct should be to connect wavelength, wave speed, and period through two fundamental relationships: c=fλc = f\lambda and T=1/fT = 1/f. Combining these gives you T=λ/cT = \lambda/c, which is all you need here. For a 500 nm wave, you must first convert correctly: 500 nm = 500×10−9500 \times 10^{-9} m. Plugging into the formula: T=500×10−93.0×108=5.0×10−73.0×108≈1.67×10−15 sT = \frac{500 \times 10^{-9}}{3.0 \times 10^8} = \frac{5.0 \times 10^{-7}}{3.0 \times 10^8} \approx 1.67 \times 10^{-15} \text{ s} This confirms D — light oscillates incredibly fast, on the order of femtoseconds, which makes physical sense given its enormous frequency (~6×10146 \times 10^{14} Hz). Choice A arrives at 1.67×10−121.67 \times 10^{-12} s by misreading 500 nm as 500×10−6500 \times 10^{-6} m (confusing nanometers with micrometers), which inflates the answer by a factor of 1000. Choice C makes the same nm-to-μm confusion but then compounds it with an extra factor of 10310^3, pushing the result to 1.67×10−91.67 \times 10^{-9} s — two orders of magnitude off from the correct answer. Choice B is a unit-analysis breakdown: expressing frequency in MHz instead of Hz makes the period appear microsecond-scale, which is completely unphysical for visible light. The big takeaway: always double-check your unit conversion for wavelength. The nm → m conversion (factor of 10−910^{-9}) is the most common stumbling block in optics calculations, and getting it wrong shifts your answer by orders of magnitude.