Properties of Radio WavesRadio Navigation — DGCA CPL practice questions
Question 1 of 12
A radio wave is:
All 12 questions — Properties of Radio Waves
Radio Navigation · DGCA CPL. The correct option is marked on each.
Q1. A radio wave is:
- A.an energy wave comprising an electrical field in the same plane as a magnetic field
- B.an electrical field alternating with a magnetic field
- C.an energy wave where there is an electrical field perpendicular to a magnetic field✓
- D.an energy field with an electrical component
Why: A radio wave is electromagnetic radiation comprising two components: an E (electrical) field and an H (magnetic) field . These two fields are always perpendicular to each other , and both are perpendicular to the direction of propagation. See Section 3 — EM Radiation . — This question tests the very definition of EM radiation. Many students confuse "alternating with" (sequential) vs "perpendicular to" (simultaneous at 90°). Remember: BOTH fields exist at the same moment, at right angles — that is what makes EM radiation self-sustaining through space.
Q2. The speed of radio waves is:
- A.300 km per second
- B.300 million metres per second✓
- C.162 NM per second
- D.162 million NM per second
Why: The speed of radio waves equals the speed of light: 300,000,000 m/s = 300 × 10⁶ m/s . In nautical miles, this is 162,000 NM/s — note 162 thousand, not 162 million. See Section 6 — Wavelength . — Two correct values to memorise: 300,000,000 m/s (or 300 × 10⁶ m/s) AND 162,000 NM/s . The NM figure appears directly in DME ranging calculations and radar questions later in the course.
Q3. The plane of polarization of an electromagnetic wave is:
- A.the plane of the magnetic field
- B.the plane of the electrical field✓
- C.the plane of the electrical or magnetic field dependent on the plane of the aerial
- D.none of the above
Why: By definition, the polarization of a radio wave is the plane of the electrical (E) field . This is a fixed definition — it is always the E field, regardless of orientation. A vertical aerial produces a vertically polarized wave because the E field is vertical. See Section 4 — Polarization . — Pure recall: polarization = plane of the E field . Option (c) is the most common wrong answer because students confuse "the definition of polarization" with "how polarization is set". The aerial sets the E field plane; the E field plane is called the polarization.
Q4. If the wavelength of a radio wave is 3.75 metres, the frequency is:
- A.80 kHz
- B.8 MHz
- C.80 MHz✓
- D.800 kHz
Why: Using the simplified formula: f (MHz) = 300 / λ (m) = 300 / 3.75 = 80 MHz . See Section 6 — Wavelength . — 3.75 m falls in the VHF band (1–10 m). Confirming the answer is in VHF (30–300 MHz) validates that 80 MHz is correct. Always sense-check your answer against the frequency band table.
Q5. The wavelength corresponding to a frequency of 125 MHz is:
- A.2.4 m✓
- B.24 m
- C.24 cm
- D.24 mm
Why: λ = 300 / f (MHz) = 300 / 125 = 2.4 m . 125 MHz is VHF, so wavelength should be in the 1–10 m range. 2.4 m fits perfectly. See Section 6 . — The frequency 125 MHz is close to the VHF comms band (118–137 MHz). Knowing that VHF comms wavelengths are approximately 2–2.5 m is useful background knowledge that cross-validates this calculation.
Q6. The frequency which corresponds to a wavelength of 6.98 cm is:
- A.4298 GHz
- B.4.298 GHz✓
- C.429.8 GHz
- D.42.98 GHz
Why: λ = 6.98 cm = 0.0698 m . f = 300 / 0.0698 = 4297.9 MHz ≈ 4.298 GHz . 6.98 cm is in the SHF band (1–10 cm), so frequency should be 3–30 GHz. 4.298 GHz fits. See Section 6 and Section 7 . — The key trap: must convert cm → m first . 6.98 cm ÷ 100 = 0.0698 m. Only then apply f = 300/0.0698. Failing to convert is the #1 error on this type of question.
Q7. The frequency band containing the frequency corresponding to 29.1 cm is:
- A.HF
- B.VHF
- C.SHF
- D.UHF✓
Why: λ = 29.1 cm = 0.291 m. f = 300/0.291 ≈ 1031 MHz ≈ 1.031 GHz . UHF band spans 300–3000 MHz (0.3–3 GHz). 1031 MHz falls squarely in UHF. Alternatively: UHF wavelengths are 10–100 cm; 29.1 cm is in that range. See Section 7 . — Alternative approach — use the wavelength ranges directly from the band table (no calculation needed): UHF = 10–100 cm. 29.1 cm is in 10–100 cm range → UHF. Knowing wavelength ranges as well as frequency ranges gives you two routes to the answer.
Q8. To carry out a phase comparison between two electromagnetic waves:
- A.both waves must have the same amplitude
- B.both waves must have the same frequency✓
- C.both waves must have the same amplitude and frequency
- D.both waves must have the same phase
Why: For a phase comparison to be meaningful, both signals must have the same frequency . If frequencies differ, the relative phase changes continuously — the comparison is meaningless. Amplitude is irrelevant to phase comparison. See Section 8 — Phase Comparison . — This is a direct recall question. The one and only requirement stated in the source: "the two signals being compared must have the same frequency, otherwise any phase comparison would be meaningless."
Q9. The phase of the reference wave is 110° as the phase of the variable wave is 315°. What is the phase difference?
- A.205°
- B.025°
- C.155°✓
- D.335°
Why: Phase difference = Reference − Variable = 110° − 315° = −205°. Negative result → add 360°: −205° + 360° = 155° . See Section 8 . — Always do Reference MINUS Variable . If the result is negative, add 360°. Never do Variable minus Reference (that gives you 205° here — a common wrong answer). The formula direction matters.
Q10. Determine the approximate phase difference between the reference wave and the variable wave: (The reference wave is the solid line and the variable wave is the dashed line)
- A.045°
- B.135°
- C.225°✓
- D.315°
Why: From the diagram, start at zero phase on the reference wave (solid line). Moving in the positive direction, the reference wave travels through approximately 225° before zero phase is reached on the variable wave (dashed line). The variable wave appears to lead the reference by about 135°, which means the reference wave must travel 360° − 135° = 225° in the positive direction. See Section 8 . — On graphical phase questions, always identify the reference wave first, locate its zero crossing, then measure how far it travels (positive direction) before the variable wave's zero crossing. Don't m…
Q11. The wavelength corresponding to a frequency of 15,625 MHz is:
- A.1.92 m
- B.19.2 m
- C.1.92 cm✓
- D.19.2 cm
Why: λ = 300 / 15,625 = 0.0192 m = 1.92 cm . 15,625 MHz = 15.625 GHz, which is in the SHF band (3–30 GHz). SHF wavelengths are 1–10 cm, so 1.92 cm is confirmed correct. See Section 6 and Section 7 . — 15,625 MHz is a typical weather radar or airborne radar frequency (SHF band). Always check: SHF → wavelengths in centimetres (1–10 cm). This eliminates options in metres immediately.
Q12. Which frequency band is a wavelength of 1200 m?
- A.UHF
- B.LF✓
- C.HF
- D.MF
Why: f = 300 / 1200 m = 0.25 MHz = 250 kHz . LF band spans 30–300 kHz . 250 kHz falls in LF. Alternatively: LF wavelengths are 1–10 km; 1200 m = 1.2 km, which is in the 1–10 km range. Both routes confirm LF. See Section 7 . — The MF/LF boundary is at 300 kHz = 1000 m wavelength. 1200 m > 1000 m, so the frequency (250 kHz) is below 300 kHz → LF. Option (d) catches students who don't check the boundary carefully. Recall: longer wavelength = lower frequency = lower band.