50 questions on this topic of Electromagnetic Waves (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.
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Answer: (C) Diagnosis (imaging) and treatment of certain cancers
X-rays penetrate soft tissue more than bone, so they are used for diagnostic imaging, and controlled doses are used to treat some cancers.
NCERT Class 12
Answer: (B) Red visible light and Infrared radiation
Pulse oximeters utilize the difference in absorption spectra between oxygenated hemoglobin (HbO₂) and deoxygenated hemoglobin (Hb). Hb absorbs more red light (around 660 nm), whereas HbO₂ absorbs more infrared light (around 940 nm). Measuring their differential transmission allows estimation of blood oxygen saturation.
Answer: (A) 1.0 × 10⁻⁸ N
Radiation pressure on a completely absorbing surface at normal incidence is P_rad = I / c. Total force exerted F = P_rad × Area = (I × A) / c = (1.5 × 10⁶ × 2.0 × 10⁻⁶) / (3.0 × 10⁸) = 3.0 / (3.0 × 10⁸) = 1.0 × 10⁻⁸ N.
Answer: (A) They are reflected back to the Earth by ions present in the ionosphere.
Short-wave radio frequencies (roughly 3 MHz to 30 MHz) undergo total internal reflection or refraction from the ionosphere, allowing them to bounce between the ionosphere and Earth surface to achieve sky-wave propagation across long distances.
Answer: (D) Radar systems and microwave ovens
Microwaves are suitable for radar used in aircraft navigation, and in microwave ovens their frequency matches resonant frequency of water molecules to heat food.
NCERT Class 12
Answer: (B) 47.4 V/m
The intensity at distance r is I = Power / (4πr²) = 150 / (4π × 2²) = 150 / (16π) ≈ 2.984 W/m². Intensity is related to the peak electric field by I = (1/2) ε₀ c E₀². Therefore, E₀ = √[2I / (ε₀ c)]. Substituting ε₀ c ≈ (8.85 × 10⁻¹²)(3.0 × 10⁸) = 2.655 × 10⁻³: E₀ = √[(2 × 2.984) / (2.655 × 10⁻³)] = √[2248] ≈ 47.4 V/m.
Answer: (B) Statement I is correct but Statement II is incorrect
Statement I is correct because infrared rays have longer wavelengths and thus scatter much less than visible light, allowing clear images through haze. Statement II is incorrect because Rayleigh scattering intensity varies inversely with the fourth power of wavelength (I ∝ 1/λ⁴), not directly with the square.
Answer: (A) Both Statement I and Statement II are true.
According to NCERT, the amplitude modulated (AM) broadcast band extends from 530 kHz to 1710 kHz. In cellular telephony, radio waves in the UHF band (typically between 800 MHz and 900 MHz or higher) are employed to transmit signals to and from mobile handsets. Hence, both statements are true.
Answer: (A) Both Statement I and Statement II are correct
Cellular mobile communications operate within the UHF band (typically 800-900 MHz to 1800-2400 MHz), which lies in the range 300-3000 MHz. The ionosphere reflects frequencies only up to the critical frequency (generally below 30-40 MHz); waves of higher frequencies, such as TV signals and UHF, penetrate through the ionosphere, making sky-wave propagation unviable.
Answer: (B) 1.0 × 10¹⁶
Energy of one photon is E = hc / λ = (6.6 × 10⁻³⁴ × 3.0 × 10⁸) / (600 × 10⁻⁹) = 1.98 × 10⁻²⁵ / 6.0 × 10⁻⁷ = 3.3 × 10⁻¹⁹ J. Number of photons n = total pulse energy / E = (3.3 × 10⁻³ J) / (3.3 × 10⁻¹⁹ J) = 1.0 × 10¹⁶.
Answer: (A) Statement I is true, but Statement II is false
Ionospheric reflection occurs for short waves (roughly 3 to 30 MHz) enabling sky-wave propagation. Frequencies above 30 MHz (such as space waves and satellite links in the UHF/microwave band) penetrate right through the ionosphere without being reflected or absorbed, which is precisely why they are used for satellite communication.
Answer: (D) Ultraviolet rays
The stratospheric ozone layer absorbs harmful ultraviolet (UV) radiation emitted by the Sun. Exposure to excessive UV radiation can cause sunburn, skin cancer, and cataracts in humans.