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#electromagnetic wave

47 public questions tagged with this topic.

Optical Phenomena and Applications

What allows light to propagate through a vacuum without requiring a traditional medium?

**Polarization requires transverse waves** because only transverse can have orientation perpendicular to propagation, longitudinal cannot be polarized, wave theory requires light transverse to explain polarization, polaroids transmit only component along pass-axis, unpolarized has random transverse orientations, after polaroid polarized. Light consists of oscillating electric and magnetic fields that sustain each other, enabling propagation without a material medium, as explained by electromagnetic theory. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n

Ref: NCERT > Physics Book > Wave Optics > Optical Phenomena and Applications

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Wave Properties, Frequency and Energy Conservation

What is the frequency of light with a wavelength of \( 595 \, \text{nm} \) in air, given the speed of light in air is \(

**Intensity not depend on speed** when enters denser medium because intensity I ∝ n E₀²? Actually Poynting vector S = E×H, energy density u =½ ε E², for same amplitude E₀ intensity proportional to n, but amplitude changes at interface due to reflection, total energy conserved incident = reflected + transmitted, interference does not destroy energy, it redistributes. Frequency nu = (c/λ) . λ = 5.95 × 10⁻⁷ m , c = 3.0 × 10⁸ m/s . nu = (3.0 × 10⁸/5.95 × 10⁻⁷) ≈ 5.04 × 10¹⁴ Hz . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ,

Ref: NCERT > Physics Book > Wave Optics > Wave Properties, Frequency and Energy Conservation

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Polarization and Malus Law

What type of wave is light, based on the direction of its electric field relative to its propagation direction?

**Polaroid rotation** intensity varies sinusoidally with angle due to Malus law I = I₀ cos²θ, when polaroid rotated 90° from initial, intensity goes from max to zero, for unpolarized light rotating polaroid does not change intensity after first polaroid because average, but second polaroid intensity depends on relative angle, explains why intensity transmitted through two polaroids drops to zero when perpendicular. Light is a transverse wave because its electric field oscillates perpendicular to the direction of propagation. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, s

Ref: NCERT > Physics Book > Wave Optics > Polarization and Malus Law

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Applications of EM Waves in Communication and Medicine

An electromagnetic wave in vacuum has an electric field amplitude of \( 60 \, \text{V/m} \). What is the magnetic field

**Radio waves** effective for long-distance communication because low frequency waves diffract around Earth curvature and reflect from ionosphere, enabling beyond line-of-sight, used in AM (≈10⁶ Hz) ground wave, short wave sky wave. Microwaves used for satellite due to high frequency penetrates ionosphere and directional beam. Using B₀ = (E₀/c) , we have B₀ = (60/3 × 10⁸) = 2 × 10⁻⁷ T . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 2 × 10⁻⁷ T, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Applications of EM Waves in Communication and Medicine

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Production of EM Waves and Hertz Experiment

An electromagnetic wave in vacuum has a frequency of \( 40 \, \text{MHz} \). What is its wavelength? (Given \( c = 3 \ti

**Hertz experiment** used induction coil connected to two rods with gap, spark produced oscillating charge, emitted EM wave, received by loop with gap sparking when E induced, measured wavelength by standing wave, demonstrated EM wave properties, validating Maxwell. Using v λ = c , we have λ = (c/v) = (3 × 10⁸/40 × 10⁶) = 7.5 m . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 7.5 m, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Production of EM Waves and Hertz Experiment

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Production of EM Waves and Hertz Experiment

An electromagnetic wave has a magnetic field amplitude of \( B_0 = 6 \times 10^{-8} \, \text{T} \). What is the electric

**Aerials produce radio waves** by rapid acceleration/deceleration of electrons in antenna driven by AC, frequency equals driving frequency, for 60 MHz, λ=c/f=3×10⁸/60×10⁶=5 m, half-wave antenna length λ/2=2.5 m, efficient radiation when antenna size comparable to λ. Using E₀ = B₀ c , we have E₀ = (6 × 10⁻⁸) × (3 × 10⁸) = 18 V/m . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 18 V/m, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Production of EM Waves and Hertz Experiment

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Production of EM Waves and Hertz Experiment

What is the general form of the electric field component of a plane electromagnetic wave propagating along the \( z \)-d

**Production of EM waves** requires accelerated charge, oscillating LC circuit produces changing E and B, antenna radiates when charge accelerates, frequency determined by L and C, f=1/(2π√(LC)). Hertz used spark gap with inductor and capacitor, produced ~10⁸ Hz radio waves, detected with loop, confirmed transverse nature, reflection, refraction, polarization, speed c. The document provides the form E_x = E₀ sin(kz - ω t) , where E_x is along the x -axis for a wave propagating along the z -direction. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yie

Ref: NCERT > Physics Book > Electromagnetic Waves > Production of EM Waves and Hertz Experiment

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Production of EM Waves and Hertz Experiment

An electromagnetic wave has a magnetic field given by \( B_y = 5 \times 10^{-8} \sin(3 \times 10^3 z - 9 \times 10^{11}

**Aerials produce radio waves** by rapid acceleration/deceleration of electrons in antenna driven by AC, frequency equals driving frequency, for 60 MHz, λ=c/f=3×10⁸/60×10⁶=5 m, half-wave antenna length λ/2=2.5 m, efficient radiation when antenna size comparable to λ. Comparing with B_y = B₀ sin(kz - ω t) , we have k = 3 × 10³ rad/m . Wavelength λ = (2 π/k) = (2 π/3 × 10³) ≈ 2.09 × 10⁻³ m . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 2.09 × 10⁻³ m, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Production of EM Waves and Hertz Experiment

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Production of EM Waves and Hertz Experiment

An electromagnetic wave in vacuum has a magnetic field amplitude of \( B_0 = 3 \times 10^{-8} \, \text{T} \). What is th

**Production of EM waves** requires accelerated charge, oscillating LC circuit produces changing E and B, antenna radiates when charge accelerates, frequency determined by L and C, f=1/(2π√(LC)). Hertz used spark gap with inductor and capacitor, produced ~10⁸ Hz radio waves, detected with loop, confirmed transverse nature, reflection, refraction, polarization, speed c. Using E₀ = B₀ c , we have E₀ = (3 × 10⁻⁸) × (3 × 10⁸) = 9 V/m . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 9 V/m, illustrating EM wave transverse nature and Maxwell's displa

Ref: NCERT > Physics Book > Electromagnetic Waves > Production of EM Waves and Hertz Experiment

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Energy, Intensity and Momentum of EM Waves

An electromagnetic wave in vacuum has an electric field amplitude of \( 90 \, \text{V/m} \). What is the magnetic field

**Energy in EM wave** equally divided between electric and magnetic fields, energy density u = ½ ε₀ E² + B²/(2μ₀) = ε₀ E² = B²/μ₀, average u_avg = ½ ε₀ E₀², intensity I = c u_avg = ½ c ε₀ E₀² = E₀ B₀/(2μ₀) = c B₀²/(2μ₀), radiation pressure p = I/c for absorption, 2I/c for reflection. Using B₀ = (E₀/c) , we have B₀ = (90/3 × 10⁸) = 3 × 10⁻⁷ T . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 3 × 10⁻⁷ T, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

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Energy, Intensity and Momentum of EM Waves

An electromagnetic wave in vacuum has a magnetic field amplitude of \( B_0 = 2 \times 10^{-8} \, \text{T} \). What is th

**Energy in EM wave** equally divided between electric and magnetic fields, energy density u = ½ ε₀ E² + B²/(2μ₀) = ε₀ E² = B²/μ₀, average u_avg = ½ ε₀ E₀², intensity I = c u_avg = ½ c ε₀ E₀² = E₀ B₀/(2μ₀) = c B₀²/(2μ₀), radiation pressure p = I/c for absorption, 2I/c for reflection. Using E₀ = B₀ c , we have E₀ = (2 × 10⁻⁸) × (3 × 10⁸) = 6 V/m . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 6 V/m, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

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Energy, Intensity and Momentum of EM Waves

An electromagnetic wave in vacuum has an angular frequency \( \omega = 4 \times 10^{11} \, \text{rad/s} \). What is its

**Poynting vector** S = E×B/μ₀ gives energy flow (W/m²), magnitude S = E B/μ₀, average = E₀ B₀/(2μ₀) = intensity, direction of propagation, showing energy transport perpendicular to E and B. Frequency v = (ω/2π) . Given ω = 4 × 10¹¹ rad/s , v = (4 × 10¹¹/2 π) ≈ 6.37 × 10¹⁰ Hz . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 6.37 × 10¹⁰ Hz, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

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