Notes on Wave Behaviour of Light and Electromagnetic Waves

Oscillating charges and electromagnetic waves

• Stationary charges produce a stationary electric field; oscillating charges produce an oscillating electric field.

• A moving charge generates a magnetic field perpendicular to its motion, in addition to the electric field; constant velocity gives a constant magnetic field; oscillating charge yields an oscillating magnetic field.

• A changing or oscillating magnetic field induces an oscillating current in a conductor.

• James Clerk Maxwell (Scottish mathematician and physicist) proposed the electromagnetic radiation theory in 1864; the wave model for light states that an oscillating electric field produces an oscillating magnetic field, and an oscillating magnetic field produces an oscillating electric field.

• When charges are forced to oscillate, they emit electromagnetic waves that consist of oscillating electric and magnetic fields; the two fields are at right angles to each other and to the direction of wave propagation.

• By definition, electromagnetic waves are transverse waves and travel with a constant speed of $c = 3.00 \times 10^{8}~\mathrm{m\,s^{-1}}$ in a vacuum.

The wave equation and electromagnetic spectrum

• The wave speed, frequency, and wavelength are related by the wave equation: $v = f \lambda$.

• The electromagnetic spectrum diagram (Figure 3.1.2) indicates magnitude ranges of wavelength and frequency for different EM waves (radio, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays).

• Key ideas (from the text):
  1) Oscillating charges produce EM waves of the same frequency as the charge oscillation; EM waves cause charges to oscillate at the wave frequency.
  2) EM waves are transverse with mutually perpendicular, oscillating electric and magnetic fields.
  3) EM waves travel at a constant speed $c = 3.00 \times 10^{8}~\mathrm{m\,s^{-1}}$ in vacuum.
  4) The wave equation $v = f \lambda$ applies to all forms of electromagnetic radiation.

• Worked Example: Frequency of X-rays with wavelength $\lambda = 2.00 \times 10^{-12}~\mathrm{m}$.

  • Use $v = f \lambda$ with $v = c = 3.00 \times 10^{8}~\mathrm{m\,s^{-1}}$.

  • Solve for frequency: $f = \frac{v}{\lambda} = \frac{3.00 \times 10^{8}}{2.00 \times 10^{-12}} = 1.50 \times 10^{20}~\mathrm{Hz}.$

Radio and television signals: Antennae

• Antennae (usually aluminium rods) convert electric currents to electromagnetic radiation and vice versa.

• Antennae are designed to transmit and/or receive signals within specific frequency ranges based on their size and shape.

• Types of antennas:

  • Dipole antenna: two metal rods.

  • Monopole antenna: a single rod with a ground plane (e.g., vehicle antenna where the vehicle roof acts as the ground plane).

• The plane of polarization: the plane in which the electric field vibrates; radio and TV waves are plane-polarised; the oscillations occur in a single plane parallel to the transmitting antenna.

• The plane of polarization is defined by the plane of the oscillating electric field (Figure 3.1.4 shows vertically and horizontally plane-polarised waves).

Transmitting a radio or television signal using an antenna

• Electrons in the transmitting antenna are forced to oscillate along the length of the antenna by an alternating potential difference (RF) connected to the antenna; this creates an oscillating electric field.

• The oscillating electric field induces an oscillating magnetic field at right angles to the electric field; the two fields are perpendicular to each other and to the direction of wave travel.

• Because these fields continuously reproduce one another, plane-polarised EM radio/TV waves radiate away from the antenna in all directions.

• The plane of polarisation is defined by the plane of the oscillating electric field.

• Example: A vertically plane-polarised wave is emitted from a vertical transmitting antenna (Figure 3.1.5).

Receiving a radio or television signal using an antenna

• An approaching EM wave exerts a force on the stationary electrons in the receiving antenna via the electric field: $F = qE$.

• The electrons in the receiving antenna oscillate at the same frequency as the transmitted wave, producing an alternating potential difference across the two rods of the dipole.

• The orientation of the receiving antenna relative to the transmitted wave matters: a signal is strongest when the receiving antenna is aligned parallel to the plane of polarization of the wave (i.e., the receiving antenna’s axis is parallel to the oscillating electric field).

• If the receiving antenna is oriented at right angles to the transmitting antenna’s plane of polarization, the electric field cannot efficiently drive electron oscillation, leading to a very weak signal (or none).

• As an example, city antennas (horizontal polarization) versus country antennas (vertical polarization) illustrate how signals are received best when the receiver’s orientation matches the signal’s polarization; cross-polarization reduces interference.

Wave terminology and coherence

• Monochromatic light: derived from Greek words mono (one) and chroma (color); consists of a single frequency (and hence wavelength) and radiates in all directions from the source.

• In-phase vs out-of-phase:

  • In phase: crests align with crests; troughs align with troughs (constructive similarity).

  • Out of phase: crests align with troughs (half-wavelength phase difference); this is often described as a phase difference of (\pi) or 1/2 wavelength.

• Coherent sources: wave sources that maintain a constant phase relationship with each other; they must have the same frequency to remain in phase as they radiate.

Light from an incandescent source

• Incandescent light is produced by heating a material until it glows (e.g., a filament in a light bulb).

• The light is often white and consists of a broad range of wavelengths; the filament’s random electron motion results in a spectrum of frequencies.

• Such light is neither monochromatic nor coherent (the emitted waves do not maintain a constant phase relationship).

• Examples: glowing coals (incandescent) and a heated filament (light bulb).

Principle of superposition and interference

• Principle of superposition: when two or more EM waves overlap, the resultant electric and magnetic fields at any point are the vector sum of the individual fields.

• Constructive interference: waves are in phase; the resultant amplitude is the sum of the amplitudes (e.g., two in-phase identical waves yield a resultant amplitude of twice the original).

• Destructive interference: waves are out of phase (e.g., half-wavelength phase difference); the resultant amplitude can cancel out (zero for identical waves).

Two-source interference

• Interference patterns from two coherent sources depend on path differences from the sources to a point on a screen.

• If sources are in phase, crests and troughs align (constructive interference) at points where the path difference is an integer multiple of the wavelength:$\Delta \mathrm{PD} = m\lambda,\quad m = 0,1,2,\dots$

• Destructive interference occurs where the path difference is an odd multiple of half a wavelength:$\Delta \mathrm{PD} = \left(m + \tfrac{1}{2}\right)\lambda,\quad m = 0,1,2,\dots$

• Figure 3.1.15 maps crests and troughs for two coherent sources; curved lines denote loci of equal path difference (PD).

Young’s double-slit experiment

• Demonstrated in 1801 by Thomas Young; shows interference fringes (bright and dark) for light, providing evidence of wave nature.

• The two-slit interference pattern consists of alternating bright and dark fringes of equal spacing; bright fringes correspond to constructive interference, dark fringes to destructive interference.

• The colour of bright fringes matches the monochromatic light used (for a single wavelength). Figure 3.1.16 shows sodium (yellow) light’s pattern.

Producing two-slit interference in the lab

• Using an incoherent source: two-slit interference can be produced by first having light pass through a single slit to create circular waves; the double slits then act as two coherent sources.

• Interferometer setup (Figure 3.1.17 and 3.1.18): a single slit S produces circular waves that reach double slits S1, S2; diffracted waves then interfere to form a pattern on a screen.

• If a coherent source (e.g., a laser) is used, a single slit is not required; the laser itself provides coherence, and a double-slit setup yields visible fringes or, with a highly coherent laser, bright spots.

• Fringe separation Ay and its dependence on slit separation d, distance to screen L, and wavelength λ: $A_y \approx \frac{\lambda L}{d}$ (for small angles).

• Practical notes: interferometer dimensions (slit width ~ 10^{-6} m; slit separation ~ 10^{-4} m; screen ~ 0.30 m away) determine fringe visibility and spacing.

Diffraction and interference with gratings

• Diffraction: bending and spreading of waves around obstacles or through openings; observable when wavelength is comparable to the slit width or obstacle size.

• Diffraction grating: a substance with many parallel, equally spaced slits; maxima are very narrow and occur at large angles; higher angular orders exist depending on wavelength and slit spacing.

• For a transmission diffraction grating with parallel slits of spacing d, the maxima occur when

  • For the m-th order: $d \sin \theta = m \lambda,\quad m = 0,1,2,\dots$

• The role of single-slit diffraction in producing a grating pattern: each slit diffracts; the superposition of many diffracted wavelets yields sharp maxima when path differences align to whole wavelengths.

• White-light diffraction: central white line at the center; different wavelengths satisfy the grating condition at different angles, resulting in violet-to-red dispersion in higher orders; bright line appears where the path difference matches integers of wavelength for all colors simultaneously only at the central maximum.

Diffraction grating specifics and spectroscopy

• Transmission diffraction gratings can have large line densities (e.g., lines per cm). Spacing between lines is

  • If a grating has N lines per meter, then the slit spacing is $d = \frac{1}{N} \quad\text{(m)}.$ For example, 3000 lines per cm = 3.0 \times 10^{5} lines/m gives $d = \frac{1}{3.0 \times 10^{5}} = 3.33 \times 10^{-6}~\text{m}.$

• The angle of a given order can be found from $d \sin \theta = m \lambda.$ For a red light with (\lambda = 680~\text{nm} = 680 \times 10^{-9}~\text{m}) and d = 3.33 × 10^{-6} m, the third-order maximum angle is
  $\sin \theta = \frac{m \lambda}{d} = \frac{3 \times 680 \times 10^{-9}}{3.33 \times 10^{-6}} \Rightarrow \theta \approx 38^\circ.$

• Distance between adjacent maxima on the screen (fringe separation) for a two-slit analogy:
  $A_y \approx \frac{\lambda L}{d}.$

• Example problems (selected):

  • A grating with d = 3.33 × 10^{-6} m illuminated by 680 nm light; compute the third-order maximum angle: $\theta \approx \sin^{-1}\left(\frac{3\lambda}{d}\right) \approx 38^\circ.$

  • A grating with a known d and a blue laser: since blue has a shorter wavelength than red, the angular position of maxima increases for the same order.

  • If the angular position of the fifth-order maximum is given as 1.04°, compute the wavelength: use $\sin\theta = \frac{m\lambda}{d}$ to solve for (\lambda).

• Worked examples (derived values):

  • Example: slit spacing from a grating of 300 lines per mm illuminated by 632.8 nm light with a screen distance; compute angle and wavelength using the grating equation.

  • Example: a 1.5 mm slit separation grating used with a laser; compute angle for a given order and wavelength.

Diffraction grating formulas and derivations

• Core formulas:

  • Grating equation: $d \sin \theta = m \lambda,\quad m = 0,1,2,\dots$

  • Fringe separation for two slits: $A_y = \frac{\lambda L}{d}$

  • For a diffraction grating: the same relation applies with multiple slits; maxima are sharper and occur at larger angles due to small d.

• Derivation note (Figure 3.1.32): parallel rays from a transmission grating pass through a lens, with path difference for the m-th order given by $d \sin \theta = m \lambda.$ The intensity maxima occur at angles satisfying this condition.

Diffraction grating in spectroscopy and practical applications

• A diffraction grating is suitable for high-precision wavelength measurements because:

  • The grating lines are close together (larger angular spacing for maxima) allowing precise angle measurement.

  • The maxima are thin and intense, reducing measurement error.

• Spectroscopy uses diffraction gratings to view atomic spectra; wavelengths can be accurately determined via the grating equation and measured angles.

• White-light spectroscopy with a diffraction grating produces a central white line, with dispersion of colors (violet to red) in successive orders; higher orders show overlapping of colors, making the pattern more complex.

Key formulas summary (two-slit and grating)

• Two-slit interference (monochromatic light):

  • Condition for maxima: $\Delta \mathrm{PD} = m \lambda,\quad m = 0,1,2,\dots$

  • Condition for minima: $\Delta \mathrm{PD} = \left(m + \tfrac{1}{2}\right) \lambda,\quad m = 0,1,2,\dots$

  • Fringe spacing (on screen): $A_y = \frac{\lambda L}{d}$, where d is the separation of the slits and L is the screen distance.

• Diffraction grating:

  • Maxima condition: $d \sin \theta = m \lambda,\quad m = 0,1,2,\dots$

  • Distance between lines on the grating: if there are N lines per metre, then $d = \dfrac{1}{N}.$ For 3000 lines per cm, $d = \dfrac{1}{3.0 \times 10^{5}} = 3.33 \times 10^{-6}~\text{m}.$

• Diffraction of white light: central maximum is white; higher orders show dispersion with violet at smaller angles than red for the same order because shorter wavelengths diffract at larger angles for a fixed d, or vice versa depending on geometry.

Worked problems (brief solutions)

• Example 1: A two-slit interferometer with slit separation d = 5.00 × 10^{-5} m, screen distance L = 0.40 m, light wavelength λ = 485 nm. To find the angle of the third-order maximum:

  • Use $d \sin \theta = m \lambda$ with m = 3 and λ = 485 × 10^{-9} m.

  • Solve for sin θ: $\sin \theta = \dfrac{3 \times 485 \times 10^{-9}}{5.0 \times 10^{-5}} \approx 0.0146$ (sample value; compute accurately for exam).

  • Then (\theta ≈ \sin^{-1}(0.0146) ≈ 0.84^{\circ}) (example value shows method).

• Example 2: Red laser of λ = 628 nm illuminates two slits with screen distance L = 3.00 m; measured distance between adjacent maxima on screen is Ay = 1.50 × 10^{-2} m. Find the slit separation d:

  • Use Ay ≈ (λ L)/d ⇒ d ≈ (λ L)/Ay = (628 × 10^{-9} × 3.00)/(1.50 × 10^{-2}) ≈ 1.26 × 10^{-6} m.

  • For a blue laser (shorter λ), the same d yields larger angle separation or smaller depending on configuration; the fringe spacing decreases as λ decreases if L and d are fixed.

• Example 3: With a diffraction grating of 300 lines per mm illuminated by a helium-neon laser, third-order maximum observed at angle 34.8°; find λ:

  • N = 300 lines/mm => d = 1/(300 × 10^3) m = 3.33 × 10^{-6} m.

  • Solve for λ from d sin θ = m λ with m = 3: $\lambda = \dfrac{d \sin\theta}{m} = \dfrac{3.33 \times 10^{-6} \times \sin(34.8^{\circ})}{3} \approx 6.33 \times 10^{-7}~\text{m}.$ (≈ 633 nm, red HeNe line.)

• Example 4: Maximum number of orders for a grating with 3000 lines per cm illuminated by λ = 680 nm:

  • d = 3.33 × 10^{-6} m; max order mmax is the largest integer with mλ ≤ d; mmax ≈ ⌊d/λ⌋ ≈ ⌊3.33 × 10^{-6} / 680 × 10^{-9}⌋ ≈ 4; hence four orders are possible.

• Example 5: White-light diffraction pattern: central white line; first-order shows violet to red; second/third orders show overlapping colors due to dispersion; higher orders are less intense and spread further apart.

Connections and real-world relevance

• Antennae physics underpins radio/TV broadcasting and reception; matching polarization between transmitter and receiver boosts signal strength and reduces interference.

• Diffraction gratings form the basis for spectroscopy, enabling precise wavelength measurements essential to chemical analysis, astronomy, and materials science.

• Understanding coherence and interference is foundational for imaging systems, holography, and optical communication.

Ethical, philosophical, and practical implications

• Accurate wavelength measurement supports safety-critical applications (e.g., chemical detection, environmental monitoring, medical diagnostics).

• The study of light’s wave nature has influenced philosophy of science, challenging particle-only models and guiding the development of quantum concepts.

• Applications like Blu-ray/discs and anti-reflective surfaces rely on precise control of interference at nanoscale, illustrating the importance of practical physics in technology.

Quick reference formulas (index)

• Speed of EM waves in vacuum: $c = 3.00 \times 10^{8}~\mathrm{m\,s^{-1}}$

• Wave equation: $v = f \lambda$

• Two-slit interference (maxima): $\Delta \mathrm{PD} = m\lambda,\quad m = 0,1,2,\dots$

• Two-slit interference (fringe spacing): $A_y = \frac{\lambda L}{d}$

• Destructive interference (half-period): $\Delta \mathrm{PD} = \left(m + \tfrac{1}{2}\right)\lambda$

• Diffraction grating (maxima): $d \sin \theta = m \lambda$

• Grating line spacing: $d = \frac{1}{N}$ where N is lines per metre

• Central white line in white light diffraction: central maximum occurs for all wavelengths at PD = 0.

• Diffraction grating spectroscopy advantages: narrow, intense lines; large angular separation increases measurement precision.