Assam Don Bosco University Electromagnetic Waves Parag Bhattacharya Department of Basic Sciences School of Engineering and Technology

Maxwell's Equations

In free space, in the absence of any charge or currents (i.e. ρ = 0, and J = 0) 1) Gauss Law for electric fields 2) Gauss Law for magnetic fields 3) Faraday's Law 4) Ampere's Law 1) Gauss Law for electric fields 2) Gauss Law for magnetic fields 3) Faraday's Law 4) Ampere's Law Integral formDifferential form

Consider the differential form of the Maxwell's Equations (1) (2) (3) (4)

(3) Taking the curl on both sides of (3), we get Therefore, (A)

(A) implies: Each component of the electric field obeys the scalar differential wave equation. Thus, any disturbance in the electric field propagates as a wave.

(4) Similarly, taking the curl on both sides of (4), we get Therefore, (B)

(B) implies: Each component of the magnetic field obeys the scalar differential wave equation. Thus, any disturbance in the magnetic field also propagates as a wave. (B)

Therefore, equations (A) and (B) implies that: (A) (B) Any disturbance in the electromagnetic field will propagate on its own, independent of the medium, in the form of an electromagnetic wave, commonly known as “light”.

Comparing (A) and (B) with the differential wave equation: where 'c': speed of light in free space Plane EM wave:Spherical EM wave:

Superposition of electromagnetic waves

Simplest case: We study the superposition of two harmonic waves with same wavelength and frequency. Consider a 1D EM wave propagating along the positive X-axis: We can separate the space and time parts of the phase by introducing a new quantity as: Thus,

By Supersposition Principle, the resultant wave is: We introduce E 0 and α such that:

Therefore, Conversely, knowing E 01, E 02, α 1 and α 2, we can find E 0, α as: Thus, the composite wave is harmonic and is of the same frequency as the components, though of different amplitude and phase.

The intensity of a wave is proportional to the square of its amplitude. Thus, (for 1st wave) (for 2nd wave) (for resultant wave) We have, where β is a constant Therefore,

Responsible for the variation of intensity due to superposition of waves Interference Term: where the total phase difference is: Thus,

Therefore, the total phase difference between the waves may arise due to:  A difference in the path lengths traversed  A difference in the initial phase angle Coherence: Waves for which 2 – 1 is constant, regardless of its value, are said to be coherent.

The above results can be obtained using the complex representation: The resultant is:

0 Im(E) Re(E) E 01 E 02 E0E0 α1α1 α2α2 α Here,

Superposition of multiple scalar EM waves Consider N harmonic waves with the same frequency and travelling in the same direction. The ith wave is: or, where, i = 0, 1, 2,..., N The resultant wave is: or,

Superposition of multiple scalar EM waves Thus, where, (A) and (B)

(A) If the sources are incoherent: The 2nd summation in (A) will average out to zero because of the random rapid nature of phase changes Note: Intensity is proportional to the time-average of the magnitude of the field squared (taken over a long interval of time) Thus, If each source emits waves of the same amplitude E 01

(A) If the sources are coherent: In the 2nd summation in (A), say the phase is same for both waves at all times, i.e., And, If each source emits waves of the same amplitude E 01 Thus,

Assignment: Two harmonic scalar waves E 1 and E 2 with the same frequency and amplitude travel in the same direction separated by a distance Δx. ΔxΔx E1E1 E2E2 E Apply the superposition principle to finally arrive at the resultant wave E. Obtain the amplitude and phase of the resultant wave in terms of the path difference Δx. Use both trigonometric and complex methods to solve the above problem.

Superposition of two harmonic scalar waves E 1 and E 2 with the same frequency travel in the same direction separated by a distance Δx. ΔxΔx E1E1 E2E2 Assuming the two waves to be travelling in the same medium, both waves have the same speed, and hence the same wavelength. Thus, the angular wave number k is same for both waves. Therefore, we can represent the two waves as:

ΔxΔx E1E1 E2E2 Thus, By superposition principle, the resultant is: where, and, (1) (2) (3)

ΔxΔx E1E1 E2E2 Since the amplitudes are equal, i.e., E 01 = E 02, equations (2) and (3) become: E How??? Prove these results. Therefore,

If Then, Thus, Constructive Interference Then, Thus, Destructive Interference

Assam Don Bosco University Thank You!