Displacement Current, EM Waves & Electromagnetic Spectrum - Physics | Agrigyan

Standard wavelength ranges in the electromagnetic spectrum:

• X-rays: $1\,\text{nm}$ to $10^{-3}\,\text{nm}$ $\rightarrow$ (A) matches (III).
• Radio waves: $\gt 0.1\,\text{m}$ $\rightarrow$ (B) matches (IV).
• Infrared waves: $1\,\text{mm}$ to $700\,\text{nm}$ $\rightarrow$ (C) matches (I).
• Microwaves: $0.1\,\text{m}$ to $1\,\text{mm}$ $\rightarrow$ (D) matches (II).
  Correct match: (A) - (III), (B) - (IV), (C) - (I), (D) - (II).

• For an electromagnetic wave propagating in a medium with propagation constant $k$ and angular frequency $\omega$, the wave speed $c$ is given by:
  $$c = \frac{\omega}{k}$$
• From Maxwell's equations, the relationship between the peak electric field amplitude $E_0$ and peak magnetic field amplitude $B_0$ is:
  $$c = \frac{E_0}{B_0}$$
• Equating the two expressions for wave velocity:
  $$\frac{E_0}{B_0} = \frac{\omega}{k}$$

In an electromagnetic wave propagating in free space:

• Average electric energy density is:
  $$u_E = \frac{1}{4}\varepsilon_0 E_0^2$$
• Average magnetic energy density is:
  $$u_B = \frac{B_0^2}{4\mu_0}$$
• The relationship between amplitudes is $E_0 = c B_0$ and the speed of light is $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$, so $c^2 = \frac{1}{\mu_0 \varepsilon_0}$:
  $$u_E = \frac{1}{4}\varepsilon_0 (c B_0)^2 = \frac{1}{4}\varepsilon_0 c^2 B_0^2 = \frac{1}{4}\varepsilon_0 \left(\frac{1}{\mu_0 \varepsilon_0}\right) B_0^2 = \frac{B_0^2}{4\mu_0} = u_B$$

The complete electromagnetic spectrum in decreasing order of wavelength (longest to shortest wavelength) is:

Comparing the given options, the correct sequence is:
Radio waves, microwaves, Infrared waves, visible waves, X-rays.