Electromagnetic Waves Questions and Answers

Answer: D

The correct relation is \(v = \frac{1}{\sqrt{\mu\varepsilon}}\).

Maxwell's equations give the speed of an electromagnetic wave in a medium as \(v = \frac{1}{\sqrt{\mu\varepsilon}}\). In vacuum this is \(c = \frac{1}{\sqrt{\mu_0\varepsilon_0}} \approx 3 \times 10^{8}\) m/s, and the refractive index is \(n = \frac{c}{v} = \sqrt{\mu_r\varepsilon_r}\).

A dimension check rules out the rest: \(\mu\varepsilon\) has units of s²/m², so only its inverse square root is a speed. \(\frac{1}{\mu\varepsilon}\) is a speed squared, \(\frac{1}{(\mu\varepsilon)^{2}}\) a fourth power and \(\frac{1}{(\mu\varepsilon)^{-2}}\) is \((\mu\varepsilon)^{2}\), not a speed at all.

Answer: E

The peak electric field is \(12 \times 10^{4}\ \text{V m}^{-1}\).

In an electromagnetic wave, \(E_{0} = cB_{0}\). Here \(B_{0} = 400 \times 10^{-6}\) T:

\[ \begin{aligned} E_{0} &= 3 \times 10^{8} \times 4 \times 10^{-4} \\ &= 1.2 \times 10^{5} = 12 \times 10^{4}\ \text{V m}^{-1} \end{aligned} \]

The angular frequency in the argument does not affect the amplitude. \(4 \times 10^{4}\) and the other options come from dropping the factor 3 in c or a power of ten.

Answer: B

Vibrations of atoms and molecules produce infrared radiation.

The energy gaps between the vibrational levels of molecules correspond to photons in the infrared region. That is why warm bodies emit infrared (heat) radiation and why infrared spectroscopy is used to identify molecular bonds: each bond absorbs at the frequency of its own vibration.

The other regions come from other sources in the standard table of the electromagnetic spectrum. Visible and ultraviolet light come from electrons moving between outer energy levels of atoms, X-rays from inner-shell electrons or from fast electrons being slowed down, and microwaves from devices such as klystrons and magnetrons; molecular rotations also lie in the microwave region.

Answer: C

The velocity is \(\frac{c}{3}\).

The speed of light in a medium is

\[ \begin{aligned} v &= \frac{c}{\sqrt{\mu_{r}\varepsilon_{r}}} \\ &= \frac{c}{\sqrt{2 \times 4.5}} \\ &= \frac{c}{\sqrt{9}} \\ &= \frac{c}{3} \end{aligned} \]

\(\frac{c}{\sqrt{3}}\) and \(\frac{c}{2}\) come from using one of the two quantities alone, and \(\frac{c}{4}\) has no basis.

The refractive index of this medium is therefore \(n = \sqrt{\mu_{r}\varepsilon_{r}} = 3\).

Answer: A

The wavelength is 4.0 m.

Compare the phase \(1.57y + 5.4t\) with \(ky + \omega t\). The coefficient of y is the wave number, k = 1.57 m\(^{-1}\), which is \(\frac{\pi}{2}\).

\[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{\pi/2} = 4\ \text{m} \]

Taking \(\lambda = \frac{1}{k}\) or \(\frac{\pi}{k}\) gives about 0.64 m or 2.0 m. The coefficient of t gives the angular frequency, not the wavelength.

Answer: C

The speed depends on the permittivity and permeability of the medium.

The speed of electromagnetic waves in a medium is

\[ v = \frac{1}{\sqrt{\mu\varepsilon}} \]

which reduces to \(c = \frac{1}{\sqrt{\mu_{0}\varepsilon_{0}}}\) in vacuum. The intensity, phase and energy of the wave do not change its speed, and reflectivity is a property of a surface, not of propagation through the medium.

Answer: C

The intensity is 5 Wm^-2.

Intensity is power per unit area. Light radiated uniformly in all directions spreads evenly over the surrounding sphere:

\[ \begin{aligned} I &= \frac{P}{A} = \frac{1000\ \text{W}}{200\ \text{m}^{2}} \\ &= 5\ \text{W m}^{-2} \end{aligned} \]

The 1 kW must be converted to 1000 W. The area is given directly, so there is no need for \(4\pi r^{2}\). 2 and 4 come from slips in the division.

Answer: B

The transverse nature is confirmed by polarization.

Only transverse waves can be polarized: their vibrations are perpendicular to the direction of travel and can be restricted to one plane. When light passes through a polaroid and then a second one is rotated, the transmitted intensity varies and can fall to zero, which shows the vibrations have a direction across the beam.

Diffraction and interference occur for all waves, including longitudinal sound waves, so they prove wave nature but not transverse nature. The photoelectric effect shows the particle nature of light, and total internal reflection is a refraction effect.

Answer: C

The transverse nature is shown by polarization of light.

Only a transverse wave has a direction of vibration at right angles to the direction of travel, and that direction can be filtered. A polaroid passes light vibrating in one plane; two crossed polaroids block it completely. A longitudinal wave, vibrating along its direction of travel, could not be polarized this way.

Interference and diffraction show that light is a wave, but longitudinal waves such as sound interfere and diffract too. Dispersion shows that the speed depends on wavelength. The photoelectric effect shows the particle (photon) nature of light.

Answer: C

The radiations of extremely short wavelength are gamma rays.

Gamma rays sit at the short-wavelength end of the electromagnetic spectrum, with wavelengths below about \(10^{-10}\) m and down to \(10^{-14}\) m. They are emitted by nuclei.

X-rays are next, with longer wavelengths (roughly \(10^{-8}\) to \(10^{-13}\) m), and ultraviolet is longer still. Alpha and beta rays are streams of particles (helium nuclei and electrons), not electromagnetic radiation, so the question of a 'short wavelength radiation' points to gamma rays.