The oscillating electric and magnetic vectors of an electromagnetic wave are oriented along

the same direction but differ in phase by 90°
the same direction and are in phase
mutually perpendicular directions and are in phase
mutually perpendicular directions and differ in phase by 90°

Solution

\(\overrightarrow{E}\) and \(\overrightarrow{B}\) are mutually perpendicular to each other and are in phase i.e., they become a zero and minimum at the same place and at the same time.

In which one of the following regions of the electromagnetic spectrum will the vibrational motion of molecules give rise to absorption ?

Ultraviolet
Microwaves
Infrared
Radio waves

Solution

Molecular spectra due to vibrational motion lie in the microwave region of EM-spectrum. Due to Kirchhoff’slaw in spectroscopy the same will be absorbed.

Radio waves and visible light in vacuum have

same velocity but different wavelength
continuous emission spectrum
band absorption spectrum
line emission spectrum

Solution

In vacuum velocity of all EM waves are same but their wavelengths are different.

An electromagnetic wave, going through vacuum is described by E = E₀ sin(kx – ωt). Which of the following is independent of wavelength ?

k
ω
k/ω
kω

Solution

The angular wave number \(k=\frac{2\pi }{\lambda }\); where λ is the wave length. The angular frequency is ω = 2πv.

A plane electromagnetic wave is incident on a material surface. If the wave delivers momentum p and energy E,then

p= 0, E = 0
p ≠ 0,E ≠ 0
p ≠ 0,E = 0
p = 0,E ≠ 0

Solution

EM waves carry momentum and hence can exert pressure on surfaces. They also transfer energy to the surface so p ≠ 0 and E ≠ 0.

It is possible to take pictures of those objects which are not fully visible to the eye using camera films sensitive to

ultraviolet rays
infrared rays
microwaves
radio waves

Solution

It is given by relation = \(\frac{1}{\sqrt{\mu _{0}\varepsilon _{0}}}\)

The transmitting antenna of a radiostation is mounted vertically. At a point 10 km due north of the transmitter the peak electric field is 10^–3Vm^–1. The magnitude of the radiated magnetic field is

3.33 × 10^–10 T
3.33× 10^–12 T
10^–3 T
3 × 10⁵ T

Solution

To double the covering range of a T.V. transmitter tower, its height should be made

2 times
4 times
√2 times
8 times

Solution

It is in accordance with relation = \(\sqrt{2hR}\)
[where h is height of antenna]
So, it must be made 4 times, to make coverage distance 2 times.

The relation between electric field E and magnetic field H in an electromagnetic wave is

E = H
\(E=\frac{\mu _{0}}{\varepsilon _{0}}H\)
\(E=\sqrt{\frac{\mu _{0}}{\varepsilon _{0}}}H\)
\(E=\sqrt{\frac{\varepsilon _{0}}{\mu _{0}}}H\)

The frequency modulated waves are

reflected by atmosphere
absorbed by atmosphere
bend by atmosphere
radio waves

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