Electromagnetic Waves Chapter-Wise Test 15

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What property of visible light enables the human eye to differentiate colors?

The human eye perceives different wavelengths within the visible spectrum as different colors, with retinal cells sensitive to specific wavelength ranges.

Uniform speed
High energy
Different wavelengths
Constant amplitude
3

An electromagnetic wave in vacuum has a wavelength of \( 4 \, \text{m} \). What is its frequency? (Given \( c = 3 \times 10^8 \, \text{m/s} \))

Using \( v \lambda = c \), we have \( v = \frac{c}{\lambda} = \frac{3 \times 10^8}{4} = 7.5 \times 10^7 \, \text{Hz} \).

\( 6 \times 10^7 \, \text{Hz} \)
\( 7.5 \times 10^7 \, \text{Hz} \)
\( 9 \times 10^7 \, \text{Hz} \)
\( 12 \times 10^7 \, \text{Hz} \)
2

What is the nature of the electric and magnetic fields in an electromagnetic wave?

The document states that in an electromagnetic wave, the electric and magnetic fields oscillate sinusoidally, are perpendicular to each other, and are perpendicular to the direction of propagation.

Perpendicular to each other and to the direction of propagation
Parallel to each other
Perpendicular to each other but parallel to propagation
Independent of each other
1

An electromagnetic wave has a wave number \( k = 5 \, \text{rad/m} \). What is its wavelength in vacuum?

The wave number \( k = \frac{2 \pi}{\lambda} \). Given \( k = 5 \, \text{rad/m} \), we have \( \lambda = \frac{2 \pi}{k} = \frac{2 \pi}{5} \approx 1.26 \, \text{m} \).

0.63 m
1.26 m
2.52 m
3.14 m
2

Which type of electromagnetic waves are produced by bombarding a metal target with high-energy electrons?

The document states that X-rays are generated by bombarding a metal target with high-energy electrons.

Gamma rays
X-rays
Ultraviolet rays
Microwaves
2

What is the frequency of an electromagnetic wave with wavelength \( \lambda = 2 \, \text{cm} \) in vacuum? (Given \( c = 3 \times 10^8 \, \text{m/s} \))

Using \( v \lambda = c \), we have \( v = \frac{c}{\lambda} = \frac{3 \times 10^8}{2 \times 10^{-2}} = 1.5 \times 10^{10} \, \text{Hz} \).

\( 1.0 \times 10^{10} \, \text{Hz} \)
\( 1.5 \times 10^{10} \, \text{Hz} \)
\( 2.0 \times 10^{10} \, \text{Hz} \)
\( 2.5 \times 10^{10} \, \text{Hz} \)
2

Which type of electromagnetic waves are used in aircraft navigation radar systems?

Microwaves, due to their short wavelengths, are suitable for radar systems used in aircraft navigation.

Radio waves
Microwaves
Infrared waves
Ultraviolet waves
2

What makes electromagnetic waves capable of carrying energy from one place to another?

The document highlights that electromagnetic waves carry energy due to their oscillating electric and magnetic fields, as seen in applications like radio signals.

Their constant wavelength
Their dependence on a medium
Their variable frequency
Their oscillating electric and magnetic fields
4

What is the general form of the magnetic field component of a plane electromagnetic wave propagating along the \( z \)-direction?

The document provides the form \( B_y = B_0 \sin(kz - \omega t) \), where \( B_y \) is along the \( y \)-axis for a wave propagating along the \( z \)-direction.

\( B_y = B_0 \sin(kz - \omega t) \)
\( B_y = B_0 \cos(kz + \omega t) \)
\( B_y = B_0 \sin(\omega t - kz) \)
\( B_y = B_0 \cos(kz - \omega t) \)
1

The amplitude of the electric field in an electromagnetic wave is \( E_0 = 120 \, \text{V/m} \). What is the amplitude of the magnetic field? (Given \( c = 3 \times 10^8 \, \text{m/s} \))

Using \( B_0 = \frac{E_0}{c} \), we have \( B_0 = \frac{120}{3 \times 10^8} = 4 \times 10^{-7} \, \text{T} \).

\( 2 \times 10^{-7} \, \text{T} \)
\( 3 \times 10^{-7} \, \text{T} \)
\( 4 \times 10^{-7} \, \text{T} \)
\( 5 \times 10^{-7} \, \text{T} \)
3

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