Physics

Electromagnetic Waves and Spectrum

659 Questions

Electromagnetic waves and spectrum questions test a candidate's understanding of radiation frequencies, wavelengths, and the properties of different rays like infrared, ultraviolet, and visible light. Concepts also cover practical applications in astronomy and the fundamental speed of light calculations. This physics topic appears regularly in general science sections of major competitive examinations.

UV rays propertiesElectromagnetic radiation frequencyWavelength identificationSpeed of light calculationsBlack body radiation

Electromagnetic Waves and Spectrum Questions

Multiple choice physics observing space: telescopes optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

In the spectrum of light from a star, the wavelength of a particular line is measured to be $4747 $, while the actual wavelength of the line is $4700 $. What is the relative velocity ?

  1. $2\times10^5 m/s$
  2. $3\times10^5 m/s$
  3. $3\times10^6 m/s$
  4. $4.5\times10^6 m/s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the Doppler effect formula for light, delta_lambda / lambda = v / c, where delta_lambda = 4747 - 4700 = 47. Assuming typical values or a misprint in the prompt numbers where delta_lambda is 47 and lambda is 4700, v = c * (delta_lambda / lambda) = (3 x 10^8) * (47 / 4700) = 3 x 10^6 m/s.

Multiple choice physics communication system commonly used terms in electronic communication system elements of communication system electromagnetic waves and communication system

Generation, propagation and detection of electromagnetic waves is the basis of

  1. lasers

  2. reactors

  3. radio and television

  4. computer
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The communication and broadcasting following the base on generation, propagation, and detection of electromagnetic waves.

The electromagnetic spectrum describes a different range of electromagnetic waves. These EM waves are a special type of wave that can travel without a medium.

Electromagnetic waves are named like this due to the fact that they have both an electric and a magnetic component. In a vacuum, EM waves always travel at the same speed i.e. the speed of light. So, other EM waves besides light are infrared, ultraviolet, radio waves, and microwaves. Therefore radio and television both are based on EM wave properties. Other options like lasers, reactors, and computers are not guided by EM waves.

Thus the correct option is C. 

Multiple choice physics communication system commonly used terms in electronic communication system elements of communication system electromagnetic waves and communication system

"Man-made" noise can come from:

  1. equipment that sparks

  2. temperature

  3. static

  4. all of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Man-made noise is a noise created due to wear and tear of electrical machinery used in transmission systems. They are due to the type of amplifiers (class B) type of pulses (wideband pulses), often caused by ignition circuits, lightning, and switching elements causing spark gaps

Multiple choice physics communication system elements of a communication system elements of communication system electromagnetic waves and communication system

In sky wave communication which of the frequency will be reflected back by the ionospheric layer having electron density $4 \times 10 ^ { 6 } / \mathrm { m } ^ { 3 }$

  1. $19 \times 10 ^ { 3 } \mathrm { Hz }$
  2. $9 \times 10 ^ { 3 } H z$
  3. $31 \times 10 ^ { 3 } \mathrm { Hz }$
  4. $20 \times 10 ^ { 3 } \mathrm { Hz }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics communication system elements of a communication system elements of communication system electromagnetic waves and communication system

An antenna is a device that converts

  1. electromagnetic energy into radio frequency

    signal

  2. radio frequency signal into electromagnetic

    energy

  3. guided electromagnetic waves into free space

    electromagnetic waves

  4. free space electromagnetic waves into guided

    electromagnetic waves
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

$Answer:-$ C ,D

An antenna (plural antennae or antennas), or aerial, is an electrical device which converts electric power (guided electromagnetic waves) into radio waves(free spaceelectromagnetic waves), and vice versa. It is usually used with a radio transmitter or radio receiver.

Multiple choice physics option a: relativity the nature of light speed of light and optical density introduction to light

Choose the correct answer from the alternatives given.
An electromagnetic wave of frequency $\nu= 3\ MHz$ passes from vacuum  into a dielectric medium with permittivity $\varepsilon= 4$. Then

  1. wavelength and frequency both become half.

  2. wavelength is doubled and frequency remains unchanged.

  3. wavelength and frequency both remain unchanged.

  4. wavelength is halved and frequency remains unchanged.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given : frequency $v =3 MHz=3\times10^6Hz$, relative permitivity $\varepsilon _r = 4$
Here the frequency of electromagnetic wave remains unchanged but the wavelength of electromagnetic wave changes when it passes from one medium to another.
The refractive index is the square root of permeability and permittivity product. 
For formula,
$c=\dfrac 1{\sqrt {\mu _0\varepsilon _0}}\\\implies c\propto \dfrac1{\sqrt{\varepsilon _0}}$
Similarly,
$v\propto\dfrac1{\sqrt{\varepsilon}}$
Therefore,
$\dfrac cv=\sqrt{\dfrac {\varepsilon}{\varepsilon _0}}=\sqrt{\dfrac 41}=2........(i)$
But
$\dfrac cv=\dfrac {\nu\lambda}{\nu\lambda'}\\\implies \dfrac cv=\dfrac{\lambda}{\lambda'}\\\implies 2=\dfrac{\lambda}{\lambda'}\\\implies \lambda'=\dfrac \lambda2$
Hence wavelength is halved.
Multiple choice physics option a: relativity the nature of light speed of light and optical density introduction to light

The electric field associated with an electromagnetic wave in vacuum is given by $|\overrightarrow { E } |= 40\ cos (kz -6\times{10}^{8}t )$, where $E$, $z$ and $t$ are in volt per meter, meter and second respectively. The value of wave vector $k$ is:

  1. $2\ {m}^{-1}$
  2. $0.5\ {m}^{-1}$
  3. $3\ {m}^{-1}$
  4. $6\ {m}^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: The electric field associated with  an electromagnetic wave in vacuum is given by $|\vec E|=40 \cos(kz−6\times 10^8t)$


To find: Value of wave vector $k$


Solution: 
We know electromagnetic wave eqution is
$|\vec E|=E _0\cos(kz-\omega t)$

And given equation is
$|\vec E|=40 \cos(kz−6\times 10^8t)$

By comparing these two, we get
$\omega=6\times10^8$ and 
$E _0=40$

We also know,
Speed of electromagnetic wave is given by:
$v=\dfrac \omega k$
where v is the speed of the light.

Hence, 
$k=\dfrac \omega v\\\implies k=\dfrac {6\times 10^8}{3\times 10^8}\\\implies k=2m^{-1}$

Option $(A)$ is correct.

Multiple choice physics option a: relativity the nature of light speed of light and optical density introduction to light

The speed of electromagnetic wave in vacuum depends upon the source of radiation. It

  1. increases as we move from $\gamma$-rays to radio waves
  2. decreases as we move from $\gamma$-rays to radio waves

  3. is same for all of them

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Answer:-$ C

speed of electromagnetic wave in vacuum  is given by:-
c(speed of light)=frequency$\times$ wavelength =$\dfrac{1}{\mu _0 \epsilon _0}$=constant
as we go from gamma rays to radio waves  frequency decreases and wavelength increases thereby maintaining the product constant.

Multiple choice physics option a: relativity the nature of light speed of light and optical density introduction to light

The formula for the velocity of electromagnetic waves in vacuum is given by

  1. $c = \sqrt{\mu _0 \varepsilon}$
  2. $c = \dfrac{1}{\sqrt{\mu _0 \varepsilon _0}}$
  3. $c = \sqrt{\dfrac{\mu _0}{\varepsilon _0}}$
  4. $c = \sqrt{\dfrac{\varepsilon _0}{\mu _0}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Maxwell deduced that the speed of propagation of an electromagnetic wave through a vacuum is entirely determined by the constants $\mu _0$ and $\epsilon _0$ as the following:
$c=\frac{1}{\sqrt{\mu _0 \epsilon _0}}$
We know that   $\mu _0 = 4\pi\times 10^{-7}\,{\rm N}\,{\rm s}^2 \,{\rm C}^{-2}$ and  $\epsilon _0 = 8.854\times 10^{-12}\,{\rm C}^2\,{\rm N}^{-1} \,{\rm m}^{-2}$ which gives:
$c=\frac{1}{\sqrt{4 \pi \times 10^{-7} 8.854 \times 10^{-12}}} = 2.998 \times 10^8$ m/s

Multiple choice physics option a: relativity the nature of light speed of light and optical density introduction to light

If a source of power $4kW$ produces $10^{20}$ photons/second, the radiation belongs to a part of the spectrum called:

  1. y-rays

  2. X-rays

  3. Ultraviolet rays

  4. Microwaves
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The correct option is B

given p=4000w


$E=\dfrac{hc}{\lambda}$

$\lambda=\dfrac{hc\times10^{20}}{4000}$

$=hc\times\dfrac{10^{17}}{4}$

$\lambda=3\times10^8\times6.6\times\dfrac{10^{-34+17}}{4}$

$=19.8\times\dfrac{10^{-9}}{4}$

$=4.9\times10^{-9}$

$=49\times10^{-10}$

$=49\dot{A}$

Since,

$0.1\dot{A}<\lambda<100\dot{A}$
 It is X-rays
Multiple choice physics option a: relativity the nature of light speed of light and optical density introduction to light

If $C=$ the velocity of light, which of the following is correct?

  1. ${\mu} _{0}{ \varepsilon } _{ 0 }=c$
  2. ${\mu} _{0}{ \varepsilon } _{ 0 }={c}^{2}$
  3. ${\mu} _{0}{ \varepsilon } _{ 0 }=\cfrac{1}{c}$
  4. ${\mu} _{0}{ \varepsilon } _{ 0 }=\cfrac{1}{{c}^{2}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
In electromagnetic wave, the speed of light is related to the permeability and permittivity constants.
$c=\dfrac 1{\sqrt {\mu _0\varepsilon _0}}\\\implies \mu _0\varepsilon _0=\dfrac1{c^2}$