Physics

Electromagnetic Waves and Spectrum

670 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 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.
$c=\dfrac 1{\sqrt {\mu _0\varepsilon _0}}\\\implies \mu _0\varepsilon _0=\dfrac1{c^2}$
Multiple choice physics option a: relativity the nature of light speed of light and optical density introduction to light

If $v _s$ , $v _x$ and $v _m$ are the velocities of soft gamma rays, X-rays and Microwaves respectively in vacuum, then

  1. $v _s < v _x < v _m$
  2. $v _s = v _x = v _m$
  3. $v _s > v _x > v _m$
  4. $v _x < v _s < v _m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Soft gamma rays, $x-$rays and microwaves are namely eletromagnetic having different wavelengths. 

But, all of them propagate through space with the same speed,
$e=3\times 10^{8}\ m/s$
 so,
$v _{s}=v _{x}=v _{m}$

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

Three observers $A,B$ and $C$ measure the speed of light coming from a source to be $v _A,v _B$ and $v _C$. The observer $A$ moves towards the source and $C$ moves away from the source at the same speed. The observer $B$ stays stationary. The surrounding space is vacuum everywhere.

  1. $\displaystyle v _A>v _B>v _C$
  2. $\displaystyle v _A < v _B < v _C$
  3. $\displaystyle v _A=v _B=v _C$
  4. $\displaystyle v _B=\frac{1}{2}(v _A+v _C)$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

Relativity states that the speed of light in vacuum will be same independent of the frame of reference. The frame of reference can be stationary or moving.
So all  three will find the speed of light to be $ c $
Also $ c=\dfrac{1}{2}(c+c) $
So, (C) and (D) are correct.

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

An electromagnetic wave of frequency $v=3.0:MHz$ passes from vacuum into a dielectric medium with permittivity $\epsilon=4.0$. Then

  1. wavelength is halved and frequency remains unchanged.

  2. wavelength is doubled and frequency becomes half.

  3. wavelength is doubled and frequency remains unchanged.

  4. wavelength and frequency both remain unchanged.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Frequency remains constant during refraction.

$\displaystyle v _{med}=\dfrac{1}{\sqrt{\mu _0\epsilon _0\times4}}=\dfrac{c}{2}$

$\displaystyle\dfrac{\lambda _{med}}{\lambda _{air}}=\dfrac{v _{med}}{v _{air}}=\dfrac{\displaystyle\dfrac{c}{2}}{c}=\dfrac{1}{2}$

$\therefore$ wavelength is halved and frequency remains unchanged.

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

The electromagnetic waves

  1. travel with the the speed of sound.

  2. travel with the the same speed in all media.

  3. travel in free space with the speed of light.

  4. do not travel through a medium.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The electromagnetic waves of all wavelengths travel with the same speed in space which is equal to velocity of light.

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

The apparent wavelength of light from a star moving away from the earth is 0.02% more than actual wavelength. What is the velocity of the star.

  1. $\displaystyle 30{ kms }^{ -1 }$
  2. $\displaystyle 60{ kms }^{ -1 }$
  3. $\displaystyle 90{ kms }^{ -1 }$

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

Answer is B.

The apparent change in the wavelength is given as follows.
$\displaystyle \frac { \Delta \lambda  }{ \lambda  } =\frac { v }{ C } ,Hence\quad V=\frac { \Delta \lambda  }{ \lambda  } C$
$\displaystyle =\frac { 0.02 }{ 100 } \times 3\times { 10 }^{ 8 }{ ms }^{ -1 }=60{ kms }^{ -1 }$.
Hence, the velocity of the star is 60 km/s.

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

An electromagnetic wave of frequency $\mathrm{v}=3.0$ MHz passes from vacuum into a dielectric medium with permittivity $\epsilon =4.0$. Then : 

  1. wavelength is doubled and frequency remains unchanged

  2. wavelength is doubled and frequency becomes half

  3. wavelength is halved and frequency remains unchanged

  4. wavelength and frequency both remain unchanged
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$f=3.0mHz$

$E=4.0$

$C=fZ$

C is the speed of light into the medium

${ C } _{ 0 }=f _{ i }{ Z } _{ i }  C=\dfrac { { C } _{ 0 } }{ n } \\ C={ f } _{ f }{ Z } _{ f }  n=\sqrt { \dfrac { \varepsilon }{ { \varepsilon } _{ 0 } } } \\ C=\dfrac { { C } _{ 0 } }{ 2 } $

Now since frequency depends on source

Hence ${ f } _{ f }={ f } _{ i }\\ { Z } _{ f }=\dfrac { { Z } _{ i } }{ 2 } $

Speed and wavelength will be halved

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

A certain color of light towards the purple end of the visible spectrum has a wavelength of $420\ nm$ in a vacuum.
What is the frequency of this light?
The speed of light in a vacuum is $3.00\times 10^8 m/s$

  1. $126 Hz$
  2. $1.26\times 10^{11}Hz$
  3. $7.1\times 10^{14}Hz$
  4. $1.4\times 10^{-15}Hz$
  5. $1.4\times 10^{-6}Hz$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given :   $v = 3.00 \times 10^8$  m/s                 $\lambda = 420$ nm $ = 420 \times 10^{-9}$  m

$\therefore$ Frequency of the light       $\nu = \dfrac{v}{\lambda} = \dfrac{3.00 \times 10^8}{420 \times 10^{-9}} = 7.1 \times 10^{14}$  $Hz$