Tag: the nature of light

Questions Related to the nature of light

The speed of electromagnetic wave is same for

  1. odd frequencies

  2. even frequencies

  3. all frequencies

  4. all intensities


Correct Option: D
Explanation:

The speed of electromagnetic wave in a region is same for all intensities but different for different frequencies.

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. $c = \sqrt{\dfrac{\varepsilon _0}{\mu _0}}$


Correct Option: B
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

The amplitudes $E _{0}$ and $B _{0}$ of electric and the magnetic component of an electromagnetic wave respectively are related to the velocity $c$ in vacuum as

  1. $E _{0}B _{0} = \dfrac {1}{c}$

  2. $E _{0} = \dfrac {c}{B _{0}}$

  3. $B _{0} = cE _{0}$

  4. $E _{0} = cB _{0}$

  5. $E _{0} = c^{3}B _{0}$


Correct Option: D
Explanation:

As we know, in electromagnetic waves, speed or light,
$c = \dfrac {E _{0}}{B _{0}} \Rightarrow E _{0} = cB _{0}$.

An electromagnetic wave passing through the space is given by equations $E=E _o\sin(wt-kx), B=B _0\sin(wt-kx)$ which of the following is true?

  1. $E _oB _0=wk$

  2. $E _ow=B _ok$

  3. $E _ok=B _ow$

  4. $E _owk=B _o$


Correct Option: C
Explanation:

As $\dfrac{E _o}{B _o}=c$ (a)

where $c=$speed of light
$c=\nu \lambda$
Also
$w=2\pi \nu$ (1)
$k=\dfrac{2\pi}{\lambda}$ (2)
Dividing (2) by (1)
$\dfrac{w}{k}=\dfrac{2\pi \nu}{\dfrac{2\pi}{\lambda}}=\nu \lambda=c$ 
Hence (a) becomes
$\dfrac{E _o}{B _o}=\dfrac{w}{k}$
$E _ok=B _ow$
Hence the correct option is (C).



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


Correct Option: B
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

For a medium with permitivity $\epsilon$ and permeability $\mu$, the velocity of light is given by:

  1. $\sqrt{\mu/\epsilon}$

  2. $\sqrt{\mu\epsilon}$

  3. $1/\sqrt{\mu\epsilon}$

  4. $\sqrt{\epsilon/\mu}$


Correct Option: C
Explanation:

The velocity of electromagnetic radiation is the velocity of light (c), i.e., 

$c=\dfrac {1}{\sqrt{\mu\epsilon}}$
where $\mu$ is the permeability and $\epsilon$ is the permitivity

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. ${\mu} _{0}{ \varepsilon } _{ 0 }=\cfrac{1}{{c}^{2}}$


Correct Option: D
Explanation:
In electromagnetic wave, the speed of light is related to the permeability and permittivity constants.
$c=\dfrac 1{\sqrt {\mu _0\varepsilon _0}}\\\implies \mu _0\varepsilon _0=\dfrac1{c^2}$

The wave function (in S.I. units) for an electromagnetic wave is given as-
$\psi (x, t) = 10^{3} \sin \pi (3\times 10^{6} x - 9\times 10^{14}t)$ The speed of the wave is:

  1. $9\times 10^{14} m/s$

  2. $3\times 10^{8} m/s$

  3. $3\times 10^{6} m/s$

  4. $3\times 10^{7} m/s$


Correct Option: B
Explanation:
Given: The wave function (in S.I. units) for an electromagnetic wave is given as- $\psi(x, t)=10^3\sin\pi(3\times 10^6x-9\times10^{14}t)$
To find the speed of the wave
Solution: 
We know electromagnetic wave eqution is
$E=E _0\cos(kz-\omega t)$
And given equation is
$\psi(x, t)=10^3\sin\pi(3\times 10^6x-9\times10^{14}t)$
By comparing these two, we get
$\omega=9\times10^{14}$ and 
$k=3\times10^6$
we also know,
Speed of electromagnetic wave, $v=\dfrac \omega k$
where v is the speed of the light
Hence, $v=\dfrac {9\times10^{14}}{3\times10^6}\\\implies v=3\times10^{8}m/s$
is the speed of the wave

The electric field part of an electromagnetic wave in a medium is represented by $ { E } _{ x }=0 $ ;
$ { E } _{ y }=2.5\frac { N }{ C } cos[(2\pi \times { 10 }^{ 6 }\frac { rad }{ s } )t-(\pi \times { 10 }^{ -2 }\frac { rad }{ m } )x] $ ;
$ { E } _{ z }=0 $.The wave is:

  1. Moving along -x direction with frequency $ { 10 }^{ 6 } $ Hz and wave length 200 m.

  2. Moving along y direction with frequency $ 2\pi \times 10^{ 6 } $ Hz and wave length 200 m.

  3. A and B both .

  4. None of them .


Correct Option: A

The velocity of electromagnetic waves in free space is $3 \times 10^6 m/sec$. The frequency of a radio wave of wavelength $150 m$, is:-

  1. $20 \ kHz$

  2. $45 \ MHz$

  3. $2 \ kHz$

  4. $2 \ MHz$


Correct Option: D