Tag: option a: relativity

Questions Related to option a: relativity

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

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

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

  2. $c = \dfrac{1}{\sqrt{\mu _0 \varepsilon _0}}$

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

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

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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

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

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

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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?

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

  1. $E _oB _0=wk$

  2. $E _ow=B _ok$

  3. $E _ok=B _ow$

  4. $E _owk=B _o$

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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:

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

  1. y-rays

  2. X-rays

  3. Ultraviolet rays

  4. Microwaves

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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:

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

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

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

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

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

Show answer
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?

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

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

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

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

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

Show answer
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:

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

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

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

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

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

Show answer
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:

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

  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 .

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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:-

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

  1. $20 \ kHz$

  2. $45 \ MHz$

  3. $2 \ kHz$

  4. $2 \ MHz$

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Correct Option: D

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

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

  1. $v _s < v _x < v _m$

  2. $v _s = v _x = v _m$

  3. $v _s > v _x > v _m$

  4. $v _x < v _s < v _m$

Show answer
Correct Option: B
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,
$v _{s}=v _{x}=v _{m}$