Tag: speed of light and optical density

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Questions Related to speed of light and optical density

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

An electromagnetic wave is propagating along x-axis. At x = 1 m and t = 10 s, its electric vector |$\overset{-}{E}|  = 6 V/m$ then the magnitude of its magnetic vector is:

  1. $2 \, \times \, 10^{-8} \, T$

  2. $3 \, \times \, 10^{-7} \, T$

  3. $6 \, \times \, 10^{-8} \, T$

  4. $5 \, \times \, 10^{-7} \, T$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Electric and magnetic compounds of an electromagnetic field are related by 

$E = CB$
$B = \dfrac{E}{C}$
$B = \dfrac{6}{3 \times 10^8}$  (when $E$ is given)
$B = 2 \times 10^{-8} T$ 

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 velocity of electromagnetic waves in a dielectric medium $\left( { \varepsilon  } _{ r }=2 \right) $ is:

  1. $3\times { 10 }^{ 8 }$ meter/second

  2. $1.5\times { 10 }^{ 8 }$ meter/second

  3. $6\times { 10 }^{ 8 }$ meter/second

  4. $7.5\times { 10 }^{ 7 }$ meter/second
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
It turns out that electromagnetic waves cannot propagate very far through a conducting medium before they are either absorbed or reflected. However, electromagnetic waves are able to propagate through transparent dielectric media without difficultly. The speed of electromagnetic waves propagating through a dielectric medium is given by 
$c'=\dfrac{c}{\epsilon _r}$
Multiple choice physics option a: relativity the nature of light speed of light and optical density introduction to light

If the relative permeability of a medium is $ \mu _r$ and its dielectric constant is  $\varepsilon _r$ then the velocity of light in that medium will be

  1. $ \sqrt{\dfrac { \mu _r }{ { { \varepsilon } _{ r } } }} $

  2. $ \dfrac { 1 }{ \sqrt { { \mu } _{ r }{ \varepsilon } _{ r } } } $

  3. $ \sqrt { { \mu } _{ r }{ \varepsilon } _{ r }/{ \mu } _{ { \varepsilon } _{ 0 } } } $

  4. $ \sqrt { { \mu } _{ 0 }{ \varepsilon } _{ 0 }/{ \mu } _{ { r } }{ \varepsilon } _{ r } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$c=\dfrac{1}{\sqrt{\epsilon _0\mu _0}}$

$v=\dfrac{1}{\sqrt{\epsilon\mu}}=\dfrac{1}{\sqrt{\epsilon _0\epsilon _r\mu _0\mu _r}}$
$=\dfrac{c}{\sqrt{\epsilon _r\mu _r}}$

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

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. $E _{0} = c^{3}B _{0}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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