Tag: the nature of light

Questions Related to the nature of light

According to Maxwell's hypothesis, changing of electric filed give rise to

physics option a: relativity maxwell's equations the nature of light introduction to electromagnetic waves

  1. magnetic field

  2. pressure gradient

  3. charge

  4. voltage

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Correct Option: A
Explanation:

According to Maxwell's hypothesis, changing of electric field gives rise to Magnetic field.

We know that $F=qE,$, where $F$ is force and $E$ is electric field.
We can relate magnetic field and force by $F=qvB$, where $v$ is velocity and $B$ is the magnetic field.
Therefore we can obtain magnetic field by changing electric field.
Therefore option $A$ is correct.

Unpolarized light falls first on polarizer $\left( P \right) $ and then on analyzer $\left( A \right) $. If the intensity of the transmitted light from the analyser is $\dfrac { 1 }{ 8 }$th of the incident unpolarized light. What will be the angle between optic axes of $P$ and $A$?

physics option a: relativity maxwell's equations the nature of light introduction to electromagnetic waves

  1. ${ 45 }^{ o }$

  2. ${ 30 }^{ o }$

  3. Zero

  4. ${ 60 }^{ o }$

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Correct Option: D
Explanation:

Given,
$I=\dfrac { { I } _{ 0 } }{ 2 } $              ....(i)
${ I }^{ ' }=I\cos ^{ 2 }{ \theta  } $                 $\left( \because { I }^{ ' }=\dfrac { { I } _{ 0 } }{ 8 }  \right) $
$\therefore \dfrac { { I } _{ 0 } }{ 8 } =\dfrac { { I } _{ 0 } }{ 2 } \cos ^{ 2 }{ \theta  } $
From the equation (i), we have
$\dfrac { 1 }{ 4 } =\cos ^{ 2 }{ \theta  } \Rightarrow \cos { \theta  } ={ 1 }/{ 2 }$
$\Rightarrow \cos { \theta  } =\cos { { 60 }^{ o } } $
$\Rightarrow \theta ={ 60 }^{ o }$

A plane electromagnetic wave with an intensity of $200 W/m^2$ is incident normal to a flat plate of radius 30 cm. If the plate absorbs $60%$ and reflect $40%$ of the incident radiation, what is the momentum transferred to it in 5 min?

physics option a: relativity maxwell's equations the nature of light introduction to electromagnetic waves

  1. $1.7 \times 10^{-3} kg ms^{-1}$

  2. $2.7 \times 10^{-4} kg ms^{-1}$

  3. $3.7 \times 10^{-4} kg ms^{-1}$

  4. $3.7 \times 10^{-3} kg ms^{-1}$

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Correct Option: A

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

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

  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.

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Correct Option: D
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.

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:

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

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

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

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

  4. $5 \, \times \, 10^{-7} \, T$

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Correct Option: A
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)
$B = 2 \times 10^{-8} T$ 

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:

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

  1. $2\ {m}^{-1}$

  2. $0.5\ {m}^{-1}$

  3. $3\ {m}^{-1}$

  4. $6\ {m}^{-1}$

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Correct Option: A
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.

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

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

  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

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Correct Option: B
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 
$c'=\dfrac{c}{\epsilon _r}$

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

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

  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 } } $

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Correct Option: B
Explanation:

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

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

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

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

  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

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Correct Option: C
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.

The electromagnetic waves travel with a velocity

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

  1. equal to velocity of sound.

  2. equal to velocity of light.

  3. less than velocity of light.

  4. None of the above.

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Correct Option: B
Explanation:

Velocity of electromagnetic waves $\displaystyle=\dfrac{1}{\mu _0\epsilon _0}=3\times{10}^8:m/s=$ velocity of light.