Electromagnetic Waves and Light Properties

Electric and magnetic compounds of an electromagnetic field are related by 

Given: The electric field associated with  an electromagnetic wave in vacuum is given by $|\vec E|=40 \cos(kz−6\times 10^8t)$

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

$Answer:-$ C

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

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

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

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

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

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

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

We know that 
$\mu=\dfrac{V _{n}}{V _{m}}$

Speed of the light is constant with respect to any observer holds only for the vacuum. In different medium (say water) if observer is moving with respect to medium towards the source, the speed observed by the observer will be more compared to a stationary observer. 
Same thing holds for a observer moving away from the source. Hence Option A. 
Since they are moving at the same speed, change in speed will be same and hence Option D. 
So answer is A and D.

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.

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.

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

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.

In a vacuum, the refractive index is 1 for all frequencies, so all colors (wavelengths) travel at the same speed c. In media like glass or water, dispersion occurs, meaning speed depends on the refractive index, which varies with wavelength.

Velocity of electromagnetic radiation is the velocity of light $\left(c\right)$, i.e.,
$c=\dfrac { 1 }{ \sqrt { { \mu  } _{ 0 }{ \varepsilon  } _{ 0 } }  } $
where ${ \mu  } _{ 0 }$ is the permeability and ${ \varepsilon  } _{ 0 }$ is the permittivity of free space.

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