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

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. $E _owk=B _o$
Reveal answer Fill a bubble to check yourself
C Correct answer
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).



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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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
Multiple choice physics option a: relativity the nature of light speed of light and optical density introduction to light

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. $\sqrt{\epsilon/\mu}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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. ${\mu} _{0}{ \varepsilon } _{ 0 }=\cfrac{1}{{c}^{2}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
In electromagnetic wave, the speed of light is related to the permeability and permittivity constants.
$c=\dfrac 1{\sqrt {\mu _0\varepsilon _0}}\\\implies \mu _0\varepsilon _0=\dfrac1{c^2}$
Multiple choice physics option a: relativity the nature of light speed of light and optical density introduction to light

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. $3\times 10^{7} m/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
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
Multiple choice physics option a: relativity the nature of light speed of light and optical density introduction to light

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 .
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics option a: relativity the nature of light speed of light and optical density introduction to light

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

  1. $v _s &lt; v _x &lt; v _m$

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

  3. $v _s &gt; v _x &gt; v _m$

  4. $v _x &lt; v _s &lt; v _m$
Reveal answer Fill a bubble to check yourself
B Correct answer
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,
$v _{s}=v _{x}=v _{m}$

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

If velocity of an electromagnetic wave in a medium is $3\times 10^8m/s$ then find refractive index of medium.

  1. $1$

  2. $2$

  3. $0.5$

  4. $0.25$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

When $\mu=$ refractive index of wave
$V _{n}=$ velocity of wave in vacuum
$V _{m}=$ velocity of wave in medium
Now, velocity of wave in vacuum $=3\times 10^{8}m/s$
$\Rightarrow V _{n}=3\times 10^{8}m/s$
velocity of wave in medium $=3\times 10^{8}m/s$ (given)
$\Rightarrow V _{m}=3\times 10^{8}m/s$
Hence, $\mu=\dfrac{3\times 10^{8}}{3\times 10^{8}}=1$
$\Rightarrow \mu=1 =$ option $A$


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

Three observers $A,B$ and $C$ measure the speed of light coming from a source as $v _A,v _B$ and $v _C$. Observer $A$ moves towards the source, $C$ moves away from source and $B$ stays stationary. The surrounding medium is water. If A and C are moving at the same speed, then

  1. $\displaystyle v _A&gt;v _B&gt;v _C$

  2. $\displaystyle v _A &lt; v _B &lt; v _C $

  3. $\displaystyle v _A=v _B=v _C$

  4. $\displaystyle v _B=\frac{1}{2}(v _C+v _A)$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

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.