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The nature of light - class-VIII

Description: the nature of light
Number of Questions: 29
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Tags: light energy refraction of light physics ray optics and optical instruments optics option a: relativity
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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.

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

Show answer
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. $6\ {m}^{-1}$

Show answer
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

Show answer
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 } } $

Show answer
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

Show answer
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.

Show answer
Correct Option: B
Explanation:

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 is same for

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

  1. odd frequencies

  2. even frequencies

  3. all frequencies

  4. all intensities

Show answer
Correct Option: D
Explanation:

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

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

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

Show answer
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. $E _owk=B _o$

Show answer
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

Show answer
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 .

Show answer
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. $2 \ MHz$

Show answer
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,
$v _{s}=v _{x}=v _{m}$

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

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

  1. $1$

  2. $2$

  3. $0.5$

  4. $0.25$

Show answer
Correct Option: A
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 $A$


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

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

  1. $\displaystyle v _A>v _B>v _C$

  2. $\displaystyle v _A < v _B < v _C $

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

  4. $\displaystyle v _B=\frac{1}{2}(v _C+v _A)$

Show answer
Correct Option: A,D
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.

Three observers $A,B$ and $C$ measure the speed of light coming from a source to be $v _A,v _B$ and $v _C$. The observer $A$ moves towards the source and $C$ moves away from the source at the same speed. The observer $B$ stays stationary. The surrounding space is vacuum everywhere.

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

  1. $\displaystyle v _A>v _B>v _C$

  2. $\displaystyle v _A < v _B < v _C$

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

  4. $\displaystyle v _B=\frac{1}{2}(v _A+v _C)$

Show answer
Correct Option: C,D
Explanation:

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.

Which of the following properties of light conclusively support wave theory of light?

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

  1. Light obeys laws of reflection

  2. Speed of light in water is smaller than the speed in vacuum

  3. Light doesn't show interferrence

  4. Light shows photoelectric effect

Show answer
Correct Option: B
Explanation:
Laws of reflection was successfully explained by both wave theory and corpuscular theory of light.
Only Speed of light in water is smaller than the speed in vacuum was was proven by wave theory of light.

An electromagnetic wave of frequency $v=3.0:MHz$ passes from vacuum into a dielectric medium with permittivity $\epsilon=4.0$. Then

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

  1. wavelength is halved and frequency remains unchanged.

  2. wavelength is doubled and frequency becomes half.

  3. wavelength is doubled and frequency remains unchanged.

  4. wavelength and frequency both remain unchanged.

Show answer
Correct Option: A
Explanation:

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

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

  1. travel with the the speed of sound.

  2. travel with the the same speed in all media.

  3. travel in free space with the speed of light.

  4. do not travel through a medium.

Show answer
Correct Option: C
Explanation:

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

The apparent wavelength of light from a star moving away from the earth is 0.02% more than actual wavelength. What is the velocity of the star.

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

  1. $\displaystyle 30{ kms }^{ -1 }$

  2. $\displaystyle 60{ kms }^{ -1 }$

  3. $\displaystyle 90{ kms }^{ -1 }$

  4. None of these

Show answer
Correct Option: B
Explanation:

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.

All colours travel with the same speed in :

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

  1. Vaccum

  2. Glass 

  3. Water

  4. All of the above

Show answer
Correct Option: A

An electromagnetic wave of frequency $\mathrm{v}=3.0$ MHz passes from vacuum into a dielectric medium with permittivity $\epsilon =4.0$. Then : 

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

  1. wavelength is doubled and frequency remains unchanged

  2. wavelength is doubled and frequency becomes half

  3. wavelength is halved and frequency remains unchanged

  4. wavelength and frequency both remain unchanged

Show answer
Correct Option: C
Explanation:

$f=3.0mHz$

$E=4.0$

$C=fZ$

C is the speed of light into the medium

${ C } _{ 0 }=f _{ i }{ Z } _{ i }  C=\dfrac { { C } _{ 0 } }{ n } \\ C={ f } _{ f }{ Z } _{ f }  n=\sqrt { \dfrac { \varepsilon }{ { \varepsilon } _{ 0 } } } \\ C=\dfrac { { C } _{ 0 } }{ 2 } $

Now since frequency depends on source

Hence ${ f } _{ f }={ f } _{ i }\\ { Z } _{ f }=\dfrac { { Z } _{ i } }{ 2 } $

Speed and wavelength will be halved

The velocity of electromagnetic radiation in a medium of permittivity ${\varepsilon} _{0}$ and permeability ${\mu} _{0}$ is given by :

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

  1. $\sqrt { \dfrac { { \varepsilon } _{ 0 } }{ { \mu } _{ 0 } } } $

  2. $\sqrt { { \mu } _{ 0 }{ \varepsilon } _{ 0 } } $

  3. $\dfrac { 1 }{ \sqrt { { \mu } _{ 0 }{ \varepsilon } _{ 0 } } } $

  4. $\sqrt { \dfrac { { \mu } _{ 0 } }{ { \varepsilon } _{ 0 } } } $

Show answer
Correct Option: C
Explanation:

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.

A certain color of light towards the purple end of the visible spectrum has a wavelength of $420\ nm$ in a vacuum.
What is the frequency of this light?
The speed of light in a vacuum is $3.00\times 10^8 m/s$

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

  1. $126 Hz$

  2. $1.26\times 10^{11}Hz$

  3. $7.1\times 10^{14}Hz$

  4. $1.4\times 10^{-15}Hz$

  5. $1.4\times 10^{-6}Hz$

Show answer
Correct Option: C
Explanation:

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

$\therefore$ Frequency of the light       $\nu = \dfrac{v}{\lambda} = \dfrac{3.00 \times 10^8}{420 \times 10^{-9}} = 7.1 \times 10^{14}$  $Hz$

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