Polarisation of light - class-XII

In any type of light whether polarised or unpolarised, the magnitude of electric field vector always varies periodically with time.
Actually the change in electric field vector gives rise to periodically changing magnetic field.

The intensity of the light after passing through the polariser
$I , = , I _0 , cos^2\phi , = , I _0 , cos^245$
$= , I _0 \left ( \dfrac{I}{\sqrt2} \right )^2 , = , \dfrac{I}{2} , \times , , \dfrac{I}{2} , = , , \dfrac{I}{4} ,, \left ( \because , I _0 , = , , \dfrac{I}{2} \right )$

Given,

A quarter-wave plate is basically used for the elliptical polarization of the unpolarized light incident over it. The quarter-wave plate produces the electric field at the various angle about the axis of the light and thereby making the light elliptically polarized.

Brewster's law,

The refractive index of water, $\mu= \dfrac{\text{Speed of light in air }}{\text{Speed of light in air}}= \dfrac{3\times10^8}{2.2\times 10^8}=1.36$

If '$i$' is polarizing angle and '$c$' is critical angle for a medium. taking other medium as air for which refractive index $=1$,

Let $I _0$ be the intensity of incident light. As the light coming from sodium lamp is unpolarised, so the intensity of the light emerging from the first polaroid is
$I _1 , = , \dfrac{I _0}{2}$
If $\theta$ is angle between two polaroids, then the intensity of the light emerging from the second polaroid is
$I _2 , = , I _1 , cos^2\theta , = , \dfrac{I _0}{2} cos^2\theta$
But $I _0 , = , 100%$ (Given)
$\therefore , I _1, = , 50% , cos^2\theta$
Since $\theta$ varies from 90 to 0, so the intensity of the outcoming light  can be varied from 0% to 50%.

Brewster's angle i_p satisfies tan(i_p) = mu, and critical angle theta_c satisfies sin(theta_c) = 1/mu. Thus, tan(i_p) = 1/sin(theta_c) = cosec(theta_c), so i_p = tan^-1(cosec(theta_c)).

When unpolarized light passes through a polarizer, the intensity becomes I_0/2. The second polarizer's orientation does not change the intensity of the already polarized light if we assume the question implies the first polarizer is the one with the vertical axis.

By definition, the plane of polarization is the plane containing the direction of propagation and the electric field vector, which is the plane of vibration.

Brewester's angle depends on the wavelength of light because Brewester's angle depends on the refractive index of medium and refractive index depends on the wavelength of light. 

Refractive index mu = tan(i_p) = tan(60) = sqrt(3). Critical angle theta_c = sin^-1(1/mu) = sin^-1(1/sqrt(3)).

The polarization phenomenon, verifies the transverse nature of light. Since sound has longitudinal nature, so it does not show polarization effect.

When the light is incident at the polarising angle on the transparent medium, the reflected light is completely polarised.

$ \beta $ are stream of particles comprising of electrons moving with very high velocity, hence it cannot be polarized, while others can as they are electromagnetic waves.

Polarisation of light establishes that light are transverse in nature, otherwise it was believed that they are longitudinal waves, like the sound waves.

Polarisation of light establishes that light are transverse in nature, otherwise it was believed that they are longitudinal waves, like the sound waves.

When angle of incidence is equal to the Brewster's angle then, the angle between reflected and refracted rays is $ { 90 }^{ o }$

Electromagnetic wave theory explains light as being composed of electric field vibrating in planes.
Polarisation of light refers to vibrating electric field in a particular plane.

$ \bf{Plane\ of\ Vibrations} $ is the plane in which the vibrations of electric vector of plane polarized light are present. 

when a beam of unpolarized light is incident on the glass block, certain amount of the incident vibration is reflected in a plane perpendicular to the plane of incidence and transmits the rest.
Thus reflected ray is completely polarized but not partially polarized.
Thus only (c) is wrong and rest options are correct.

Different kinds of glass have different refractive index. Thus as polarising angle depends on refractive index the polarising angle is different for different kinds of glass. $tan \ i = \mu $

In 1669, another Danish scientist, Erasmus Bartholinus discovered the polarization of light by double refraction in Iceland spar (calcite).

Brewster's angle is an angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface,
with no reflection.
$\theta _{B}=tan^{-1}(\dfrac{n _{2}}{n _{1}})$
$n _{2}$ refractive index of final medium
$n _{1}$ refractive index of initial medium
As different wavelength have different refractive indices the brewster angle is also different.

pile of plates is used to produce plane polarized light by refraction.
By refraction through multiple plates vibrations in the plane of incidence are transmitted.
Thus we obtain a polarized light.

At polarizing angle $\theta _{B}$
$\dfrac{sin \theta _{B}}{cos \theta _{B}}=\dfrac{sin \theta _{B}}{sin \gamma }$
we get $sin \gamma =cos \theta _{B}$
$sin \gamma =sin(90-\theta _{B})$
Thus $\gamma =90-\theta _{B}$

The refractive index of glass is n
If the light is incident on the surface with an angle of incidence i given by tan i $=\mu $ using the Brewster's law the reflected light is completely polarized.
Here tan i $=n$
i$=tan^{-1}(n)$

Angle of incidence $=$ polarising angle
$tan i = \mu $
$tan i = 1.732$
$i = 60^{0}$
From snell's law
$\dfrac{sin i}{sin \gamma }=\mu $
$sin \gamma =\dfrac{sin 60^{0}}{1.732}$
$\gamma =30^{0}$

If polarizing angle is $\theta $
Then tan $\theta =\dfrac{1}{sin c}$
$tan \theta =\dfrac{1}{sin 45}=\sqrt{2}$
$\theta =54^{0}{44}'$

Huygen's theory failed to explain polarisation because according to the Huygen's theory light is a longitudinal waves and longitudinal wave does not shows polarisation.

The experiments on stationary light waves establish that most of the optical properties of light depends on the electric vector, which is also known as light vector.

Polarizers are used in industry to reveal stress patterns in machinery and tools. Sunglasses are used to protect the eyes by polarizing the light to reduce glare using the tailor-made material of the glasses.

Only the transverse waves can be polarized.

Light in the form of a plane wave in space is said to be linearly polarized. Light is a transverse electromagnetic wave, but natural light is generally unpolarized, all planes of propagation being equally probable. 

When ordinary (unpolarised) light is reflected from a surface , it gets partial polarisation , it means some of the electric vectors in some planes are cut-off . When this partially polarized light is incident on polarized glasses , glass acts as an analyser and therefore block out some reflected light .

The angle between incident ray and reflected is $21°$ and the angle between reflected and refracted ray is $90°$.

The wavelength in a medium is lambda_medium = lambda_vacuum / mu. Since mu = tan(phi) at the polarizing angle, lambda_medium = lambda / tan(phi).

Brewster's angle i_p = tan^-1(mu) = tan^-1(4/3) = 53 degrees. The angle with the horizon is 90 - 53 = 37 degrees.

Malus's Law: I = I_0 * cos^2(theta). We want I = I_0/2, so cos^2(theta) = 1/2, which means cos(theta) = 1/sqrt(2), so theta = 45 degrees.

Given,

sin(theta_c) = 1/mu = sin(45) = 1/sqrt(2), so mu = sqrt(2) = 1.414. Polarizing angle i_p = tan^-1(1.414) = 54.7 degrees. Angle of refraction r = 90 - 54.7 = 35.3 degrees.

Given i_p = 53 degrees. Angle of refraction r = 90 - 53 = 37 degrees. Refractive index mu = tan(53) = 1.327.

At Brewster's angle, the reflected and refracted rays are perpendicular (A). The reflected light is polarized perpendicular to the plane of incidence (C is incorrect). The refracted light is partially polarized (D is correct).

When unpolarized light of intensity I0 passes through the first polaroid A, its intensity becomes I1 = I0 / 2. When it passes through the second polaroid B at an angle of 45 degrees, Malus's law applies: I2 = I1 * cos^2(45) = (I0 / 2) * (1 / 2) = I0 / 4.

At the polarizing angle, i_p + r = 90 degrees. r = 90 - 53 degrees 4 minutes = 36 degrees 56 minutes.

Both light and sound waves exhibit interference, refraction, and reflection because these are general wave properties. However, light is a transverse electromagnetic wave and can be polarized, whereas sound is a longitudinal pressure wave and cannot be polarized.

In transverse wave propagation, the direction of vibration of the electric field vector is perpendicular to the direction of wave travel, while the plane of polarization contains the direction of propagation and the vibration vector. Thus, the angle between the direction of vibration and the plane of polarization is 90 degrees.

Polarization is a property of transverse waves. Ultrasonic waves are longitudinal waves and cannot be polarized.

Refractive index mu = tan(i_p) = tan(60) = sqrt(3).

Refractive index mu = tan(i_p) = tan(60) = sqrt(3).

Initial intensity I = I_0. After turning 30 degrees, I' = I_0 * cos^2(30) = I_0 * 3/4. The change is I_0 - 0.75I_0 = 0.25I_0, which is 25%.

When unpolarized light is incident at Brewster angle from one medium to another, the light that is reflected from the surface is perfectly polarized

Dichromism is the selective absorption of one orthogonal polarization component of an incident beam over the other. this phenomenon is due to anisotropy of the material, with one polarization component experiencing preferential absorption.

$\theta _B=tan^{-1}\dfrac{n _2}{n _1}$

Reflect and refracted rays are completely polarized with their planes of polarization perpendicular to each other.
option $B$ is correct 

Phenomenon of polarisation helps in establishing the fact that light waves are transverse in nature, otherwise it was believed that they are longitudinal in nature like sound waves.

Brewster's angle or polarizing angle is given by $\tan\theta=\dfrac{n _{medium}}{n _{air}}=\dfrac{\lambda _a}{\lambda _m}$

• Only transverse waves can be polarised. Longitudinal waves (sound waves in air) cannot be polarised. The reason that only transverse wave can be polarised is that their vibrations can  be potentially occur in all directions perpendicular to the direction of travel.
  
• Polarisation of "light wave" is possible only because they can oscillate in more than one orientation because they are transverse in Nature.
• It has no dependence  on wavelength and frequencies
• Hence option D is the right answer

At Brewster angle, only the reflected light is completely plane polarised, but the transmitted (refracted) light is partially polarised.

When the reflected light is completely polarised, then the angle between reflected and refracted ray are $ { 90 }^{ o }$

As reflected and refracted rays are perpendicular to each other, therefore, i$ _p$ = i = 60$^{o}$, where i$ _p$ is called angle of polarisation.
$\displaystyle\mu = tan i _{p} = tan 60^{o} = \sqrt{3}$ = 1.732.

Using Brewster law, $ \tan { { i } _{ p } } =\mu$

To have polarized waves, they first need to be transverse waves - the disturbance needs to be at right angles to the direction of propagation. Therefore sound waves in air (the usual sort) or in other gases and liquids can't be polarized because they're purely compressive.
Hence, option D is correct.

If the light in incident on the surface with an
angle of incidence i given by tan i $=\mu $,
the reflected light is completely polarized.
Thus $\mu =tan 58^{0}$
refractive $\mu =1.6$
index
From snell's law
$\dfrac{sin i}{sin \gamma }=\mu $
$\dfrac{sin 58^{0}}{1.6}=sin \gamma $
$\gamma =32^{0}$

$tan  p =\dfrac{1}{sinc}$
where p is polarising angle
c is critical angle
$p=tan^{-1}(\dfrac{1}{sin 45^{0}})$
$=tan^{-1}(\dfrac{1}{1/\sqrt{2}})     (\because sin 45^{0}=\sqrt{2})$
$=tan^{-1}(\sqrt{2})$

$I={ I } _{ 0 }\cos ^{ 2 }{ \theta  } $
Intensity of polarized light $=\dfrac { { I } _{ 0 } }{ 2 } $
$\therefore$ Intensity of untransmitted light $={ I } _{ 0 }-\dfrac { { I } _{ 0 } }{ 2 } =\dfrac { { I } _{ 0 } }{ 2 } $

From Brewster's law
$\mu=\tan{{i} _{p}}$
$\Rightarrow$ $\cfrac{c}{v}=\tan{{60}^{o}}$
$=\sqrt {3}$
$\Rightarrow$ $v=\cfrac{c}{\sqrt{3}}$
$=\cfrac{3\times {10}^{8}}{\sqrt {3}}=\sqrt {3}\times {10}^{8}m/s$

A

The idea behind polarized sunglasses is to eliminate the glare. If the solar glare is all at a $90^o$ angle to the normal line, sunglasses polarized at a $ 0^o$ angle to this normal will not allow any of the glare to pass. Most other light is not polarized, so it will still be possible to see the road and other cars, but the distracting glare will cease to be a problem. 

Polarising angle for water, $i _p = 53^o4'$

$I=I _o (cos^2\phi)^4$

Relation between refractive index of medium and polarising angle is given by
$n = \tan i _p$
Given :  Polarising angle  $i _p = 60^o$
So, refractive inedx  $n = \tan (60^o) = 1.732$

well resolved (definition of qualia)

$\Delta \theta  = \left( {\dfrac{{1.22\lambda }}{d}} \right)$
$d = \left( {\dfrac{{1.22 \times \lambda }}{{\Delta \theta }}} \right)$
$ \Rightarrow ,,{10^{ - 4}}, \to ,\dfrac{\lambda }{{180}} \times {10^{ - 4}}$
$ = 1.74 \times {10^{ - 6}},rad.$
$d = \left( {\dfrac{{1.22 \times 5000 \times {{10}^{ - 10}}}}{{1.74 \times {{10}^{ - 6}}}}} \right)$
$ = 0.35,m,,or,,35,cm$

A transverse wave is a wave that oscillates perpendicular to its direction of propagation.  Polarization of light proved that light is a transverse wave.

Polarised light is used to studied helical structures of nucleic acids using different methods such as optical rotatory dispersion. In this method due to refractive indices of the medium for left and right circularly polarised light reflects the structural details of such molecular complexes.

Because of birefringence the ordinary and extra ordinary ray are completely polarized.

Because the speed of light waves in a medium is equal to their speed in a vacuum divided by the index of refraction for that wavelength, an extra ordinary ray can be either faster or slower than an ordinary ray. The extra ordinary ray does not obey snell's law. It is because refractive index of the e-ray varies according to the direction taken because it has components that are both parallel and perpendicular to the crystal's optic axis.

Only for the ordinary ray the refractive index is observed to be constant in all directions

Nucleic acid is an optically active medium. When circularly polarized light passes through an absorbing optically active medium, the speeds between right and left polarizations differ as well as their wavelengths. This can be used to study the helical structure of nucleic acids.
An optical axis is a line along which there is some degree of rotational symmetry in an optical system. Thus it is a particular direction and not any
random line.

polaroid has absorption of O- Ray as the phenomenon behind its polarising ability.

Polaroid consists of needle - like crystals aligned in a particular direction.
Thus option C is correct answer.

calcite crystal display an optical property called birefringence. This causes objects viewed through a clear piece of calcite to appear doubled.
This is the reason one will see one dot rotating about the other.

In double - refraction the refractive index of the extra - ordinary ray varies according to the direction taken because it has components that  are both parallel and perpendicular to the crystal's optic axis. Thus (A) is correct.
The ordinary ray and the extraordinary ray are polarized in planes vibrating at right angles to each other. Thus (B) is correct.

For quartz crystal $\mu _{0}=1.544  ;  \mu _{e}=1.553$
For calacite crystal $\mu _{0}=1.658  ;  \mu _{e}=1.486$
Thus both a and b are true.

Huygens waves theory could well explain reflection, refraction and interference of light but could not explain the double refraction because in huygens principle light is considered as mechanical wave which could not explain the double refraction.

Some transparent crystals, such as tourmaline and calcite, have the property that when one views an object through them one sees two images of the object. If one passes a narrow beam of light through them, the refracted beam is split into two parts which travel through the crystal and emerge as two separate beams. One of the beams obeys the ordinary laws of refraction and is called the ordinary ray. The other beam is called the extraordinary ray. The extraordinary ray does not always lie in the plane of incidence. Its speed, and consequently its index of refraction, depends on its direction of propagation through the crystal.

When unpolarized light is incident at Brewster's angle, the reflected light is plane-polarized. The electric field component perpendicular to the plane of incidence remains in the reflected light, while the parallel component is refracted.

The first polarizer reduces intensity to I0/2. The second polarizer is at angle theta to the first, so intensity becomes (I0/2) * cos^2(theta). The third polarizer is at 90 degrees to the first, meaning it is at (90-theta) to the second. Intensity = (I0/2) * cos^2(theta) * cos^2(90-theta) = (I0/2) * cos^2(theta) * sin^2(theta) = (I0/2) * (sin(2theta)/2)^2 = (I0/8) * sin^2(2theta).

Brewster's law states that the angle of incidence for which reflected and refracted rays are perpendicular is given by tan(i) = n. Here, n = 4/3, so i = tan^-1(4/3).

A polarizer is an optical filter that lets light waves of a specific polarization pass while blocking light waves of other polarizations, effectively producing polarized light from unpolarized light.

Polarizability increases down a group in the periodic table as the atomic radius increases and the outer electrons are further from the nucleus, making them easier to displace. Xenon (Xe) is the largest noble gas listed.

Unpolarized light passing through the first polarizer has its intensity halved (I1 = I0 / 2). The second polarizer is at an angle of 60 degrees to the first, so by Malus's law, the transmitted intensity is I2 = I1 * cos^2(60) = (I0 / 2) * (1 / 4) = I0 / 8.

If a polarizer makes an angle of 60 degrees with the electric vector of polarized light, the intensity is reduced by a factor of cos^2(60) = (1/2)^2 = 1/4.

The derivation is identical to question 527898. The intensity is (I0/8) * sin^2(2*theta).

This is a repeat of the previous question. The intensity is (I0/8) * sin^2(2*theta).

A dot which is shifted by some distance is appeared due to normal incidence and another dot is appeared which is rotating along with the crystal, when we see at a angle to the normal.
Hence option $D$ is correct .

The ability of a substance to rotate plane-polarized light is called optical activity and these substances are called optically active

In Nicol Prism, ordinary ray is eliminated and Extra-ordinary ray, which is plane polarized, is transmitted through the prism.

In general equation for electric field vector for linearly polarized light is given as $\left| \vec { E }  \right| ={ E } _{ o }\sin { \left( { \omega t }-{ kx } \right)  } $ which indicate that its varies periodically with respect to time.

Optically active substance by definition are substances which can rotate the plane of polarization of polarized light.

In calcite crystal, only along the optical axis, the refractive indices of the crystal are equal for O-ray and E-ray

By definition, optically active substances are those which can rotate the plane of polarisation of polarised light.

Polaroid glass cut glare and haze by reducing the light intensity to half on account of polarisation, so that eyes are more comfortable and viewer can see better, therefore they are used in sun glasses.

When a ray of light enters nicol prism, it splits into two plane polarised light in mutually perpendicular direction. One of this light undergoes total reflection and absorption whereas other comes out as a plane polarised light.

A Nicol prism is use to produce and analyse the polarised light.

Brewster's law states that the angle of incidence i for which reflected light is totally polarized is given by tan(i) = n, where n is the refractive index.