Polarization of light - class-XII

As sound waves are longitudinal waves, they don't exhibit Polarization phenomenon, which is exhibited by the transverse wave only.

Using Malus's Law, the intensity after the first polarizer is I1 = I0/2 = 16. After the second, I2 = 16cos^2(theta). After the third (perpendicular to the first), I3 = I2cos^2(90-theta) = 16cos^2(theta)sin^2(theta) = 16(sin(2theta)/2)^2 = 4sin^2(2theta). Setting 4sin^2(2theta) = 3 gives sin^2(2theta) = 3/4, so sin(2theta) = sqrt(3)/2, meaning 2*theta = 60 degrees, or theta = 30 degrees.

The intensity after the first polarizer is I0/2. The intensity after the second polarizer is I = (I0/2)cos^2(30). Since cos(30) = sqrt(3)/2, cos^2(30) = 3/4. Thus, I = (I0/2)(3/4) = 3/8 * I0 = 0.375 * I0, which is 37.5%.

Let I0 be the initial intensity. After the polarizer, intensity is I1 = I0/2. After the analyzer, I2 = I1cos^2(60) = (I0/2)(1/4) = I0/8. Since intensity is proportional to the square of amplitude (I = kA^2), A_emergent^2 = (A_incident^2)/8. Thus, A_incident = Asqrt(8) = 2*sqrt(2)*A.

Intensity after the first polarizer is I1 = I0/2. Applying Malus's Law for the second polarizer at 45 degrees: I2 = I1cos^2(45) = (I0/2)(1/2) = I0/4.

The Intensity of light coming out of a polaroid is
given by $I=I _{0}  cos^{2}\theta $
when $\theta $ is changed to $\theta +2\pi $  by rotation
$cos^{2}\theta$ becomes 1 twice at $\pi $ and $2\pi $ and
o twice at $\pi /2$ and $3\pi /2$
Thus option B is correct.

In an unpolarised light the light has Intensity distributed in all polarizing directions.
When it is polarized half the Intensity is polarized in a particular direction and the other half in a perpendicular direction is not transmitted.

The unpolarized light of Intensity $I _{0}$ is incident on a polarizer then a polarized light at intensity $I _{0}/2$ comes out.
Intensity of transmitted beam
$I _{t}=I _{0}/2 cos^{2} 45$ from the formula $I=I _{0} cos^{2} \theta $
$I _{t}=\dfrac{I _{0}}{2}\times \dfrac{1}{2}=\dfrac{I _{0}}{4}$

After the first polarizer, intensity is I0/2. After the second (at 45 degrees), I2 = (I0/2)cos^2(45) = I0/4. After the third (crossed with the first, so 45 degrees to the second), I3 = I2cos^2(45) = (I0/4)*(1/2) = I0/8 = 12.5%.

Brewster's Law states tan(ip) = n. Given ip = 60 degrees, n = tan(60) = sqrt(3). The critical angle c is given by sin(c) = 1/n = 1/sqrt(3). c = arcsin(1/sqrt(3)) is approximately 35.26 degrees, which is 35 degrees 16 minutes.

Intensity I = kA^2. After the polarizer, the amplitude is A/sqrt(2). After the analyzer at 60 degrees, the amplitude becomes (A/sqrt(2))cos(60) = (A/sqrt(2))(1/2) = A/(2sqrt(2)).

Intensity after the first polarizer is I0/2. The second polarizer is at 90 degrees to the first. The third polarizer is at angle theta to the first (so 90-theta to the second). The intensity after the third is (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).

Let the analyzer be at angle alpha. Intensity I = I_Acos^2(alpha) + I_Bcos^2(alpha+90) = I_Acos^2(alpha) + I_Bsin^2(alpha). At alpha = 30 degrees, I_Acos^2(30) = I_Bsin^2(30). I_A*(3/4) = I_B*(1/4), so I_A/I_B = 1/3.

Using theta = a + b/lambda^2: 30 = a + b/(5000^2) and 50 = a + b/(4000^2). Subtracting the equations: 20 = b*(1/1610^6 - 1/2510^6) = b*(9/40010^6). b = 20 * 40010^6 / 9. Substituting back: a = 30 - (20 * 40010^6 / 9) / 2510^6 = 30 - 320/9 = (270-320)/9 = -50/9.

 initial unpolarized intensity is $2a^{2}$ 
the intensity of light transmitted by the first polarizered will be  $\dfrac{I _{unpolarized}}{2}=a^{2}$
option $B$ is correct 

$I _{I}=Icos^{2} \theta =Icos^{2}45=\dfrac{I}{2}$
option $C$ is correct 

 initial unpolarized intensity is $32Wm^{-2}$
the intensity of light transmitted by the first polarizered will be  $\dfrac{I _{unpolarized}}{2}=16Wm^{-2}$

When the unpolarized light falls on the first tourmaline crystal, the intensity of the light halves and becomes polarized.

Intensity of light after passing through an analyzer which is at an angle $\theta$ with the beam,

Since the natural light is unpolarized, the first polaroid reduces the intensity to half. Therefore, percentage of incident light transmitted by first polarized is 50%

For ordinary unpolarized light and circularly polarized light, there is no change in intensity of illumination on passing it through rotating Nicol prism.

Intensity of polarised light transmitted from 1st polariser, 
$I _1$ = $I _0$ cos$^{2} \theta$
but (cos$^{2}\theta) _{av} = \displaystyle\frac{1}{2}$

So $I _1$ = $\displaystyle\frac{1}{2} I _0 = \frac{32}{2} = 16Wm^{-2}$

According to law of Malus,
$\displaystyle I = I _0 cos^{2}\theta =I _0(cos45^{o})^{2} = I _0 \left ( \frac{1}{\sqrt{2}} \right )^{2} = \frac{I _0}{2}$

For unpolarized light, the outgoing intensity from polaroid is $I \cos ^2\theta$
Since $\cos ^2\theta \lt 1$, after passing through each polaroid the intensity will get mlultiplied by $\cos ^2\theta$ where $\theta$ is the angle between successive polaroids.
Hence, final intensity will be less than the original incident intensity.

The correct answer is option(B).

Intensity of unpolarized light $I'=\cfrac { I }{ 2 } \cos ^{ 2 }{ \theta  } $
$\therefore \cfrac { I }{ 2 } \cos ^{ 2 }{ \theta  } =\cfrac { I }{ 6 } $
$\Rightarrow \cos ^{ 2 }{ \theta  } =\cfrac { 1 }{ 3 } $
$\Rightarrow \cos { \theta  } =\cfrac { 1 }{ \sqrt { 3 }  } $
$\therefore \theta ={ 55 }^{ o }$

Intensity of unpolarised light  $I =20 \ W/m^2 $
Intensity of light passing through first polariser  $I _1 = \dfrac{I}{2} = \dfrac{20}{2}= 10 \ W/m^2$
Angle between transmission axis of two poalriser  $\theta = 60 - 30 =30^o$
Intensity of light passing through second polariser  $I _2 = I _1\cos^2\theta$
$I _2 = 10\times \dfrac{3}{4} =7.5 \ W/m^2$

Let $I _0$ be the intensity of unpolarized light, then intensity of light from first transparent thin plate of a polaroid is $I=\dfrac{I _0}2$

Let $\theta$ be angle between the axis of the first two polarisers then obviously $(90^o - \theta)$ is the angle between $2^{nd} , and , 3^{rd}$ polarisers. 
$\therefore I = \dfrac{I _0}{2} , cos^2 \theta , cos^2 (90^o - \theta)$ 

According to Malus's Law, when unpolarized light of intensity I_0 passes through the first polarizer, its intensity becomes I_0 / 2. Each subsequent polarizer reduces the intensity by cos^2(pi/3) = (1/4). With four polarizers in total, there are three successive transmissions, yielding (I_0 / 2) * (1/4)^3 = I_0 / 128.

Initially crossed (90 degrees), I=0. Rotating one by 45 degrees means the angle between them is 45 degrees. Intensity I = (I0/2) * cos^2(45) = I0/4 = 25%.

Let the initial intensity of the light be  $I _o$.
Final intensity of the light   $I = I _o\cos^2\theta$
where  $\theta= 60^o$
$\implies \ I = I _o\times \cos^260^o = 0.25 I _o$
Percentage change in the initial and final intensities  $ = \dfrac{I _o - I}{I _o}\times 100 = \dfrac{I _o-0.25I _o}{I _o}\times 100 = 75$ %
So option C is correct.