Tag: polarisation of light

Questions Related to polarisation of light

Ordinary light passes through two polarizing filters. The filters have been rotated so that their polarizing axes are oriented at $90^{\circ}$ to each other, and no light gets through both of them.
By adding a third polarizing filter so that there are three in a row, how might one cause light to pass through the three filters?

physics oscillations and waves polarization of light polarisation of light polarisation

  1. Orient the third filter so that its polarizing axis is rotated $45^{\circ}$ clockwise relative to the first and place it in front of the first

  2. Orient the third filter so that its polarizing axis is rotated $45^{\circ}$ counter-clockwise relative to the seconds and place it in back of the second

  3. Orient the third filter so that its polarizing axis is rotated $45^{\circ}$ clockwise relative to the first and place it in between the two filters

  4. Orient the third filter so that its polarizing axis is rotated $90^{\circ}$ clockwise relative to the first and place it in front of the first

  5. Both A and B will work to allow light through

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Correct Option: A
Explanation:
The intensity of the light passing through a polarizer is given by Malus' Law, $I=I _0 cos^2(theta)$. When randomly polarized light (unpolarized light) passes through the first polarizer, the intensity is cut in half because there is an even distribution of incident angles, and the mean value of $cos^2(theta)$ is $\dfrac 12$ from 0 to $\dfrac \pi 2$. When it passes through the second polarizer, our "new $I _0$" (for Malus' Law) is $\dfrac 12$ of the original$ I _0$, and applying Malus' Law, we see that $cos(45°)=\dfrac 12$. In other words, the intensity is halved again and the orientation of the polarization is shifted by 45°. This then happens again on the final polarizer, resulting in a final intensity of $\dfrac 18 I _0$.

If the polarizers had been set to anything other than 45°, this would not be the case.

Polarizing filter # $1$ Is oriented so that its polarizing axis is vertical.
Polarizing filter # $2$ is oriented so that its polarizing axis is rotated clockwise $45^{\circ}$ from filter # $1$
Polarizing filter # $3$ is oriented so that its polarizing filter is rotated $90^{\circ}$ from filter #$1$
Polarizing filter # $4$ oriented so that its polarizing is rotated $135^{\circ}$ from filter # $1$
Which sequence of filters-front to back-will block out all light that starts through the front filter?

physics oscillations and waves polarization of light polarisation of light polarisation

  1. $1, 2, 3$

  2. $3, 2, 1$

  3. $4, 1, 2$

  4. $1, 3, 4$

  5. All of the combinations will block our all the light

Show answer
Correct Option: B
Explanation:

The correct answer is option(B).

The intensity of the light passing through a polarizer is given by Malus' Law, I=I_0 cos(theta)2. When randomly polarized light (unpolarized light) passes through the first polarizer, the intensity is cut in half because there is an even distribution of incident angles, and the mean value of cos(theta)2 is 1/2 from 0 to pi/2. When it passes through the second polarizer, our "new I_0" (for Malus' Law) is 1/2 of the original I_0, and applying Malus' Law, we see that cos(45deg.)2=1/2. In other words, the intensity is halved again and the orientation of the polarization is shifted by 45 degrees. This then happens again on the final polarizer, resulting in a final intensity of 1/8 I_0.
So only case possible for no light is option(B).

Unpolarized light falls on two polarizing sheets placed one on top of the other. What must be the angle between the characteristic directions of the sheets if the intensity of the final transmitted light is one-third the maximum intensity of the first transmitted beam

physics oscillations and waves polarization of light polarisation of light polarisation

  1. ${ 15 }^{ o }$

  2. ${ 35 }^{ o }$

  3. ${ 55 }^{ o }$

  4. ${ 75 }^{ o }$

Show answer
Correct Option: C
Explanation:

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 }  } $
$\therefore \theta ={ 55 }^{ o }$

Two circularly shaped linear polarisers are placed coaxially. The transmission axis of the first polarizer is at $30^o$ from the vertical while the second one is at $60^o$, both in the clockwise sense. If an unpolarised beam of light of intensity $I=20$ $W/m^2$ is incident on this pair of polarisers, then the intensities $I _1$ and $I _2$ transmitted by the first and the second polarisers, respectively, will be close to.

physics oscillations and waves polarization of light polarisation of light polarisation

  1. $I _1=10.0W/m^2$ and $I _2=7.5W/m^2$

  2. $I _1=20W/m^2$ and $I _2=15W/m^2$

  3. $I _1=10.0W/m^2$ and $I _2=8.6W/m^2$

  4. $I _1=15.0W/m^2$ and $I _2=0.0W/m^2$

Show answer
Correct Option: A
Explanation:

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  $I _2 = I _1\cos^2\theta$
$I _2 = 10\times \dfrac{3}{4} =7.5 \ W/m^2$

A transparent thin plate of a polaroid is placed on another similar plate such that the angle between their axes is $30^\circ$. The intensities of the emergent and the unpolarized incident light will be in the ratio of

physics oscillations and waves polarization of light polarisation of light polarisation

  1. $1 : 4$

  2. $1 : 3$

  3. $3 : 4$

  4. $3 : 8$

Show answer
Correct Option: D
Explanation:

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


Now this light will pass through the second similar plate whose axis is inclined at an angle of $30^o$ to that of first plate.
According to Malus law, the intensity of emerging light is:-


$I'= Icos^2 30^o=\dfrac{I _0}2\bigg(\dfrac{\sqrt 3}{2}\bigg)^2=\dfrac 38 I _0$

$\therefore \dfrac{I'}{I _0}= \dfrac 38$

Unpolarised light of intensity $32\ W\ m^{-2}$ passes through three polarizers such that transmission axis of first is crossed with third. If intensity of emerging light is $2\ W\ m^{-2}$, what is the angle of transmission axis between the first two polarisers? 

physics oscillations and waves polarization of light polarisation of light polarisation

  1. $30^{\circ}$

  2. $45^{\circ}$

  3. $22.5^{\circ}$

  4. $60^{\circ}$

Show answer
Correct Option: C
Explanation:

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. 
$\therefore I = \dfrac{I _0}{2} \, cos^2 \theta \, cos^2 (90^o - \theta)$ 

or $2 = \dfrac{32}{2} \, cos^2 \, \theta \, cos^2 \,  (90^o - \theta) = 16 cos^2 \theta \, sin^2 \theta$
$(2sin \theta \, cos \theta)^2 = \dfrac{1}{2} \, or \, sin^2 2 \theta = \dfrac{1}{\sqrt{2}}$
$\therefore 2 \theta = 45^o \, \ or \, \theta = 22.5^o$

On unpolarised beam of light is incident on a set of four polarising plates, such that each plate makes an angle of $\dfrac{\pi}{3}$ with preceding sheet. The light transmitted through the combination is:-

physics oscillations and waves polarization of light polarisation of light polarisation

  1. $\dfrac{1}{128}$

  2. $\dfrac{1}{256}$

  3. $\dfrac{1}{64}$

  4. $\dfrac{1}{32}$

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Correct Option: C

Two polaroids are kept crossed to each other. Now one of the polaroids is rotated through an angle $45^0$. The percentage of incident light now transmitted through the system is: 

physics oscillations and waves polarization of light polarisation of light polarisation

  1. 15%

  2. 25%

  3. 50%

  4. 60%

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Correct Option: B

A plane polarized light is incident normally on a tourmaline plate. Its $\vec { E } $ vectors make an angle of ${ 60 }^{ o }$ with the optic axis of the plate. Find the percentage difference between initial and final intensities.

physics oscillations and waves polarization of light polarisation of light polarisation

  1. $25$%

  2. $50$%

  3. $75$%

  4. $90$%

Show answer
Correct Option: A
Explanation:

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.

In case of linearly polarised light, the magnitude of the electric field vector.

polarisation of light polarisation wave optics optics physics

  1. Does not change with time

  2. Varies periodically with time

  3. Increases and decreases linearly with time

  4. Is parallel to the direction of propagation

Show answer
Correct Option: B
Explanation:

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.