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Polarization of light - class-XII

Description: polarization of light
Number of Questions: 39
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Tags: physics wave optics oscillations and waves optics
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Which one is not produced by sound waves in air?

  1. Polarisation

  2. Diffraction

  3. Refraction

  4. Reflection


Correct Option: A
Explanation:

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

A point source of monochromatic light is situated at the centre of a circle, what is the phase difference between the light waves passing through the end points of any diameter

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

  2. $\pi$

  3. $\dfrac{3\pi}{2}$

  4. $zero$


Correct Option: D

An unpolarised light of intensity $32  \mathrm{W} / \mathrm{m}^{2}  $ passes through three polarisers, such that the transmission axis of last polarizer is perpendicular with the first. If the intensity of emergent light is $3  \mathrm{Wh}  $ Im $ ^{2} $ then the angle between the transmission axes of the first two polarisers is:

  1. $ 30^{\circ} $

  2. $ 19^{\circ} $

  3. $ 45^{\circ} $

  4. $ 90^{\circ} $


Correct Option: A

Two polaroids are oriented with their transmision axes making angle of $30^{\circ}$ with each other. The fraction of indicent unpolarised light is transmitted.

  1. $37\%$

  2. $37.5\%$

  3. $3.36\%$

  4. $36.5\%$


Correct Option: B

The axes of the polarizer and analyzer are inclined to each other at an angle of $60^{o}$. If the amplitude of polarized light emerging through the analyzer is $A$, the amplitude of unpolarized light incident on the polarizer is

  1. $A/2$

  2. $A$

  3. $2A$

  4. $2\sqrt{2}\ A$.


Correct Option: D

Unpolarised light of intensity $I _{0}$, is passed through a polarizer. What is the intensity of the transmitted light ?

  1. $I _{0}\cos^2\theta$

  2. $2I _{0}\cos^2\theta$

  3. $I _{0}\cos^2\theta/2$

  4. $2I _{0}\cos^2\theta2$


Correct Option: A

Two polorides are placed having their transmission axis at an angle of $45^0$. If unpolarised light is incident on first polorid acting as polarizer then, calculate intensity of emergent light from second polariser:-

  1. $I _0$

  2. $\frac{I _0}{4}$

  3. $\frac{I _0}{2}$

  4. $2I _0$


Correct Option: B

Plane polarized light is passed through a Polaroid. Now the Polaroid is given one complete rotation about the direction of light propagation. When viewed through another Polaroid (analyser), one of the following is observed:

  1. The intensity of light gradually decreases to zero and then remains zero

  2. The intensity of light becomes twice maximum and twice zero

  3. The intensity of light becomes maximum and stays maximum

  4. The intensity of light does not change


Correct Option: B
Explanation:

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.

When an unpolarised light is polarized, then the intensity of light of the polarized wave :

  1. remains the same

  2. gets doubled

  3. gets halved

  4. depends on the colour of the light.


Correct Option: C
Explanation:

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.

Unpolarized light of intensity $I _{0}$ is incident on a polarizer and the emerging light strikes a second polarizing filter with its axis at 45$^{\circ}$ to that of the first. The intensity of the emerging beam :

  1. $\dfrac{I _{}o}{2}$

  2. $\dfrac{I _{}o}{4}$

  3. $I _{o}$

  4. $\dfrac{I _{}o}{3}$


Correct Option: B
Explanation:

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 $
$I _{t}=\dfrac{I _{0}}{2}\times \dfrac{1}{2}=\dfrac{I _{0}}{4}$

Two Polaroids $P _1$ and $P _2$ are placed with their axis perpendicular to each other. Unpolarized light $l _0$ is incident on $P _1$. A third polaroid $P _3$ is kept in between $P _1$ and $P _2$ such that its axis makes an angle $45^{\circ}$ with that of $P _1$. The intensity of transmitted light through $P _2$ is 

  1. $\frac {I _0 }{2 }$

  2. $\frac {I _0 }{4 }$

  3. $\frac {I _0 }{8 }$

  4. $\frac {I _0 }{16 }$


Correct Option: C
Explanation:
$I=I _0 .\cos^2 \theta$
$\theta =$ angle made by $E$ vector with transmission axis.wherein
$I=$ Intensity of transmitted light after polarisation.
$I _0=$Intensity of incident light.

Intensity of light after crossing $P _1=\dfrac {I _0}{2}$
Intensity of light after crossing $P _3=\dfrac {I _0}{2}.\cos^2 45^o =\dfrac {I _0}{4}$
Intensity of light after crossing $P _2=\dfrac {I _0}{4}.\cos^2 45^o$
$I=\dfrac {I _0}{8}$

Three polaroides are placed one above other, such that the first and the last polaroids are crossed with each other. If the angle between the transmission axis of the first two polaroids is $45$, then what is the percentage of incident light transmitted through the combination of three polaroids?

  1. 0%

  2. 12.50%

  3. 50%

  4. 100%


Correct Option: B

For a given medium, the polarising angle is $60^o$. What will be the critical angle for this medium?

  1. $i = 35^o16'$

  2. $i = 45^o16'$

  3. $i = 55^o16'$

  4. $i = 65^o16'$


Correct Option: A

A light has amplitude A and angle between analyzer and polarizer is.$60 ^ { \circ }$ Light is transmitted by analyzer has amplitude. 

  1. $\mathrm { A } \sqrt { 2 }$

  2. $\frac { A } {2 \sqrt { 2 } }$

  3. $\frac { \sqrt { 3 } \mathrm { A } } { 2 }$

  4. $\frac { A } { 2 }$


Correct Option: B

The polaroids are placed in the path of unpolarized beam of intensity $I _{0}$ such that no light is emitted from the second polaroid. If a third polaroid whose polarization axis makes an angle $\theta$ with the polarization axis of first polaroid, is placed between these polariods then the intensity of light emerging from the last polaroid will be

  1. $\left (\dfrac {I _{0}}{8}\right )\sin^{2} 2\theta$

  2. $\left (\dfrac {I _{0}}{4}\right )\sin^{2} 2\theta$

  3. $\left (\dfrac {I _{0}}{2}\right )\sin^{2} 2\theta$

  4. $I _{0}\cos^{4}\theta$


Correct Option: A

A Polaroid examines two adjacent plane polarised beams $A$ and $B$ whose planes of polarisation are mutually perpendicular. In the first position of the analyser, beam $B$ shows zero intensity. From this position a rotation of $30^{o}$ shows that the two beams have same intensity. The ratio of intensities of the two beams $I _{A}$ and $I _{B}$ will be

  1. $1:3$

  2. $3:1$

  3. $\sqrt{3}:1$

  4. $1:\sqrt{3}$


Correct Option: A

In, the visible region of the spectrum the rotation of the plane of polarization is given by $\displaystyle\theta=a+\frac{b}{\lambda^2}$. The optical rotation produced by a particular material is found to be $30^0$ per $mm$ at $\lambda=5000A^o$ and $50^0$ per $mm$ at $\lambda=4000A^o$. The value of constant $a$ will be

  1. $\displaystyle +\frac{50^0}{9}$ per $mm$

  2. $\displaystyle -\frac{50^0}{9}$ per $mm$

  3. $\displaystyle +\frac{9^0}{50}$ per $mm$

  4. $\displaystyle -\frac{9^0}{50}$ per $mm$


Correct Option: B

An unpolarized beam of intensity $2a^2$ passes through a thin Polaroid. Assuming zero absorption in the Polaroid, the intensity of emergent planes polarized light will be  

  1. $2a^2$

  2. $a^2$

  3. $\displaystyle\sqrt2a$

  4. $\displaystyle\frac{a^2}{\sqrt2}$


Correct Option: B
Explanation:

 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 

A beam of unpolarized light is passed first through a tourmaline crystal $A$ and then through another tourmaline crystal $B$ oriented so that its principal plane is parallel to that of $A$. The intensity of final emergent light is $I$. If $A$ is rotated by $45^0$ on a plane, perpendicular to the direction of incident ray, then intensity of emergent light will be

  1. $\displaystyle\frac{I}{8}$

  2. $\displaystyle\frac{I}{4}$

  3. $\displaystyle\frac{I}{2}$

  4. none of these


Correct Option: C
Explanation:

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

Unpolarized light of intensity $32Wm^{-2}$ passes through three polarizes such that the transmission axis of the last polarizers is crossed with that of the first. The intensity of final emerging light is $3Wm^{-2}$.The intensity of light transmitted by the first polarizered will be 

  1. $32Wm^{-2}$

  2. $16Wm^{-2}$

  3. $8Wm^{-2}$

  4. $4Wm^{-2}$


Correct Option: B
Explanation:

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

A beam of unpolarized light is passed first through a tourmaline crystal $A$ and then through another tourmaline crystal $B$ oriented so that its principal plane is parallel to that of $A$. The intensity of final emergent light is $I$. The value of the $I$ is 

  1. $\displaystyle\frac{I _o}{2}$

  2. $\displaystyle\frac{I _o}{4}$

  3. $\displaystyle\frac{I _o}{8}$

  4. none of these


Correct Option: A
Explanation:

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

Thus $I _1=\dfrac{I _0}{2}$
When this polarized light falls on the next tourmaline crystal at an angle $\theta$, the intensity of light becomes,
$I _2=I _1cos^2\theta=I _1 cos^20^{\circ}$
$=\dfrac{I _0}{2}$

A beam of unpolarized light is passed first through a tourmaline crystal $A$ and then through another tourmaline crystal $B$ oriented so that its principal plane is parallel to that of $A$. The intensity of final emergent light is $I$. Flux of energy of the incident ray is $10^{-3}W$, the percentage of incident light transmitted by the second polarizered will be____

  1. $12.5\%$

  2. $25\%$

  3. $37.5\%$

  4. $50\%$


Correct Option: C

A beam of unpolarized light is passed first through a tourmaline crystal $A$ and then through another tourmaline crystal $B$ oriented so that its principal plane is parallel to that of $A$. The intensity of final emergent light is $I$. The intensity of the emergent beam, if flux of energy of the incident ray is $10^{-3}W$, will be (in $W/m^2$)

  1. $\displaystyle\frac{I}{3}$

  2. $\displaystyle\frac{2I}{3}$

  3. $\displaystyle\frac{4I}{3}$

  4. $\displaystyle\frac{5I}{3}$


Correct Option: D

A Plane polarized light is incidents on an analyzer. The intensity then becomes three-fourth. The angle of the axis of the analyzer with the beam is

  1. $30^0$

  2. $45^0$

  3. $60^0$

  4. zero


Correct Option: A
Explanation:

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

$I=I _0\cos^2\theta$
$\implies \dfrac{3}{4}I _0=I _0\cos^2\theta$
$\cos\theta=\dfrac{\sqrt{3}}{2}$
$\implies \theta=30^{\circ}$

An unpolarized beam of light is incidents on a group of four polarizing sheets, which are arranged in such a way, that of the characteristic direction of each polarizing sheet makes an angle of $30^0$ with that of the preceding sheet. The percentage of incident light transmitted by the first polarizered will be :

  1. $100\%$

  2. $50\%$

  3. $25\%$

  4. $12.5\%$


Correct Option: B
Explanation:

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%

Intensity observed in an interference pattern is $I={ I } _{ 0 }\sin ^{ 2 }{ \theta  } $. At $\theta={30}^{o}$, intensity $I=5\pm 0.002$. The percentage error in angle if $I _0=20w/m^2$is

  1. $4\sqrt { 3 } \times { 10 }^{ -2 }$%

  2. $\cfrac { 4 }{ \pi } \times { 10 }^{ -2 }$%

  3. $\cfrac { 4\sqrt { 3 } }{ \pi } \times { 10 }^{ -2 }$%

  4. $\sqrt 3\times { 10 }^{ -2 }$%


Correct Option: C
Explanation:

$\sin\theta =\sqrt{\dfrac{I}{I _{0}}}$

Differentiating the above equation,
$\cos\theta  d\theta=\dfrac{1}{2}\dfrac{1}{I^{3/2}I _{0}^{1/2}}$
Thus $d\theta=\dfrac{1}{2I}\tan\theta dI$
$\implies \dfrac{d\theta}{\theta}=\dfrac{\tan\theta dI}{2\theta I}$

Put $\theta=30\times \dfrac{\pi}{180}radians$, $dI=0.002,I=5$,
Percentage error in angle $=\dfrac{d\theta}{\theta}\times 100$%
$=\dfrac{4\sqrt{3}}{\pi}\times 10^{-2}$%

When light passing through rotating nicol is observed, no change in intensity is seen. What inference can be drawn ?

  1. The incident light is unpolarized.

  2. The incident light is circularly polarized.

  3. The incident light is unpolarized or circularly polarized.

  4. The incident light is unpolarized or circularly polarized or combination of both.


Correct Option: C
Explanation:

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

Unpolarised light of intensity 32 W m$^{-2}$ passes through three polarizes is crossed with that of the first. The intensity of final emerging light is 3 W m$^{-2}$. The intensity of light transmitted by first polarizer will be

  1. 32 W m$^{-2}$

  2. 16 W m$^{-2}$

  3. 8 W m$^{-2}$

  4. 4 W m$^{-2}$


Correct Option: B
Explanation:

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

A beam of unpolarised light passes through a tourmaline crystal $A$ and then through another such crystal $B$ oriented so that the principal plane is parallel to $A$. The intensity of emergent light is $\displaystyle I$. If $A$ now rotated by $45^{o}$ in a plane perpendicular to direction of the incident ray. The emergent light will have intensity.

  1. $\displaystyle \dfrac{I}{2}$

  2. $\displaystyle \dfrac{I}{\sqrt 2}$

  3. $\displaystyle I$

  4. $\displaystyle \dfrac{I}{4}$


Correct Option: A
Explanation:

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

Three or more number of polaroids ($n$) kept in the path of unpolarized light of intensity $I$ such that angle between any two successive polaroids is other than $90^{\circ}$, then the intensity of emergent light is :

  1. less than I

  2. more than I

  3. I/n

  4. zero


Correct Option: A
Explanation:

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.

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?

  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


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?

  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


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

  1. ${ 15 }^{ o }$

  2. ${ 35 }^{ o }$

  3. ${ 55 }^{ o }$

  4. ${ 75 }^{ o }$


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.

  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 $I _2=0.0W/m^2$


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

  1. $1 : 4$

  2. $1 : 3$

  3. $3 : 4$

  4. $3 : 8$


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? 

  1. $30^{\circ}$

  2. $45^{\circ}$

  3. $22.5^{\circ}$

  4. $60^{\circ}$


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:-

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

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

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

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


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: 

  1. 15%

  2. 25%

  3. 50%

  4. 60%


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.

  1. $25$%

  2. $50$%

  3. $75$%

  4. $90$%


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.

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