Tag: oscillations and waves

Questions Related to oscillations and waves

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 :

physics oscillations and waves polarization of light polarisation of light polarisation

  1. $100\%$

  2. $50\%$

  3. $25\%$

  4. $12.5\%$

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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

physics oscillations and waves polarization of light polarisation of light polarisation

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

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

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

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

Show answer
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 ?

physics oscillations and waves polarization of light polarisation of light polarisation

  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.

Show answer
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

physics oscillations and waves polarization of light polarisation of light polarisation

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

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

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

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

Show answer
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.

physics oscillations and waves polarization of light polarisation of light polarisation

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

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

  3. $\displaystyle I$

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

Show answer
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 :

physics oscillations and waves polarization of light polarisation of light polarisation

  1. less than I

  2. more than I

  3. I/n

  4. zero

Show answer
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?

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

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