Tag: polarisation

Quizzes (0) Questions (165)

Questions Related to polarisation

Multiple choice polarisation of light polarisation wave optics optics physics

If the angle between the pass axis of the polarizer and the analyzer is 45, the ratio of the intensities of original light and the transmitted light after passing through the analyzer is 

  1. $\dfrac{I}{2}$
  2. $\dfrac{I}{3}$

  3. I
  4. $\dfrac{I}{4}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

Multiple choice polarisation of light polarisation wave optics optics physics

The angle between the pass axis of polarizer and analyzer is $45^{\circ}$. The percentage of polarised light passing through analyzer is:

  1. 75%

  2. 25%

  3. 50%

  4. 100%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,

$\theta=45^0$

By malus law,
$I\propto cos^2\theta$
$I\propto (cos45^0)^2$

$I\propto \left(\dfrac{1}{\sqrt{2}}\right)^2$

$I\propto (0.5)$
The percentage of polarized light passing through analyzer is $50$%.
The correct option is C. 

Multiple choice polarisation of light polarisation wave optics optics physics

When ordinary light is made incident on a quarter wave plate, the emergent light is:

  1. linearly polarised

  2. circularly polarised

  3. unpolarised

  4. elliptically polarised
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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.


In case the electric field are at an angle of $90^\circ$ from one another, it is a special case of the circularly polarized light.

Multiple choice polarisation of light polarisation wave optics optics physics

Unpolarized light is incident on a plane glass surface. The angle of incidence so that reflected and refracted rays are perpendicular to each other, then:

  1. $tan \, i _\beta \, = \, \dfrac{\mu}{2}$
  2. $tan \, i _\beta \, = \, \mu$
  3. $sin \, i _\beta \, = \, \mu$
  4. $cos \, i _\beta \, = \, \mu$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Brewster's law,

    tan i= $\mu$
i is angle of incidence at which we get perfectly polarised reflected light.

Multiple choice polarisation of light polarisation wave optics optics physics

The velocity of light in air is $3 \, \times \, 10^8 \, m \, s^{-1}$ and that in water is $2.2 \, \times \, 10^8 \, m \, s^{-1}$. The polarising angle of incidence is:

  1. $45^{\circ}$
  2. $50^{\circ}$
  3. $53.74^{\circ}$
  4. $63^{\circ}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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


From Brewster's Law:-
$tan\ i _p= \mu=1.36$
$\therefore i _p= tan^{-1}(1.36)= 53.74^o$

So correct option is $C$

Multiple choice polarisation of light polarisation wave optics optics physics

At what angle of incidence will the light reflected from glass $( \mu \, = \, 1.5)$ be completely polarised

  1. $72.8^{\circ}$
  2. $51.6^{\circ}$
  3. $40.3^{\circ}$
  4. $56.3^{\circ}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Brewster's angle(Angle between Reflected and Refrected ray from the glass is $90^o$ ) is given as
$tan \theta = \dfrac{n _2}{n _1}$
$n _2$=1.5
$n _1=1$ for air 
$\theta=56.30$

Multiple choice polarisation of light polarisation wave optics optics physics

The critical angle of a certain medium is sin−1(35)sin−1(35) The polarizing angle of the medium is:

  1. $\sin^{-1} \, \left(\dfrac{4}{5}\right)$
  2. $\tan^{-1} \, \left(\dfrac{5}{3}\right)$
  3. $\sin^{ -1} \, \left(\dfrac{3}{4}\right)$
  4. $\tan^{ -1} \, \left(\dfrac{4}{3}\right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

then the two related through expression.
$\tan i = \dfrac{1}{\sin c}$     ...(i)
$\Rightarrow i = \tan^{-1} \left(\dfrac{1}{\sin c}\right)$     ...(ii)
Here, $c = \sin ^{-1}\left(\dfrac{3}{5}\right)$
Now, putting values in the eqn (ii) we get,
$i = \tan^{-1} \left(\dfrac{1}{\sin(\sin^{-1}(\frac{3}{5}))}\right)$   
$i = \tan^{-1} \left(\dfrac{5}{3}\right)$
Correct option is B

Multiple choice polarisation of light polarisation wave optics optics physics

In the case of linearly polarized light, the magnitude of the electric field vector

  1. is parallel to the direction of propagation

  2. does not change with time

  3. increases linearly with time

  4. varies periodically with time
Reveal answer Fill a bubble to check yourself
D Correct answer
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.
Multiple choice polarisation of light polarisation wave optics optics physics

Light from sodium lamp is made to pass through two polaroids placed one after the other in the path of light. Taking the intensity of the incident light as 100%, the intensity of the out coming light that can be varied in the range: 

  1. 0% to l00%

  2. 0% to 50%

  3. 0% to 25%

  4. 0% to 75%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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%.

Multiple choice polarisation of light polarisation wave optics optics physics

If the critical angle be $ \theta$ , then the Brewster's angle is

  1. $\sin^{-1}[\cot \theta]$
  2. $90-\theta$
  3. $\tan^{-1}[cosec \theta]$
  4. $\sin^{-1}[\tan \theta]$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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)).