Tag: polarisation of light

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Multiple choice polarisation of light polarisation wave optics optics physics

A ray of light is incident on the surface of  a glass plate at an angle of incidence equal to Brewster's angle $\phi$. If $\mu$ represents the  refractive index of glass with respect to air, then the angle between the reflected and the refracted rays is

  1. 90$^o$ + $\phi$
  2. $sin^{-1}$( $ \mu cos \phi$)
  3. 90$^o$
  4. 90 $-sin^{-1}$ $\left (\displaystyle \frac{sin\phi }{\mu}\right )$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When angle of incidence is equal to the Brewster's angle then, the angle between reflected and refracted rays is $ { 90 }^{ o }$

Multiple choice polarisation of light polarisation wave optics optics physics

Human eye:

  1. can detect polarized light

  2. cannot detect polarization of light

  3. can detect only circularly polarized light

  4. can detect only linearly polarized light
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Polarization changes when plane of vibration of polarized light changes.
Human eye is insensitive to change in polarization and hence, cannot detect polarization of light.

Multiple choice polarisation of light polarisation wave optics optics physics

Polarisation of light was first successfully explained by:

  1. Corpuscular theory

  2. Huygens' wave theory

  3. Electromagnetic wave theory

  4. Planck's theory
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Electromagnetic wave theory explains light as being composed of electric field vibrating in planes.
Polarisation of light refers to vibrating electric field in a particular plane.

Multiple choice polarisation of light polarisation wave optics optics physics

Plane of polarisation is:

  1. the plane in which vibrations of the electric vector takes place

  2. a plane perpendicular to the plane in which vibrations of the electric vector takes place

  3. perpendicular to the plane of vibration

  4. horizontal plane
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

$ \bf{Plane\ of\ Vibrations} $ is the plane in which the vibrations of electric vector of plane polarized light are present. 


$ \bf{Plane\ of\ Polarization} $ is the plane perpendicular to the plane of vibration and in this plane, the vibrations of the electric vector are absent.

Hence, the correct answers are OPTIONS B and C. 

Multiple choice polarisation of light polarisation wave optics optics physics

When light is incident on a glass block at polarizing angle
a) reflected ray is plane polarized
b) reflected and refracted rays are perpendicular
c) reflected and refracted rays are partially polarized
d) refracted ray is partially polarised

  1. a, c and d are correct

  2. a, b and d are correct

  3. b, c and d are correct

  4. a, b and c are correct
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

when a beam of unpolarized light is incident on the glass block, certain amount of the incident vibration is reflected in a plane perpendicular to the plane of incidence and transmits the rest.
Thus reflected ray is completely polarized but not partially polarized.
Thus only (c) is wrong and rest options are correct.

Multiple choice polarisation of light polarisation wave optics optics physics

The polarising angle for glass is :

  1. same for different kinds of glass

  2. different for different kinds of glass

  3. same for lights of all colours

  4. varies with time
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Different kinds of glass have different refractive index. Thus as polarising angle depends on refractive index the polarising angle is different for different kinds of glass. $tan \ i = \mu $

Multiple choice polarisation of light polarisation wave optics optics physics

Bartholinus discovered :

  1. Interference by splitting the wave front

  2. Polarisation by reflection

  3. Polarisation by refraction

  4. Polarisation by double refraction
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In 1669, another Danish scientist, Erasmus Bartholinus discovered the polarization of light by double refraction in Iceland spar (calcite).

Multiple choice polarisation of light polarisation wave optics optics physics

Choose the correct statements among the following given options.

  1. Brewster's angle is independent of wavelength of light.

  2. Brewster's angle is independent of the nature of reflecting surface.

  3. Brewster's angle is different for different wavelengths.

  4. Brewsters angle depends on wavelength but not on the nature of reflecting surface.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Brewster's angle is an angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface,
with no reflection.
$\theta _{B}=tan^{-1}(\dfrac{n _{2}}{n _{1}})$
$n _{2}$ refractive index of final medium
$n _{1}$ refractive index of initial medium
As different wavelength have different refractive indices the brewster angle is also different.

Multiple choice polarisation of light polarisation wave optics optics physics

Pile of plates can be used to produce completely polarised light due to :

  1. Reflection

  2. Refraction

  3. Double refraction

  4. A and B
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

pile of plates is used to produce plane polarized light by refraction.
By refraction through multiple plates vibrations in the plane of incidence are transmitted.
Thus we obtain a polarized light.

Multiple choice polarisation of light polarisation wave optics optics physics

At the polarising angle $(\theta _{B})$, angle of refraction is given by : 

  1. $90^{\circ}$
  2. $90^o+\theta _{B}$
  3. $90^o-\theta _{B}$
  4. $\dfrac{90^o}{\theta _B }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

At polarizing angle $\theta _{B}$
$\dfrac{sin \theta _{B}}{cos \theta _{B}}=\dfrac{sin \theta _{B}}{sin \gamma }$
we get $sin \gamma =cos \theta _{B}$
$sin \gamma =sin(90-\theta _{B})$
Thus $\gamma =90-\theta _{B}$