Tag: wave optics

Questions Related to wave optics

Plane of polarisation is:

polarisation of light polarisation wave optics optics physics

  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

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Correct Option: B,C
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. 

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

polarisation of light polarisation wave optics optics physics

  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

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Correct Option: B
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.

The polarising angle for glass is :

polarisation of light polarisation wave optics optics physics

  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

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Correct Option: B
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 $

Bartholinus discovered :

polarisation of light polarisation wave optics optics physics

  1. Interference by splitting the wave front

  2. Polarisation by reflection

  3. Polarisation by refraction

  4. Polarisation by double refraction

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Correct Option: D
Explanation:

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

Choose the correct statements among the following given options.

polarisation of light polarisation wave optics optics physics

  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.

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Correct Option: C
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.

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

polarisation of light polarisation wave optics optics physics

  1. Reflection

  2. Refraction

  3. Double refraction

  4. A and B

Show answer
Correct Option: B
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.

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

polarisation of light polarisation wave optics optics physics

  1. $90^{\circ}$

  2. $90^o+\theta _{B}$

  3. $90^o-\theta _{B}$

  4. $\dfrac{90^o}{\theta _B }$

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Correct Option: C
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 $\gamma =90-\theta _{B}$

The angle of incidence at which reflected light is totally polarised for reflection from air to glass (refractive index n) is :

polarisation of light polarisation wave optics optics physics

  1. $sin^{-1}(n)$

  2. $sin^{-1}(1/n)$

  3. $ tan^{-1}(n)$

  4. $ tan^{-1}(1/n)$

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Correct Option: C
Explanation:

The refractive index of glass is n
If the light is incident on the surface with an angle of incidence i given by tan i $=\mu $ using the Brewster's law the reflected light is completely polarized.
Here tan i $=n$
i$=tan^{-1}(n)$

A light ray is incident on a transparent medium of $\mu =$ 1.732 at the polarizing angle. The angle of refraction is :

polarisation of light polarisation wave optics optics physics

  1. 60$^{\circ}$

  2. 30$^{\circ}$

  3. 45$^{\circ}$

  4. 90$^{\circ}$

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Correct Option: B
Explanation:

Angle of incidence $=$ polarising angle
$tan i = \mu $
$tan i = 1.732$
$i = 60^{0}$
From snell's law
$\dfrac{sin i}{sin \gamma }=\mu $
$sin \gamma =\dfrac{sin 60^{0}}{1.732}$
$\gamma =30^{0}$

The critical angle for total internal reflection for a substance is $45^{\circ}$. The polarizing angle for this substance is ($\tan 54^{\circ}44'=\sqrt{2}$) :

polarisation of light polarisation wave optics optics physics

  1. $46^{\circ}16'$

  2. $54^{\circ}44'$

  3. $46^{\circ}44'$

  4. $54^{\circ}16'$

Show answer
Correct Option: B
Explanation:

If polarizing angle is $\theta $
Then tan $\theta =\dfrac{1}{sin c}$
$tan \theta =\dfrac{1}{sin 45}=\sqrt{2}$
$\theta =54^{0}{44}'$