Polarisation of Light - JEE MCQ

 Suppose the angle between the transmission axes of the analyser and the polarizer is θ. The completely plane polarized light form the polarizer is incident on the analyser. If E₀ is the amplitude of the electric vector transmitted by the polarizer, then intensity I₀ of the light incident on the analyser is
I ∞ E₀²
 
The electric field vector E₀ can be resolved into two rectangular components i.e E₀ cosθ and E₀ sinθ. The analyzer will transmit only the component ( i.e E₀ cosθ ) which is parallel to its transmission axis. However, the component E₀sinθ will be absorbed by the analyser. Therefore, the intensity I of light transmitted by the analyzer is,
 
I ∞ ( E₀ x cosθ )²
 
I / I₀ = ( E₀ x cosθ )² / E₀² = cos2θ
 
I = I₀ x cos²θ
when light passes from polarizer it's intensity becomes half and when passed through analyser it becomes,
I = I₀ x cos²θ/2
 

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. When unpolarized light is incident at this angle, the light that is reflected from the surface is therefore perfectly polarized.
So the angle of incidence would be 60°

An analyser or analyzer is a person or device that analyses given data. It examines in detail the structure of the given data and tries to find patterns and relationships between parts of the data. An analyser can be a piece of hardware or a computer program running on a computer.

The polarization angle (also called Brewster's angle)is defined as the angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection.
Brewster's Law states that the tangent of the angle of polarization is numerically equal to the refractive index of the medium.
μ = tan i ...................... (1)
From Snell's law:
μ = sinr sini​ ....................(2)
Comparing (1) and (2)
cos i = sin r = cos (2π​ - r)
i.e. i = (2π​ - r)
So, i + r = 2π​
So, as seen from the diagram, if i + r = 90^o Then the angle between reflected ray and refracted ray will be 180^o - 90^o = 90^o

According to Malus’s law Iαcos²θ, when θ=0^o or 180^o, I=I_ocos²0^o = I_o
i.e the intensity of the light transmitted by the analyser is maximum when the transmission axes of analyser and polarizer are parallel.

When the light is incident at Brewster’s angle, the angle between the reflected ray and refracted ray is 90o, irrespective of the value of the refractive index.

Given from figure angle of incidence i=600
angle of refraction r=30⁰
We know μ= sin i​/sin r
μ= sin60/sin30 ​=((√3)/2) ​​×2=√3

Initial unpolarized intensity is 2a² 
the intensity of light transmitted by the first polarizered will be I_unpolarized/2​​=a²
option A is correct

In calcite crystal there is a phenomenon of double refraction. So, we see one dot as a stationary object whereas the other being refracted at a little displaced position seems to be rotating about the first dot.