Polarization NAT - IIT JAM MCQ

Light through the first filter is 
I₀= Incident intensity.
Through the second filter 

Through the third filter, intensity is 

The correct answer is: 0.125

If  'n' is refractive index and ic is critical velocity,

   n = 1/Sin ic = 1 / Sin 45 = 1.414 

Now, According to Brewster's law , if  ' ip ' is polarising angle  then 

n = tan ip 

So , ​tan ip  = 1.414 

  i.e. ​ ip =  54.7⁰

Let incident intensity be I
Intensity after first polarizer = 
Intensity after the second polarizer = I₁ cos² Ø

where Ø s the angle between the polarizer pass axis of the second polarizer and first polarizer
Intensity of light through the 3rd polarizer is 
Intensity of light through the 3rd polarizer is I₃ = I₂ cos² θ
where θ is the angle between pass axis polarizer 2 and 3. For I₃ to be maximum.

Put  because
the first and third polarizers are orthogonal

Put 

Fraction is 0.125
The correct answer is: 0.125

= 54.75°
The correct answer is: 54.75^o

If θ is the angle of polarizer’s pass axis with the x-axis and / is the intensity of the light beam between the polarizer and the analyzer.

The correct answer is: 30

Light incident = l₀
Intensity of light through first polarizer = 
Intensity of light through second polarizer =  

Percentage = 
= 25%
The correct answer is: 25

Using the polarized light would change the beam intensity / between the polarizer and analyser but it would not change the ratio 

The correct answer is: 3

Polarizing angle is the angle of incidence for which the refracted and reflected rays are orthogonal.
θ_i = 60°
⇒ θ_reflected = 60°
⇒ θ_reflected = 90°- 60° = 30°
The correct answer is: 30

Let, this happens when angle of incidence = i and angle of refraction = r. We know, angle of reflection = angle of incidence = i.

So, (i + r) + 90° = 180°

or, r = (90° - i).

So, refractive index = sin i / sin (90° - i) = sin i / cos i = tan i

1.732 = tan i

tan i = √3 = tan 60°

i = 60°.