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Polarization NAT - Physics MCQ


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10 Questions MCQ Test Topic wise Tests for IIT JAM Physics - Polarization NAT

Polarization NAT for Physics 2024 is part of Topic wise Tests for IIT JAM Physics preparation. The Polarization NAT questions and answers have been prepared according to the Physics exam syllabus.The Polarization NAT MCQs are made for Physics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Polarization NAT below.
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*Answer can only contain numeric values
Polarization NAT - Question 1

Unpolarized light of intensity. I0,  is incident on a series of 3 polarizing filters, the axis of the second filter is oriented at 45° to that of the first filter, while the axis of the third filter is oriented at 90° to that of the first filter. What fraction of the incident light that is transmitted through the third filter?


Detailed Solution for Polarization NAT - Question 1

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

*Answer can only contain numeric values
Polarization NAT - Question 2

The critical angle of light in a certain substance is 45°. What is the polarising angle?(upto one decimal place)


Detailed Solution for Polarization NAT - Question 2

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

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*Answer can only contain numeric values
Polarization NAT - Question 3

Unpolarized light is incident on two ideal polarizers in series, the polarizers are oriented so that no light emerges through the second polarizer. A third polarizer is now inserted between the first two and its orientation direction is continuously rotated through 180°.
What is the maximum fraction of the incident power transmitted through all three polarizers?


Detailed Solution for Polarization NAT - Question 3

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

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 I= 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

*Answer can only contain numeric values
Polarization NAT - Question 4

The critical angle of a medium is 45° Find the polarizing angle.


Detailed Solution for Polarization NAT - Question 4


= 54.75°
The correct answer is: 54.75o

*Answer can only contain numeric values
Polarization NAT - Question 5

A beam of light polarized in the x direction is travelling in the z direction. It passess through a polarizer, travels some distance, and reaches an analyser you can orient the pass axis of the analyzer only in the x or the y direction. You find that the intensity is three times smaller when the pass axis of the analyser is in the y direction than it was when the pass axis was in the x direction Find the angle between the pass axis of the polarizer and the x-axis (in degrees).


Detailed Solution for Polarization NAT - Question 5

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

*Answer can only contain numeric values
Polarization NAT - Question 6

Unpolarized light is incident on a pair of ideal linear polarizers whose transmission axis make an angle of 45° with each other, the transmitted light intensity through both polarizers in what percentage of the incident intensity?


Detailed Solution for Polarization NAT - Question 6

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

Percentage = 
= 25%
The correct answer is: 25

*Answer can only contain numeric values
Polarization NAT - Question 7

A beam of light polarized in the x direction is travelling in the z direction. It passess through a polarizer, travels some distance, and reaches an analyser you can orient the pass axis of the analyzer only in the x or the y direction. You find that the intensity is three times smaller when the pass axis of the analyser is in the y direction than it was when the pass axis was in the x direction Suppose that the original light beam were polarized in the Y direction, everything else remaining the same. What value of  would you observe?


Detailed Solution for Polarization NAT - Question 7

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

*Answer can only contain numeric values
Polarization NAT - Question 8

The refractive indices of glass and water are 1.54 and 1.33. Find the polarising angle for a beam incident from water to glass.


Detailed Solution for Polarization NAT - Question 8

*Answer can only contain numeric values
Polarization NAT - Question 9

If the polarizing angle of a medium is 60°. Then find the angle of refraction, (in degrees)


Detailed Solution for Polarization NAT - Question 9

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

*Answer can only contain numeric values
Polarization NAT - Question 10

If the refractive index of a medium is 1.732, then find the polarizing angle (in degrees) of the medium.


Detailed Solution for Polarization NAT - Question 10

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°.

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