Test: Types Of Polarization


12 Questions MCQ Test Electromagnetic Theory | Test: Types Of Polarization


Description
This mock test of Test: Types Of Polarization for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 12 Multiple Choice Questions for Electrical Engineering (EE) Test: Types Of Polarization (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Types Of Polarization quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Test: Types Of Polarization exercise for a better result in the exam. You can find other Test: Types Of Polarization extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
QUESTION: 1

When the phase angle between the Ex and Ey component is 00 or 1800, the polarisation is 

Solution:

Answer: c
Explanation: The phase angle between the Ex and Ey component is 00 and 1800 for linearly polarised wave. The wave is assumed to be propagating in the z direction.

QUESTION: 2

The magnitude of the Ex and Ey components are same in which type of polarisation?

Solution:

Answer: b
Explanation: In circular polarisation, the magnitude of the Ex and Ey components are the same. This is a form of the elliptical polarisation in which the major and minor axis are the same.

QUESTION: 3

When the Ex and Ey components of a wave are not same, the polarisation will be

Solution:

Answer: b
Explanation: In elliptical polarisation, the magnitude of Ex and Ey components are not same. This is due to the variation in the major and minor axes of the waves representing its magnitude.

QUESTION: 4

Identify the polarisation of the wave given, Ex = Exo cos wt and Ey = Eyo sin wt. The phase difference is +900.

Solution:

Answer: c
Explanation: The magnitude of the Ex and Ey components are not same. Thus it is elliptical polarisation. For +90 phase difference, the polarisation is left handed. In other words, the rotation is in clockwise direction. Thus the polarisation is left hand elliptical.

QUESTION: 5

Identify the polarisation of the wave given, Ex = 2 cos wt and Ey = sin wt. The phase difference is -900

Solution:

Answer: d
Explanation: The magnitude of the Ex and Ey components are not same. Thus it is elliptical polarisation. For -90 phase difference, the polarisation is right handed. In other words, the rotation is in anti-clockwise direction. Thus the polarisation is right hand elliptical.

QUESTION: 6

Identify the polarisation of the wave given, Ex = 2 cos wt and Ey = 2 sin wt. The phase difference is +900

Solution:

Answer: a
Explanation: The magnitude of the Ex and Ey components are the same. Thus it is circular polarisation. For +90 phase difference, the polarisation is left handed. In other words, the rotation is in clockwise direction. Thus the polarisation is left hand circular.

QUESTION: 7

Identify the polarisation of the wave given, Ex = cos wt and Ey = sin wt. The phase difference is -900

Solution:

Answer: b
Explanation: The magnitude of the Ex and Ey components are the same. Thus it is circular polarisation. For -90 phase difference, the polarisation is right handed. In other words, the rotation is in anti-clockwise direction. Thus the polarisation is right hand circular.

QUESTION: 8

For a non-zero Ex component and zero Ey component, the polarisation is

Solution:

Answer: a
Explanation: When the Ex is non-zero and the Ey is zero, the polarisation is parallel. The parallel polarisation is classified under the linear polarisation type.

QUESTION: 9

Identify the polarisation of the wave given that, Ex = 2 cos wt and Ey = cos wt.

Solution:

Answer: d
Explanation: The magnitude of the Ex and Ey components are not the same. Thus it cannot be circular polarisation. For a phase difference of 0, the polarisation is linear. In other words, the waves are in phase. Thus the polarisation is linear.

QUESTION: 10

The Snell law is applicable for perpendicular polarisation and the Brewster law is applicable for parallel polarisation. State True/False.

Solution:

Answer: a
Explanation: The Snell law is calculated from the oblique incidence media. Thus it is applicable for perpendicular polarisation. The Brewster law is applicable for perpendicular polarisation.

QUESTION: 11

When the polarisation of the receiving antenna is unknown, to ensure that it receives atleast half the power, the transmitted wave should be

Solution:

Answer: c
Explanation: The polarisation of the transmitting and receiving antenna has to be the same. This is the condition for maximum power transfer to occur. This is possible only when the polarisation is circular.

QUESTION: 12

Identify the polarisation of the wave given that, Ex = 2 sin wt and Ey = 3 sin wt.

Solution:

Answer: a
Explanation: The magnitude of the Ex and Ey components are not the same. Thus it cannot be circular polarisation. For a phase difference of 0, the polarisation is linear. In other words, the waves are in phase. Thus the polarisation is linear.

Similar Content

Related tests