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In S polarisation, the electric field lies in the plane perpendicular to that of the interface. State True/False
a Explanation: In the EM wave propagation, the electric and magnetic fields are perpendicular to each other. The S polarised wave is similar to the transverse magnetic (TM) wave, the electric field lies in the plane perpendicular to that of the interface.
a
See lessExplanation: In the EM wave propagation, the electric and magnetic fields are
perpendicular to each other. The S polarised wave is similar to the transverse magnetic (TM) wave, the electric field lies in the plane perpendicular to that of the interface.
The resultant electric field of a wave with Ex = 3 and Ey = 4 will be
d Explanation: The resultant electric field of two electric components Ex and Ey is E = √(Ex2 + Ey2). On substituting for Ex = 3 and Ey = 4, we get E = 5 units.
d
See lessExplanation: The resultant electric field of two electric components Ex and Ey is E =
√(Ex2 + Ey2). On substituting for Ex = 3 and Ey = 4, we get E = 5 units.
Identify the polarisation of the wave given that, Ex = 2 sin wt and Ey = 3 sin wt.
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.
a
See lessExplanation: 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.
When the polarisation of the receiving antenna is unknown, to ensure that it receives atleast half the power, the transmitted wave should be
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.
c
See lessExplanation: 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.
The Snell law is applicable for perpendicular polarisation and the Brewster law is applicable for parallel polarisation. State True/False.
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.
a
See lessExplanation: 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.
Identify the polarisation of the wave given that, Ex = 2 cos wt and Ey = cos wt.
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.
d
See lessExplanation: 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.
For a non-zero Ex component and zero Ey component, the polarisation is
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.
a
See lessExplanation: 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.
Identify the polarisation of the wave given, Ex = cos wt and Ey = sin wt. The phase difference is -900 .
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 otherwords, the rotation is in anti-clockwise direction. Thus the polarisation is right hand circular.
b
See lessExplanation: 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 otherwords, the rotation is in anti-clockwise direction. Thus the polarisation is right hand circular.
Identify the polarisation of the wave given, Ex = 2 cos wt and Ey = 2 sin wt. The phase difference is +900 .
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.
a
See lessExplanation: 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.
Identify the polarisation of the wave given, Ex = 2 cos wt and Ey = sin wt. The phase difference is -900 .
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 otherwords, the rotation is in anti-clockwise direction. Thus the polarisation is right hand elliptical.
d
See lessExplanation: 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 otherwords, the rotation is in anti-clockwise direction. Thus the polarisation is right hand elliptical.