Question 1 |
The electric field of a plane electromagnetic wave is
E=a_{x} C_{1 x} \cos (\omega t-\beta z)+a_{y} C_{1 y} \cos (\omega t-\beta z+\theta) \mathrm{V} / \mathrm{m}
Which of the following combination(s) will give rise to a left handed elliptically polarized (LHEP) wave?
E=a_{x} C_{1 x} \cos (\omega t-\beta z)+a_{y} C_{1 y} \cos (\omega t-\beta z+\theta) \mathrm{V} / \mathrm{m}
Which of the following combination(s) will give rise to a left handed elliptically polarized (LHEP) wave?
C_{1 x}=1, C_{1 y}=1, \theta=\pi / 4 | |
C_{1 x}=2, C_{1 y}=1, \theta=\pi / 2 | |
C_{1 x}=1, C_{1 y}=2, \theta=3 \pi / 2 | |
C_{1 x}=2, C_{1 y}=1, \theta=3 \pi / 4 |
Question 1 Explanation:
Given, \quad \vec{E}=\hat{a}_{x} C_{1 x} \cos (\omega t-\beta z)+\hat{a}_{y} C_{1 y} \cos (\omega t-\beta z+\theta)
at z=0
\vec{E}=C_{1 x} \cos \omega t \hat{a}_{x}+C_{1 y} \cos (\omega t+\theta) \hat{a}_{y}
Going by options,
Option (A) \quad \vec{E}=\cos \omega t \hat{a}_{x}+\cos (\omega t+\pi / 4) \hat{a}_{y}
at t=0, \omega t=0, \vec{E}=\hat{a}_{x}+\frac{1}{\sqrt{2}} \hat{a}_{y}
at t=T / 4, \omega t=\pi / 2, \vec{E}=0-\frac{1}{\sqrt{2}} \hat{a}_{y}
\Rightarrow Hence, it is LHEP.
Option (B) \quad \vec{E}=2 \cos \omega t \hat{a}_{x}+\cos (\omega t+\pi / 2) \hat{a}_{y}
at t=0, \omega t=0, \vec{E}=2 \hat{a}_{x}
at t=T / 4, \omega t=\pi / 2, \vec{E}=-1 \hat{a}_{y}
\Rightarrow Hence, it is LHEP.
Option (C) \quad \vec{E}=\cos \omega t \hat{a}_{x}+2 \cos (\omega t+3 \pi / 2) \hat{a}_{y}
at t=0, \quad \quad \omega t=0, \vec{E}=\hat{a}_{x}
at t=T / 4, \omega t=\pi / 2, \vec{E}=2 \hat{a}_{y}
\Rightarrow Hence, it is RHEP.
Option (D) \quad \vec{E}=2 \cos \omega t \hat{a}_{x}+\cos (\omega t+3 \pi / 4) \hat{a}_{y}
at t=0, \omega t=0, \vec{E}=2 \hat{a}_{x}-\frac{1}{\sqrt{2}} \hat{a}_{y}
at t=T / 4, \omega t=\pi / 2, \vec{E}=0-\frac{1}{\sqrt{2}} \hat{a}_{y}=\frac{-1}{\sqrt{2}} \hat{a}_{y}
\Rightarrow Hence, it is LHEP.
\therefore Option (A), (B) and (D) are correct.
at z=0
\vec{E}=C_{1 x} \cos \omega t \hat{a}_{x}+C_{1 y} \cos (\omega t+\theta) \hat{a}_{y}
Going by options,
Option (A) \quad \vec{E}=\cos \omega t \hat{a}_{x}+\cos (\omega t+\pi / 4) \hat{a}_{y}
at t=0, \omega t=0, \vec{E}=\hat{a}_{x}+\frac{1}{\sqrt{2}} \hat{a}_{y}
at t=T / 4, \omega t=\pi / 2, \vec{E}=0-\frac{1}{\sqrt{2}} \hat{a}_{y}
\Rightarrow Hence, it is LHEP.
Option (B) \quad \vec{E}=2 \cos \omega t \hat{a}_{x}+\cos (\omega t+\pi / 2) \hat{a}_{y}
at t=0, \omega t=0, \vec{E}=2 \hat{a}_{x}
at t=T / 4, \omega t=\pi / 2, \vec{E}=-1 \hat{a}_{y}
\Rightarrow Hence, it is LHEP.
Option (C) \quad \vec{E}=\cos \omega t \hat{a}_{x}+2 \cos (\omega t+3 \pi / 2) \hat{a}_{y}
at t=0, \quad \quad \omega t=0, \vec{E}=\hat{a}_{x}
at t=T / 4, \omega t=\pi / 2, \vec{E}=2 \hat{a}_{y}
\Rightarrow Hence, it is RHEP.
Option (D) \quad \vec{E}=2 \cos \omega t \hat{a}_{x}+\cos (\omega t+3 \pi / 4) \hat{a}_{y}
at t=0, \omega t=0, \vec{E}=2 \hat{a}_{x}-\frac{1}{\sqrt{2}} \hat{a}_{y}
at t=T / 4, \omega t=\pi / 2, \vec{E}=0-\frac{1}{\sqrt{2}} \hat{a}_{y}=\frac{-1}{\sqrt{2}} \hat{a}_{y}
\Rightarrow Hence, it is LHEP.
\therefore Option (A), (B) and (D) are correct.
Question 2 |
Consider the following wave equation,
\frac{\partial^2 f(x,t)}{\partial t^2}=10000\frac{\partial^2 f(x,t)}{\partial x^2}
Which of the given options is/are solution(s) to the given wave equation?
\frac{\partial^2 f(x,t)}{\partial t^2}=10000\frac{\partial^2 f(x,t)}{\partial x^2}
Which of the given options is/are solution(s) to the given wave equation?
f(x,t)=e^{-(x-100t)^2}+e^{-(x+100t)^2} | |
f(x,t)=e^{-(x-100t)}+0.5e^{-(x+1000t)} | |
f(x,t)=e^{-(x-100t)}+\sin (x+100t) | |
f(x,t)=e^{j100 \pi(-100x+t)}+e^{j100 \pi(100x+t)} |
Question 2 Explanation:
As we know, wave equation is given by
\frac{\partial^2 f(x,t)}{\partial x^2}=\frac{C^2d^2f(x,y)}{dt^2}
Here, option (A) and (C) are satisfying the above standard wave equation.
\frac{\partial^2 f(x,t)}{\partial x^2}=\frac{C^2d^2f(x,y)}{dt^2}
Here, option (A) and (C) are satisfying the above standard wave equation.
Question 3 |
The magnetic field of a uniform plane wave in vacuum is given by
\vec{H}(x,y,z,t)=(\hat{a}_x+2\hat{a}_y+b\hat{a}_z) \cos (\omega t+3x-y-z)
The value of b is _____
\vec{H}(x,y,z,t)=(\hat{a}_x+2\hat{a}_y+b\hat{a}_z) \cos (\omega t+3x-y-z)
The value of b is _____
0 | |
1 | |
-1 | |
-2 |
Question 3 Explanation:
For uniform plane wave
\hat{a}_{H}\cdot \hat{a}_{\rho }=0
\hat{a}_{H}is unit vector in magnetic field direction \hat{a}_{\rho } is unit vector in power flow direction
\hat{a}_{H }=\frac{1\hat{a}_{x}+2\hat{a}_{y}+b\hat{a}_{z}}{\sqrt{1^{2}+2^{2}+b^{2}}}
\hat{a}_{\rho }=\frac{-3\hat{a}_{x}+\hat{a}_{y}+\hat{a}_{z}}{\sqrt{3^{2}+1^{2}+1^{2}}}
\hat{a}_{H}\cdot \hat{a}_{\rho }=0 (\hat{a}_{x}+2\hat{a}_{y}+b\hat{a}_{z})\cdot (-3\hat{a}_{x}+\hat{a}_{y}+\hat{a}_{z})=0
-3+2+b=0
b=1
\hat{a}_{H}\cdot \hat{a}_{\rho }=0
\hat{a}_{H}is unit vector in magnetic field direction \hat{a}_{\rho } is unit vector in power flow direction
\hat{a}_{H }=\frac{1\hat{a}_{x}+2\hat{a}_{y}+b\hat{a}_{z}}{\sqrt{1^{2}+2^{2}+b^{2}}}
\hat{a}_{\rho }=\frac{-3\hat{a}_{x}+\hat{a}_{y}+\hat{a}_{z}}{\sqrt{3^{2}+1^{2}+1^{2}}}
\hat{a}_{H}\cdot \hat{a}_{\rho }=0 (\hat{a}_{x}+2\hat{a}_{y}+b\hat{a}_{z})\cdot (-3\hat{a}_{x}+\hat{a}_{y}+\hat{a}_{z})=0
-3+2+b=0
b=1
Question 4 |
A uniform plane wave traveling in free space and having the electric field
\vec{E}=(\sqrt{2}\hat{a}_{x}-\hat{a}_{z})cos[6\sqrt{3}\pi \times 10^{8}t-2\pi (x+\sqrt{2}z)]V/m
is incident on a dielectric medium (relative permittivity \gt1, relative permeability = 1) as shown in the figure and there is no reflected wave.
The relative permittivity (correct to two decimal places) of the dielectric medium is___________.
\vec{E}=(\sqrt{2}\hat{a}_{x}-\hat{a}_{z})cos[6\sqrt{3}\pi \times 10^{8}t-2\pi (x+\sqrt{2}z)]V/m
is incident on a dielectric medium (relative permittivity \gt1, relative permeability = 1) as shown in the figure and there is no reflected wave.
The relative permittivity (correct to two decimal places) of the dielectric medium is___________.
2 | |
2.9 | |
1 | |
3 |
Question 4 Explanation:
The wave has E_{i} direction and propagation direction both in same plane (ZX).
The wave is plane of incidence (P) polarized.
\begin{aligned} \beta_{x} &=1, \quad \beta_{2}=\sqrt{2} \\ \tan \theta_{i} &=\frac{\beta_{z}}{\beta_{x}}=\sqrt{2} \end{aligned}
No reflection in P polarized wave under Brewster's angle of incidence.
\begin{aligned} \theta_{i} &=\theta_{B} \\ \tan \theta_{B} &=\sqrt{2}=\sqrt{\epsilon_{r}} \\ \epsilon_{r} &=2 \end{aligned}
The wave is plane of incidence (P) polarized.
\begin{aligned} \beta_{x} &=1, \quad \beta_{2}=\sqrt{2} \\ \tan \theta_{i} &=\frac{\beta_{z}}{\beta_{x}}=\sqrt{2} \end{aligned}
No reflection in P polarized wave under Brewster's angle of incidence.
\begin{aligned} \theta_{i} &=\theta_{B} \\ \tan \theta_{B} &=\sqrt{2}=\sqrt{\epsilon_{r}} \\ \epsilon_{r} &=2 \end{aligned}
Question 5 |
The distance (in meters) a wave has to propagate in a medium having a skin depth of 0.1 m
so that the amplitude of the wave attenuates by 20 dB, is
0.12 | |
0.23 | |
0.46 | |
2.3 |
Question 5 Explanation:
Attenuation constant,
\begin{aligned} \alpha &=\frac{1}{\text { skin depth }}=10 \mathrm{Np} / \mathrm{m} \\ \text { 20log }_{10}\left(\frac{E_{0}}{E_{x}}\right) &=20 \mathrm{dB} \\ \frac{E_{o}}{E_{x}} &=10 \Rightarrow E_{x}=\frac{E_{0}}{10} \\ E_{x} &=E_{0} e^{-\alpha x}=E_{0} e^{-10 x}=\frac{E_{0}}{10} \\ e^{-10 x} &=\frac{1}{10} \\ x &=\frac{1}{10} \ln (10)=0.23 \mathrm{m} \end{aligned}
\begin{aligned} \alpha &=\frac{1}{\text { skin depth }}=10 \mathrm{Np} / \mathrm{m} \\ \text { 20log }_{10}\left(\frac{E_{0}}{E_{x}}\right) &=20 \mathrm{dB} \\ \frac{E_{o}}{E_{x}} &=10 \Rightarrow E_{x}=\frac{E_{0}}{10} \\ E_{x} &=E_{0} e^{-\alpha x}=E_{0} e^{-10 x}=\frac{E_{0}}{10} \\ e^{-10 x} &=\frac{1}{10} \\ x &=\frac{1}{10} \ln (10)=0.23 \mathrm{m} \end{aligned}
There are 5 questions to complete.