Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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 4 |
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 4 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}
Question 5 |
The permittivity of water at optical frequencies is 1.75 \varepsilon_{0}. It is found that an isotropic light source at a distance d under water forms an illuminated circular area of radius 5m, as shown in the figure. The critical angle is \theta_{c}.
The value of d (in meter) is _____________
The value of d (in meter) is _____________
1.3 | |
2.8 | |
4.3 | |
5.8 |
Question 5 Explanation:
Critical angle,
\begin{aligned} \theta_{c} &=\sin ^{-1} \sqrt{\frac{\epsilon_{2}}{\epsilon_{1}}} \\ &=\sin ^{-1} \sqrt{\frac{\epsilon_{0}}{1.75 \epsilon_{0}}}=49.1^{\circ} \end{aligned}
\begin{aligned} \tan 49.1 &=1.547=\frac{5}{d} \\ \Rightarrow \quad d &=\frac{5}{1.1547}=4.33 \mathrm{m} \end{aligned}
\begin{aligned} \theta_{c} &=\sin ^{-1} \sqrt{\frac{\epsilon_{2}}{\epsilon_{1}}} \\ &=\sin ^{-1} \sqrt{\frac{\epsilon_{0}}{1.75 \epsilon_{0}}}=49.1^{\circ} \end{aligned}
\begin{aligned} \tan 49.1 &=1.547=\frac{5}{d} \\ \Rightarrow \quad d &=\frac{5}{1.1547}=4.33 \mathrm{m} \end{aligned}
Question 6 |
The expression for an electric field in free space is E=E_{0}=(\hat{x}+\hat{y}+j2\hat{z})e^{-j(\omega t-kx+ky)}, where x, y,
z represent the spatial coordinates, t represents time, and \omega, k are constants. This electric field
does not represent a plane wave | |
represents a circular polarized plane wave propagating normal to the z-axis | |
represents an elliptically polarized plane wave propagating along x-y plane. | |
represents a linearly polarized plane wave |
Question 6 Explanation:
\vec{E} field direction \Rightarrow\left(\hat{a}_{x}+\hat{a}_{y}+j 2 \hat{a}_{z}\right)
Propagation direction \Rightarrow k \hat{a}_{x}-k \hat{a}_{y}
\vec{E} is perpendicular to propagation
\vec{E} \cdot \vec{P}=0
Component in \hat{a}_{z} has magnitude of 2.
Component in X-Y plane has magnitude of \sqrt{2}.
These two components are out of phase by 90^{\circ} and have unequal amplitudes. So, it is elliptically polarized wave.
Propagation direction \Rightarrow k \hat{a}_{x}-k \hat{a}_{y}
\vec{E} is perpendicular to propagation
\vec{E} \cdot \vec{P}=0
Component in \hat{a}_{z} has magnitude of 2.
Component in X-Y plane has magnitude of \sqrt{2}.
These two components are out of phase by 90^{\circ} and have unequal amplitudes. So, it is elliptically polarized wave.
Question 7 |
Faraday's law of electromagnetic induction is mathematically described by which one of the following equations?
\bigtriangledown \cdot \vec{B}=0 | |
\bigtriangledown \cdot \vec{D}=\rho _{v} | |
\bigtriangledown \times \vec{E}=-\frac{\partial \vec{B}}{\partial t} | |
\bigtriangledown \times \vec{H}=\sigma \vec{E}+\frac{\partial \vec{D}}{\partial t} |
Question 7 Explanation:
Rate of change of magnetic field results in induced voltage,
\nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t}
\nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t}
Question 8 |
If a right-handed circularly polarized wave is incident normally on a plane perfect conductor, then the reflected wave will be
right-handed circularly polarized | |
left-handed circularly polarized | |
elliptically polarized with a tilt angle of 45^{\circ} | |
horizontally polarized |
Question 8 Explanation:
Left circularly polarized, due to direction change
after the reflection from conductor.
Question 9 |
Let the electric field vector of a plane electromagnetic wave propagating in a homogenous medium be expressed as E=\hat{x}E_{x}e^{-j(\omega t-\beta z)}, where the propagation constant \beta is a function of the angular frequency \omega. Assume that \beta (\omega)
and E_{x} are known and are real. From the information available, which one of the following CANNOT be determined?
The type of polarization of the wave. | |
The group velocity of the wave | |
The phase velocity of the wave. | |
The power flux through the z = 0 plane. |
Question 9 Explanation:
\beta(\omega) is known so V_{g} can be calculated.
v_{p}=\omega / \beta can be calculated.
Polarization can be identified.
\mu_{r} and \epsilon_{r} cannot be found, due to which power flux cannot be calculated as power flux
P=\frac{1}{2} \frac{|E|^{2}}{\eta}, \text{ where }\eta=120 \pi \times \sqrt{\frac{\mu_{r}}{\epsilon_{r}}}
v_{p}=\omega / \beta can be calculated.
Polarization can be identified.
\mu_{r} and \epsilon_{r} cannot be found, due to which power flux cannot be calculated as power flux
P=\frac{1}{2} \frac{|E|^{2}}{\eta}, \text{ where }\eta=120 \pi \times \sqrt{\frac{\mu_{r}}{\epsilon_{r}}}
Question 10 |
The electric field of a plane wave propagating in a lossless non-magnetic medium is given by the following expression
E(z,t)=a_{x}5 cos(2\pi \times 10^{9}t+\beta z)+ a_{y} 3 cos(2\pi \times 10^{9}t+\beta z-\pi/2)
The type of the polarization is
E(z,t)=a_{x}5 cos(2\pi \times 10^{9}t+\beta z)+ a_{y} 3 cos(2\pi \times 10^{9}t+\beta z-\pi/2)
The type of the polarization is
Right Hand Circular. | |
Left Hand Elliptical | |
Right Hand Elliptical | |
Linear. |
Question 10 Explanation:
\vec{E}(z, t)=\hat{a}_{x} 5 \cos \left(2 \pi \times 10^{9} t+\beta z\right)
+\hat{a}_{y} 3 \cos \left(2 \pi \times 10^{9} t+\beta z-\frac{\pi}{2}\right)
\Rightarrow Wave is travelling in -\hat{a}_{z} direction.
\Rightarrow Wave has orthogonal components with unequal amplitudes and checking the time trace.
\therefore Wave is left hand elliptically polarized.
+\hat{a}_{y} 3 \cos \left(2 \pi \times 10^{9} t+\beta z-\frac{\pi}{2}\right)
\Rightarrow Wave is travelling in -\hat{a}_{z} direction.
\Rightarrow Wave has orthogonal components with unequal amplitudes and checking the time trace.
\therefore Wave is left hand elliptically polarized.
There are 10 questions to complete.