Question

A plane wave whose electric field is given by \(\vec{E} = 100\cos(\omega t - 6\pi x)\hat{z}\) passes normally from a material 'A' having \(\epsilon_r=4, \mu_r=1\) and \(\sigma=0\) to a material 'B' having \(\epsilon_r=9, \mu_r=4\) and \(\sigma=0\). The values of some of its parameters are given below:
A. \(6\pi\),
B. \(80\pi\),
C. \(1/7\),
D. \(8/7\).
Choose the correct sequence: intrinsic impedance of medium B, Reflection coefficient, Transmission Coefficient and Phase shift constant of medium B, respectively, from the options given below:

• A. A, C, D, B
• B. A, C, B, D
• C. B, A, D, C
• D. C, B, D, A

Hint:

When a wave passes between two media, first calculate the intrinsic impedance of each medium (\(\eta = \eta_0\sqrt{\mu_r/\epsilon_r}\)). Use these to find the reflection (\(\Gamma = (\eta_2-\eta_1)/(\eta_2+\eta_1)\)) and transmission (\(\tau=1+\Gamma\)) coefficients.

Answer 1

The Correct Option is C

Step 1: Intrinsic Impedance of Medium B
The intrinsic impedance \( Z \) of a medium is given by:
\[ Z = \sqrt{\frac{\mu}{\epsilon}} \] Where \( \mu = \mu_0 \mu_r \) and \( \epsilon = \epsilon_0 \epsilon_r \).
Using the given values for material B, we have:
\[ Z_B = 377 \times \sqrt{\frac{4}{9}} = 377 \times \frac{2}{3} = 251.33 \, \Omega \] So, the intrinsic impedance of medium B is approximately \( 251.33 \, \Omega \), which corresponds to option B \( 80\pi \, \Omega \).

Step 2: Reflection Coefficient
The reflection coefficient \( R \) is given by:
\[ R = \frac{Z_2 - Z_1}{Z_2 + Z_1} \] Where \( Z_1 = Z_A = 188.5 \, \Omega \) and \( Z_2 = Z_B = 251.33 \, \Omega \).
\[ R = \frac{251.33 - 188.5}{251.33 + 188.5} = 0.143 \] Thus, the reflection coefficient is approximately 0.143, which corresponds to option C \( \frac{1}{7} \).

Step 3: Transmission Coefficient
The transmission coefficient \( T \) is given by:
\[ T = 1 + R = 1 + 0.143 = 1.143 \] So, the transmission coefficient is 1.143, which corresponds to option D \( \frac{8}{7} \).

Step 4: Phase Shift Constant of Medium B
The phase shift constant \( \beta \) is given by:
\[ \beta = \omega \sqrt{\mu \epsilon} \] For medium B, we have:
\[ \beta_B = 6\pi \sqrt{\frac{4}{9}} = 6\pi \times \frac{2}{3} = 4\pi \] Thus, the phase shift constant of medium B is \( 4\pi \), which corresponds to option A \( 6\pi \).

Final Answer:

The correct sequence is:
B, A, D, C