Question

When a plane wave travelling in free-space is incident normally on a medium having ε_r = 9.0 and μ_r = 1.0, the fraction of power transmitted into the medium is 
1. 4/3
2. 3/4
3. 1/2
4. 1/3

Answer 1

Correct Answer - Option 2 : 3/4

Concept:

For normal incidence:

1 + Γ_E = τ_E

Where,

Γ_E: Reflection co-efficient of the electric field 

\({{\rm{Γ }}_E} = \frac{{{E_{reflected}}}}{{{E_{incident}}}} \)

\({{\rm{Γ }}_E} = \frac{{{\eta _2} - {\eta _1}}}{{{\eta _2} + {\eta _1}}}\)

τE = transmission coefficient of the electric field

\({\tau _E} = \frac{{{E_{transmitted}}}}{{{E_{incident}}}}\)

\(\eta = \sqrt {\frac{\mu }{\varepsilon }} \left( {for\;losses\;media} \right)\)

\( {{\rm{Γ }}_E} = \frac{{\sqrt {{\varepsilon _{r1}}} - \sqrt {{\varepsilon _{r2}}} }}{{\sqrt {{\varepsilon _{r1}}} + \sqrt {{\varepsilon _{r2}}} }}\)

Reflected power is given as:

 \({{\rm{Γ }}_P} = - {\rm{Γ }}_E^2\)

Transmitted power is given as:

 \({\tau _P} = 1 + {{\rm{Γ }}_P} = 1 - {\rm{Γ }}_E^2\)  

Solution:

In this question:

ε_r1 = 1 (Air)

ε_r2 = 9

\({{\rm{Γ }}_E} = \frac{{\sqrt 1 - \sqrt 9 }}{{\sqrt 1 + \sqrt 9 }}\)

\(= \frac{{1 - 3}}{{1 + 3}} = \frac{{ - 1}}{2}\)  

\({\tau _P} = 1 - {\rm{Γ }}_E^2\)

\(= 1 - {\left( { - \frac{1}{2}} \right)^2}\)

\(= 1 - \frac{1}{4}\)

\({\tau _P} = \frac{3}{4}\)