Correct Answer - Option 2 : 3/4
Concept:
For normal incidence:
1 + ΓE = τE
Where,
ΓE: 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:
εr1 = 1 (Air)
εr2 = 9
\({{\rm{Γ }}_E} = \frac{{\sqrt 1 - \sqrt 9 }}{{\sqrt 1 + \sqrt 9 }}\)
\(= \frac{{1 - 3}}{{1 + 3}} = \frac{{ - 1}}{2}\)
\({\tau _P} = 1 - {\rm{Γ }}_E^2\)
\(= 1 - {\left( { - \frac{1}{2}} \right)^2}\)
\(= 1 - \frac{1}{4}\)
\({\tau _P} = \frac{3}{4}\)