The angle of incidence θ i and the expression for E^{\to}_{i} are

Correct Answer is : 45° and E02(a^x−a^z)e−jπ×104(x+z)32V∕m

60° and {E_{0}}/{\sqrt{2}}(a^{\^}_{x}−a^{\^}_{z})e^{−j {\pi \times 10^{4}(x + z)}/{3\sqrt{2}}}V\/ m

45° and {E_{0}}/{\sqrt{2}}(a^{\^}_{x} + a^{\^}_{z})e^{−j\frac{\pi \times 10^{4}z}{3}}V\/ m

45° and {E_{0}}/{\sqrt{2}}(a^{\^}_{x}−a^{\^}_{z})e^{−j {\pi \times 10^{4}(x + z)}/{3\sqrt{2}}}V\/ m

60° and {E_{0}}/{\sqrt{2}}(a^{\^}_{x}−a^{\^}_{z})e^{−j\frac{\pi \times 10^{4}z}{3}}V\/ m

GATE Electronics and Communications (EC) 2013 Solved Paper

Question : 62 of 65

Marks: +1, -0

The angle of incidence θ_i and the expression for

E↖{→}_{i}

are

60° and $\;{E_{0}}/{{√{2}}}(a↖{\^}_{x}−a↖{\^}_{z})e^{−j\;{π\;\;×\;10^{4}(x\;+\;z)}/{3{√{2}}}}V\/ m$
45° and $\;{E_{0}}/{{√{2}}}(a↖{\^}_{x} + a↖{\^}_{z})e^{−j\;{π\;\;×\;10^{4}z}/{3}}V\/ m$
45° and $\;{E_{0}}/{{√{2}}}(a↖{\^}_{x}−a↖{\^}_{z})e^{−j\;{π\;\;×\;10^{4}(x\;+\;z)}/{3{√{2}}}}V\/ m$
60° and $\;{E_{0}}/{{√{2}}}(a↖{\^}_{x}−a↖{\^}_{z})e^{−j\;{π\;\;×\;10^{4}z}/{3}}V\/ m$

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