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Question #113432
For normal incidennce at the air dielectric interface with dielectri constant Er=4. Find the fraction of energy reflected into the air?
Expert's answer
R
=
(
n
1
−
n
2
n
1
+
n
2
)
2
R=\left(\frac{n_1-n_2}{n_1+n_2}\right)^2
R
=
(
n
1
+
n
2
n
1
−
n
2
)
2
n
1
=
1
,
n
2
=
E
r
=
2
n_1=1, n_2=\sqrt{E_r}=2
n
1
=
1
,
n
2
=
E
r
=
2
R
=
(
1
−
2
1
+
2
)
2
=
1
9
R=\left(\frac{1-2}{1+2}\right)^2=\frac{1}{9}
R
=
(
1
+
2
1
−
2
)
2
=
9
1
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