Editing Fresnel equations (section)

=== ''s'' components ===
For the ''s'' polarization, the {{math|'''E'''}} field is parallel to the {{math|''z''}} axis and may therefore be described by its component in the {{math|''z''}}&nbsp;direction. Let the reflection and transmission coefficients be {{math|''r''<sub>s</sub>}} and {{math|''t''<sub>s</sub>}}, respectively. Then, if the incident {{math|'''E'''}} field is taken to have unit amplitude, the phasor form ({{EquationNote|3}}) of its {{math|''z''}}-component is
{{NumBlk|:|<math>E_\text{i}=e^{i\mathbf{k}_\text{i}\mathbf{\cdot r}},</math>|{{EquationRef|8}}}}
and the reflected and transmitted fields, in the same form, are
{{NumBlk|:|<math>\begin{align}
  E_\text{r} &= r_{s\,} e^{i\mathbf{k}_\text{r}\mathbf{\cdot r}}\\
  E_\text{t} &= t_{s\,} e^{i\mathbf{k}_\text{t}\mathbf{\cdot r}}.
\end{align}</math>|{{EquationRef|9}}}}

Under the sign convention used in this article, a positive reflection or transmission coefficient is one that preserves the direction of the ''transverse'' field, meaning (in this context) the field normal to the plane of incidence. For the ''s'' polarization, that means the {{math|'''E'''}} field. If the incident, reflected, and transmitted {{math|'''E'''}} fields (in the above equations) are in the {{math|''z''}}-direction ("out of the page"), then the respective {{math|'''H'''}} fields are in the directions of the red arrows, since {{math|'''k''', '''E''', '''H'''}} form a right-handed orthogonal triad. The {{math|'''H'''}} fields may therefore be described by their components in the directions of those arrows, denoted by {{math|''H''<sub>i</sub>, ''H''<sub>r</sub>, ''H''<sub>t</sub>}}. Then, since {{math|''H'' {{=}} ''YE''}},
{{NumBlk|:|<math>\begin{align}
  H_\text{i} &=\, Y_1 e^{i\mathbf{k}_\text{i}\mathbf{\cdot r}}\\
  H_\text{r} &=\, Y_1 r_{s\,} e^{i\mathbf{k}_\text{r}\mathbf{\cdot r}}\\
  H_\text{t} &=\, Y_2 t_{s\,} e^{i\mathbf{k}_\text{t}\mathbf{\cdot r}}.
\end{align}</math>|{{EquationRef|10}}}}

At the interface, by the usual [[interface conditions for electromagnetic fields]], the tangential components of the {{math|'''E'''}} and {{math|'''H'''}} fields must be continuous; that is,
{{NumBlk|:|<math>\left.\begin{align}
  E_\text{i} + E_\text{r} &= E_\text{t}\\
  H_\text{i}\cos\theta_\text{i} - H_\text{r}\cos\theta_\text{i} &= H_\text{t}\cos\theta_\text{t}
\end{align}~~\right\}~~~\text{at}~~ y=0\,.</math>|{{EquationRef|11}}}}
When we substitute from equations ({{EquationNote|8}}) to ({{EquationNote|10}}) and then from ({{EquationNote|7}}), the exponential factors cancel out, so that the interface conditions reduce to the simultaneous equations
{{NumBlk|:|<math>\begin{align}
  1 + r_\text{s} &=\, t_\text{s}\\
  Y_1\cos\theta_\text{i} - Y_1 r_\text{s}\cos\theta_\text{i} &=\, Y_2 t_\text{s}\cos\theta_\text{t} \,,
\end{align}</math>|{{EquationRef|12}}}}
which are easily solved for {{math|''r''<sub>s</sub>}} and {{math|''t''<sub>s</sub>}}, yielding
{{NumBlk|:|<math>r_\text{s}=\frac{Y_1\cos\theta_\text{i}-Y_2\cos\theta_\text{t}}{Y_1\cos\theta_\text{i}+Y_2\cos\theta_\text{t}}</math>|{{EquationRef|13}}}}
and
{{NumBlk|:|<math>t_\text{s}=\frac{2Y_1\cos\theta_\text{i}}{Y_1\cos\theta_\text{i}+Y_2\cos\theta_\text{t}}\,.</math>|{{EquationRef|14}}}}
At ''normal incidence'' {{math|(''θ''<sub>i</sub> {{=}} ''θ''<sub>t</sub> {{=}} 0)}}, indicated by an additional subscript 0, these results become
{{NumBlk|:|<math>r_\text{s0}=\frac{Y_1-Y_2}{Y_1+Y_2}</math>|{{EquationRef|15}}}}
and
{{NumBlk|:|<math>t_\text{s0}=\frac{2Y_1}{Y_1+Y_2}\,.</math>|{{EquationRef|16}}}}
At ''grazing incidence'' {{math|(''θ''<sub>i</sub> → 90°)}}, we have {{math|cos{{tsp}}''θ''<sub>i</sub> → 0}}, hence {{math|''r''<sub>s</sub> → −1}} and {{math|''t''<sub>s</sub> → 0}}.