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Light Intensity


 The light intensity is expresses as "bright" and "dark". How is the light intensity quantitatively expressed?
 The light intensity is derived from the energy passing through the unit area perpendicular to its propagation direction in the unit time. The energy passing through the unit area perpendicular to the propagation direction of the electromagnetic wave in unit time is given as the Poynting vector. The Poynting vector \( {\bf S} \) of the electromagnetic wave is expressed as follows, \begin{eqnarray} {\bf S} = {\bf E} \times {\bf H}\ \ \ \ \ \ \ \ \ \ \ (1) \end{eqnarray} Note that this Poynting vector is not equal to the light intensity. By using the Maxwell equations, we get \begin{eqnarray} H = \sqrt{\frac{\epsilon}{\mu}}\ E\ \ \ \ \ \ \ \ \ \ \ (2) \end{eqnarray} where \( \epsilon \) and \( \mu \) are the dielectric constant and the magnetic permeability, respectively. By using equation 3, equation 1 can be written as \begin{eqnarray} S = \sqrt{\frac{\epsilon}{\mu}}\ E^2\ \ \ \ \ \ \ \ \ \ \ (3) \end{eqnarray} We define that the \(z\) axis is the propagation direction of light and its perpendicularly components are \( x \) and \( y \) axis. Thus, the \( x \) and \( y\) component of the electric field is expressed as \begin{eqnarray} E^2 &=& E_x^2 + E_y^2 \\ E_x^2 &=& A_x^2 \cos^2 \left( \omega t + \phi_x \right) \\ E_y^2 &=& A_y^2 \cos^2 \left( \omega t + \phi_y \right) \end{eqnarray} where \( A_x \), \( A_y \), \( \phi_x \), and \( \phi_y \) are the electric field and its the phase in the direction of x and y axes, respectively. \( \omega \) is the angular frequency. We substitute \( E_x \) and \( E_y \) for Equation (3) and calculate an average of Equation (3) over a time (\(T\)) sufficiently longer than the frequency of light (\( \nu = \frac{ \omega } {2 \pi} \)). Then we get the energy passing through the unit area perpendicular to the propagation direction of the electromagnetic wave in unit time as follows \begin{eqnarray} \overline{S} &=& \sqrt{\frac{\epsilon}{\mu}}\ \left( \int_0^T A_x^2 \cos^2 \left( \omega t + \phi_x \right)\ dt + \int_0^T A_y^2 \cos^2 \left( \omega t + \phi_y \right)\ dt \right) \\ &=& \sqrt{\frac{\epsilon}{\mu}}\ \left( \frac{1}{2}A_x^2 + \frac{1}{2}A_y^2 \right) + \frac{c}{T} \sqrt{\frac{\epsilon}{\mu}}\ \left( \frac{1}{4\omega} \sin\left( \omega T + \phi_x \right) + \frac{1}{4\omega} \sin\left( \omega T + \phi_y \right) \right)\ \ \ \ \ \ \ \ \ \ \ (4) \end{eqnarray} The second term of equation 4 become negligible because \( T \) is sufficiently larger than \( \omega \). We can get the light intensity (\(I \)) by multiplying the velocity of light (\( v = c/n \) \(n\) is refraction factor) to equation 4. (The light intensity is the energy density) \begin{eqnarray} I &=& c \overline{S} = \frac{1}{2} \frac{c}{n} \sqrt{\frac{\epsilon}{\mu}} E_0^2 \\ &=& \frac{1}{2} \epsilon E_0^2 \ \ \ \ \ \ (5) \end{eqnarray} where \( E_0 = \sqrt{ A_x^2 + A_y^2 } \) indicating the amplitude of the electric field. Thus the light intensity is proportional to the square of the amplitude of the electric field.

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