Because the speed of light in vacuum (or air) has such a large value, it is very difficult to measure directly. To measure this speed, you conduct an experiment in which you measure the amplitude of the electric field in a laser beam as you change the intensity of the beam. Figure P32.49 is a graph of the intensity I that you measured versus the square of the amplitude Emax of the electric field. The best-fit straight line for your data has a slope of $\mathbf{1}\mathbf{.}\mathbf{33}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{J}\mathbf{/}{\mathbf{V}}^{\mathbf{2}}\mathbf{s}$.(a) Explain why the data points plotted this way lie close to a straight line. (b) Use this graph to calculate the speed of light in air.

The expression for the intensity radiated by sun in terms of electric field is$\mathbf{I}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{c}{\mathbf{E}}_{\mathbf{m}\mathbf{a}\mathbf{x}}}^{\mathbf{2}}$Emax is the maximum value of electric field in the sinusoidal wave,${\mathbf{\epsilon }}_{\mathbf{0}}$is the permittivity in the vacuum, c is the speed of light.

Refer equation $\mathrm{I}=\frac{1}{2}{\mathrm{\epsilon }}_0{{\mathrm{cE}}_{\max }}^2$The speed of light and the permittivity in the vacuum is constant for vacuum.

Figure I

The graph is draw between the intensity radiated by sun and the square of the maximum electric field. Thus, the intensity is directly proportional to the square of the maximum electric field and it is depends only on electric field.

$\mathrm{I}\propto {{\mathrm{E}}_{\max }}^2$and can be concluded asthe intensity increases square of the electric field also increases.

Therefore, the graph should be a straight line and the slope is equal to$\frac{1}{2}{\mathrm{\epsilon }}_0\mathrm{c}$

The best fit straight line has a slope$1.33\times {10}^{-3}\mathrm{J}/{\mathrm{V}}^2\mathrm{s}$

Refer part (a), the slope of the straight line is$\frac{1}{2}{\mathrm{\epsilon }}_0\mathrm{c}$.Thus,

$\frac{1}{2}{\mathrm{\epsilon }}_0\mathrm{c}=1.33\times {10}^{-3}\mathrm{J}/{\mathrm{V}}^2\mathrm{s}$

Rearrange the equation to find the speed of light in air, is

$$\begin{gathered} {\mathrm{\epsilon }}_0\mathrm{c}=1.33\times {10}^{-3}\mathrm{J}/{\mathrm{V}}^2\mathrm{s}\times 2 \\ \mathrm{c}=\frac{2.66\times {10}^{=3}\mathrm{J}/{\mathrm{V}}^2\mathrm{s}}{{\mathrm{\epsilon }}_0} \\ =\frac{2.66\times {10}^{-3}\mathrm{J}/{\mathrm{V}}^2\mathrm{s}}{8.854\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2} \\ =3.00\times {10}^8\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of light in the air is $3.00\times {10}^8\mathrm{m}/\mathrm{s}$