Unpolarized light of intensity ${\mathbf{\text{10\hspace{0.17em}mW/m}}}^{\mathbf{\text{2}}}$is sent into a polarizing sheet as in Fig.33-11. What are

(a) the amplitude of the electric field component of the transmitted light and

(b) the radiation pressure on the sheet due to its absorbing some of the light?

Unpolarized light intensity is,$I_0=10{\text{\hspace{0.17em}mW/m}}^{\text{2}}$.

Figure 33-11 is the polarization of light.

Use the concept of the intensity of light related to electric field amplitude. Using the equation of intensity, find the electric field amplitude. From radiation pressure related to intensity and speed of light, find the radiation pressure on the sheet.

Formulae are as follows:

$$\begin{gathered} E_m=\sqrt{2c{\mu }_{0}I} \\ {P}_r=\frac{I}{c} \end{gathered}$$

Where,$P_r$ is the radiation pressure,l is the intensity, and c is the speed of light.

The amplitude of the electric field component of the transmitted light:

In polarization of incident light, half intensity is absorbed, and half intensity is transmitted. Write the transmitted intensity as,

$\begin{array}{c}I=\frac{I_0}{2}\\ =\frac{10\times {10}^{-3}{\text{\hspace{0.17em}W/m}}^{\text{2}}}{2}\\ =5\times {10}^{-3}{\text{\hspace{0.17em}W/m}}^{\text{2}}\end{array}$

Using the equation of amplitude of the electric field,

$E_m=\sqrt{2c{\mu }_{0}I}$

Plugging the values, and we get,

$\begin{array}{l}{E}_m=\sqrt{2(3\times {10}^8)(4\pi \times {10}^{-7})(5.0\times {10}^{-3})}\\ E_m=1.9\text{\hspace{0.17em}V/m}\end{array}$

Therefore, the amplitude of the electric field component of the transmitted light is $E_m=1.9\text{\hspace{0.17em}V/m}$.

The radiation pressure on the sheet due to its absorptionof some light:

Using the equation of radiation pressure,

$P_r=\frac{I}{c}$

Plugging the values, and we get,

$\begin{array}{l}{P}_r=\frac{5\times {10}^{-3}}{3\times {10}^8}\\ P_r=1.7\times {10}^{-11}\text{Pa}\end{array}$

Therefore, the radiation pressure on the sheet due to its absorbing some of the light is .

$P_r=1.7\times {10}^{-11}\text{\hspace{0.17em}Pa}$