In Fig. 33-41, a beam of light, with intensity${\mathbf{\text{43\hspace{0.17em}W/m}}}^{\mathbf{\text{2}}}$and polarization parallel to a y-axis, is sent into a system of two polarizing sheets with polarizing directions at angles of ${\mathbf{\theta }}_1\mathbf{\text{=70°}}$and ${\mathbf{\theta }}_2\mathbf{\text{=90°}}$to the y axis. What is the intensity of the light transmitted by the two-sheet system?

The intensity of unpolarized light is$I_0={\text{43\hspace{0.17em}W/m}}^{\text{2}}$.

The angle of the first polarizing sheet is${\theta }_1={70}^{\circ }$.

The angle of the second polarizing sheet is${\theta }_2={90}^{\circ }$.

Figure 33-41 is the two polarizing sheets.

Use the concept of the intensity of the wave. Using the equation, we find the intensity of transmitted light. For the second sheet, we find the difference of given angles and the intensity of the first sheet.The intensity (I) of a wave is defined as the rate at which it transfers energy divided by the area over which the energy is spread. In other words, intensity is the rate of energy flow per unit area.

The intensity of light transmitted through the first sheet as,

$I_1=I_0{\cos }^2{\theta }_1$

Plugging the values, we get,

$\begin{array}{l}{I}_1=\left(43\right)\left({\cos }^2{70}^{\circ }\right)\\ I_1=5.03{\text{\hspace{0.17em}W/m}}^{\text{2}}\end{array}$

Therefore, the intensity of light transmitted by the first sheet system is${\text{5.03\hspace{0.17em}W/m}}^{\text{2}}$.

The intensity of light transmitted by the second sheet:

The transmitted light from the first sheet is at${70}^{\circ }$with vertical, so$\theta ={90}^{\circ }-{70}^{\circ }={20}^{\circ }$with horizontal. In the second sheet, the polarizing angle is$\theta ={20}^{\circ }$. We get,

$I_2=I_1\left({\cos }^2{20}^{\circ }\right)$

Plugging the values, we get,

$\begin{array}{l}{I}_2=\left(5.03\right)\left({\cos }^2{20}^{\circ }\right)\\ I_2=4.4{\text{\hspace{0.17em}W/m}}^{\text{2}}\end{array}$

Therefore, the Intensity of light transmitted by the second sheet is ${\text{4.4\hspace{0.17em}W/m}}^{\text{2}}$.