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

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

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 polarization of the wave. Using the equation of the one-half rule and cosine-squared rule, we can find the intensity of the transmitted light.Polarization is a property applying to transverse waves that specify the geometrical orientation of the oscillations. In a transverse wave, the direction of the oscillation is perpendicular to the direction of motion of the wave.

Find the angle as,

$\begin{array}{c}\theta ={\theta }_2-{\theta }_1\\ ={90}^{\circ }-{70}^{\circ }\\ ={20}^{\circ }\end{array}$

The intensity of light from the first sheet is,

$I_1=\frac{1}{2}{I}_0$

The intensity of light from the second sheet is,

$I_2=\frac{1}{2}{I}_0{\cos }^2\theta$

Plugging the values,and we get,

$\begin{array}{l}{I}_2=\frac{1}{2}\left(43\right)\left({\cos }^2{20}^{\circ }\right)\\ I_2=18.99\\ I_2~19{\text{W/m}}^{\text{2}}\end{array}$

Therefore, the intensity of the light transmitted by the system is ${\text{19\hspace{0.17em}W/m}}^{\text{2}}$.