91Ó°ÊÓ

The intensity of unpolarized light is$I_0=43{\text{ W/³¾}}^{\text{2}}$.

The angle of the first polarizing sheet is${\mathrm{θ}}_1={70}^{∘}$.

The angle of the second polarizing sheet is${\mathrm{θ}}_2={90}^{∘}$.

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,

$\begin{array}{c}\mathrm{θ}={\mathrm{θ}}_2−{\mathrm{θ}}_1\\ ={90}^{∘}−{70}^{∘}\\ ={20}^{∘}\end{array}$

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,

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

Plugging the values,and we get,

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

Therefore, the intensity of the light transmitted by the system is ${\text{19 W/³¾}}^{\text{2}}$.