91Ó°ÊÓ

Figure 33.45 of the graph of ${\mathrm{θ}}_2$ vs. Intensity.

Here, first, calculate the direction of polarization of the first sheet. It can be calculated using the information from the graph considering the point of zero intensity. After knowing the direction of polarization of the first sheet, the intensity transmitted by the system using the equations of intensity can be calculated.

The formula is as follows:

$I=I_0{\cos }^2\mathrm{θ}$

Where,

I = radiant intensity,

cosθ = the angle θ between the direction of the incident light and the surface normal

From the graph, the intensity of transmitted light is zero${\mathrm{θ}}_2={160}^{°}$.

So, in this direction, two polarizing sheets must be crossed.

Hence,theangle for the first sheet can be calculated as,

${\mathrm{θ}}_1={160}^{°}−{90}^{°}={70}^{°}$

Now, consider the case of${\mathrm{θ}}_2={90}^{°}$${\mathrm{θ}}_2={90}^{°}$${\mathrm{θ}}_2={90}^{°}$.

As the incident polarization direction from the first sheet is at ${\mathrm{θ}}_1={70}^{°}$, the angle between polarization and the polarizing sheet for the second sheet will be,$\begin{array}{c}{\mathrm{θ}}_2={90}^{°}−{70}^{°}\\ ={20}^{°}\end{array}$

So, using the equation of the one-half rule and cosine-squared rule, it can be written,

$I_2=\frac{1}{2}{I}_0{\cos }^2\left({20}^{°}\right)$

So, the percentage of light’s initial intensity transmitted from the system is,

$\begin{array}{l}=\frac{I_2}{I_0}×100\%\\ =\frac{1}{2}{\cos }^2{20}^{°}×100\%\\ =44\%\end{array}$

Hence,the percentage of light’s initial intensity transmitted is$44\%$.