91Ó°ÊÓ

1. Polarizing direction of the second sheet $\text{90°}$.
2. The fraction of intensity transmitted by the two sheets is 0.10.

Cosine squared law states that when a plane of polarized light is passed through an analyzer, the intensity of the transmitted beam from the analyzer is directly propositional to the cosine of the angle between the transmission axes of the polarizer and the analyzer.

The formula is as follows:

$I=I_{0}co{s}^2\mathrm{θ}$

where,

I is the radiant intensity,

$cos\mathrm{θ}$is the angle between the direction of the incident light and the surface normal,

The angle of incidence of the beam with the second sheet is $\text{90°}$.

The second angle is as follows:

${\mathrm{θ}}_2=90−\mathrm{θ}$

The intensity of light fromthefirst polarizing sheet is as follows:

$I_1=I_0{\cos }^2\mathrm{θ}$

The intensity of light fromthesecond polarizing sheet is as follows:

$I_2=I_1{\cos }^2{\mathrm{θ}}_2$

Now plug the value $I_1$ and ${\mathrm{θ}}_2$ in the above equation.

$\begin{array}{c}{I}_2=I_0{\cos }^2\mathrm{θ}{\cos }^2(90−\mathrm{θ})\\ I_2=I_0{\cos }^2\mathrm{θ}{\sin }^2\mathrm{θ}\\ I_2=I_0\frac{1}{4}{\sin }^2\left(2\mathrm{θ}\right)\end{array}$

Since the intensity of the transmitted beam is $0.1$of the incident beam,

$I_2=0.1I_0$

Substitute all the value in the above equation.

$\begin{array}{c}{I}_0\frac{1}{4}{\sin }^2\left(2\mathrm{θ}\right)=0.10I_0\\ \frac{1}{4}{\sin }^2\left(2\mathrm{θ}\right)=0.10\\ {\sin }^2\left(2\mathrm{θ}\right)=0.10×4\\ {\sin }^2\left(2\mathrm{θ}\right)=0.4\\ \sin \left(2\mathrm{θ}\right)=0.63245\\ \left(2\mathrm{θ}\right)={\sin }^{-1}\left(0.63245\right)\\ 2\mathrm{θ}=40°\\ \mathrm{θ}=20°\end{array}$

Hence, polarizing direction of the first sheet $\mathrm{θ}=20°$ .

The value of the required angle can be calculated using the cosine squared law.