91Ó°ÊÓ

The intensity of the un-polarized light,${\text{I}}_{\text{0}}=25{\text{W/m}}^{\text{2}}$

The four polarizing sheets with polarizing directions at angles are ${\mathrm{θ}}_1=40°,{\mathrm{θ}}_2=20°,{\mathrm{θ}}_3=20°,\text{and}{\mathrm{θ}}_4=30°$.

We can use the concept of the polarizing sheet. When the unpolarized light is transmitted, then the intensity is half the original intensity.

Formula:

The maximum intensity of an unpolarized light that is transmitted according to Malus’ law, $\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{θ}$.....(1)

According to the given figure, ${\mathrm{θ}}_1$and ${\mathrm{θ}}_4$are measured from the vertical axes and${\mathrm{θ}}_2$and ${\mathrm{θ}}_3$are measured from horizontal axes. We can measure all angles concerning the x-axis.

From this figure, the angle of the first polarizer measures from the x-axis is given by:

$\begin{array}{rcl}{\mathrm{θ}}_1& =& 90°+40°\\ & =& 130°\\ & & \end{array}$

The angle difference between the first polarizer struck by the light and the second is given as:

$\begin{array}{rcl}\mathrm{Δ}{\mathrm{θ}}_1& =& {\mathrm{θ}}_1-{\mathrm{θ}}_2\\ & =& 130°-20°\\ & =& 110°\\ & & \end{array}$

Similarly, the angle difference between the second and the third polarizer is given as:

$\begin{array}{rcl}\mathrm{Δ}{\mathrm{θ}}_2& =& {\mathrm{θ}}_2-{\mathrm{θ}}_3\\ & =& 20°-\left(-20°\right)\\ & =& 40°\end{array}$

Similarly, the angle difference between the third and the fourth polarizer is given as:

$\begin{array}{rcl}\mathrm{Δ}{\mathrm{θ}}_3& =& {\mathrm{θ}}_4-{\mathrm{θ}}_3\\ & =& {60}^0-{20}^0\\ & =& {40}^0\\ & & \end{array}$

When the un-polarized light passes through any polarizing sheet, then there is a reduction of intensity by a factor of one-half. Now, using equation (1), we can get the net intensity of the light after passing through four polarizing sheets as follows:

$$\begin{gathered} I=\frac{1}{2}{I}_0{\cos }^2\mathrm{Δ}{\mathrm{θ}}_1{\cos }^2\mathrm{Δ}{\mathrm{θ}}_2{\cos }^2\mathrm{Δ}{\mathrm{θ}}_3 \\ =\frac{1}{2}\left(25{\text{W/m}}^{\text{2}}\right){\cos }^2\left(110°\right){\cos }^2\left(40°\right){\cos }^2\left(40°\right) \\ =0.50{\text{W/m}}^{\text{2}} \end{gathered}$$

Hence, the intensity of the light that emerges from the system is$0.50{\text{W/m}}^{\text{2}}$