In Fig. 33-66, unpolarized light with an intensity of $\mathbf{25}\mathbf{}{\mathbf{\text{W/m}}}^{\mathbf{\text{2}}}$is sent into a system of four polarizing sheets with polarizing directions at angleslocalid="1662978686975" ${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{40}\mathbf{°}\mathbf{}\mathbf{,}{\mathit{\theta }}_{\mathbf{2}}\mathbf{=}\mathbf{20}\mathbf{°}\mathbf{,}{\mathit{\theta }}_{\mathbf{3}}\mathbf{=}\mathbf{20}\mathbf{°}\mathbf{}\mathbf{,}\mathbf{}\mathbf{\text{and}}\mathbf{}\mathbf{}{\mathit{\theta }}_{\mathbf{4}}\mathbf{=}\mathbf{30}\mathbf{°}$. What is the intensity of the light that emerges from the system?

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 ${\theta }_1=40°,{\theta }_2=20°,{\theta }_3=20°,\text{and}{\theta }_4=30°$.

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, $\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }$.....(1)

According to the given figure, ${\theta }_1$and ${\theta }_4$are measured from the vertical axes and${\theta }_2$and ${\theta }_3$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:

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

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

$\begin{array}{rcl}\Delta {\theta }_1& =& {\theta }_1-{\theta }_2\\ & =& 130°-20°\\ & =& 110°\\ & & \end{array}$

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

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

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

$\begin{array}{rcl}\Delta {\theta }_3& =& {\theta }_4-{\theta }_3\\ & =& {60}^0-{20}^0\\ & =& {40}^0\\ & & \end{array}$

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:

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

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