Question

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?

Short answer

The intensity of the light that emerges from the system is$0.5{\text{W/m}}^{\text{2}}$.

Step-by-step solution

Given Information

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

Understanding the concept of the intensity of polarization

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)

Calculation of the intensity of the light emerging from the system

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$0.50{\text{W/m}}^{\text{2}}$