Question

In Fig. 33-75, unpolarized light is sent into a system of three polarizing sheets, where the polarizing directions of the first and second sheets are atangles${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{20}\mathbf{°}$and${\mathit{\theta }}_{\mathbf{2}}\mathbf{=}\mathbf{40}\mathbf{°}$. What fraction of the initial light intensity emerges from the system?

Short answer

The fraction of the initial light intensity that emerges from the system is.$0.024$

Step-by-step solution

Given data

Polarizing directions of the first and second sheet are:

${\theta }_1=20°$

${\theta }_2=40°$

Understanding the concept of intensity of polarization

Light can be polarized by passing it through a polarizing filter or other polarizing material. The intensity of an unpolarized light after passing through a polarized filter is given by the cosine-square rule. Here, we need to use the concept of polarization using polarizing sheets along with the equations for the intensity of light passing through the polarizing sheet.

Formulae:

If the incident light is un-polarized, then the intensity of the merging light using one-half rule is given by,$I=0.5I_0\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

If the incident light is already polarized, then the intensity of the emerging light is cosine –squared of the intensity of incident light,$I=I_0{\cos }^2\theta \cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

Here,$\theta$is the angle between polarization of the incident light and the polarization axis of the sheet.

Calculation of the fraction of the initial incident intensity

If the original light is initially unpolarized,the transmitted intensity is given using equation (1) as follows:

$I_1=\frac{1}{2}{I}_0$

Now, the intensity from the second sheet due to polarized light intensity from sheet 1 is given using equation (2) as follows:

$I_2=I_1{\cos }^2\theta {\text{'}}_1$

And again the intensity from the third sheetdue to polarized light intensity from sheet 2 is given using equation (ii) as follows:

$\begin{array}{c}{I}_3=I_2{\cos }^2\theta {\text{'}}_2\\ =I_1{\cos }^2\theta {\text{'}}_1{\cos }^2\theta {\text{'}}_2\\ =\frac{I_0}{2}{\cos }^2\theta {\text{'}}_1{\cos }^2\theta {\text{'}}_2\cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)\end{array}$

Where${\theta }_1$,is the relative angle between the first and the second polarizing sheet andthe${\theta }_2$is the relative angle between the second and the third polarizing sheet.

Thus, from the figure, the polarizing angle between first and second sheet is given by,

$\begin{array}{c}\theta {\text{'}}_1=(90°-{\theta }_1)+{\theta }_2\\ =(90°-20)+40°\\ =110°\end{array}$

Similarly, the polarizing angle between third and second sheet is given by,

$\begin{array}{c}\theta {\text{'}}_2=90-{\theta }_2\\ =90-40\\ =50°\end{array}$

Now, substituting the values in equation (1), we can get the intensity from the third sheet is given as follows:

$I_3=\frac{I_0}{2}{\cos }^2(110°){\cos }^2(50°)$

Now, the fraction of the initial incident intensity is given as:

$\begin{array}{c}\frac{I_3}{I_0}=\frac{1}{2}{\cos }^2(110°){\cos }^2(50°)\\ =\frac{1}{2}\left(0.1169\right)\left(0.4132\right)\\ =0.024\end{array}$

Hence, the value of the fraction of the intensity is.$0.024$