Question

In Fig. 33-41, unpolarized light is sent into a system of two polarizing sheets. The anglesof the polarizing directions of the sheets are measured counterclockwise from the positive direction of the y-axis (they are not drawn to scale in the figure). The angle ${\mathit{\theta }}_{\mathbf{1}}$ is fixed but the angle ${\mathit{\theta }}_{\mathbf{2}}$can be varied. Figure 33-45 gives the intensity of the light emerging from sheet 2 as a function of ${\mathit{\theta }}_{\mathbf{2}}$ . (The scale of the intensity axis is not indicated.) What percentage of the light’s initial intensity is transmitted by the two-sheet system when?

a.

b.

Short answer

The percentage of light’s initial intensity transmitted is$44\%$ .

Step-by-step solution

Step 1: Given data

Figure 33.45 of the graph of ${\theta }_2$ vs. Intensity.

Determining the concept

Here, first, calculate the direction of polarization of the first sheet. It can be calculated using the information from the graph considering the point of zero intensity. After knowing the direction of polarization of the first sheet, the intensity transmitted by the system using the equations of intensity can be calculated.

The formula is as follows:

$I=I_0{\cos }^2\theta$

Where,

I = radiant intensity,

cosθ = the angle θ between the direction of the incident light and the surface normal

Determining the percentage of light’s initial intensity transmitted by two sheet system when  θ2=90°

From the graph, the intensity of transmitted light is zero${\theta }_2={160}^{°}$.

So, in this direction, two polarizing sheets must be crossed.

Hence,theangle for the first sheet can be calculated as,

${\theta }_1={160}^{°}-{90}^{°}={70}^{°}$

Now, consider the case of${\theta }_2={90}^{°}$${\theta }_2={90}^{°}$${\theta }_2={90}^{°}$.

As the incident polarization direction from the first sheet is at ${\theta }_1={70}^{°}$, the angle between polarization and the polarizing sheet for the second sheet will be,$\begin{array}{c}{\theta }_2={90}^{°}-{70}^{°}\\ ={20}^{°}\end{array}$

So, using the equation of the one-half rule and cosine-squared rule, it can be written,

$I_2=\frac{1}{2}{I}_0{\cos }^2\left({20}^{°}\right)$

So, the percentage of light’s initial intensity transmitted from the system is,

$\begin{array}{l}=\frac{I_2}{I_0}\times 100\%\\ =\frac{1}{2}{\cos }^2{20}^{°}\times 100\%\\ =44\%\end{array}$

Hence,the percentage of light’s initial intensity transmitted is$44\%$.