(a), unpolarized light is sent into a system of three polarizing sheets. The angles ${\theta }_1$,${\mathit{\theta }}_{\mathbf{2}}$ and ${\mathit{\theta }}_3$of the polarizing directions are measured counterclockwise from the positive direction of theyaxis (they are not drawn to scale).Angles ${\mathit{\theta }}_{\mathbf{1}}$and ${\mathit{\theta }}_3$are fixed, but angle${\mathit{\theta }}_{\mathbf{2}}$ can be varied. Figure

(b) gives the intensity of the light emerging from sheet 3 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 three-sheet system when${\mathit{\theta }}_{\mathbf{2}}\mathbf{=}\mathbf{90}\mathbf{°}$ ?

Figure:

Figure 33-44 is the graph of l vs$\theta$ .

Use the concept of polarization of light intensity. First, find the angles for three sheets. Using the equation find the percentage of light transmitted by the three-sheet system.

Formulae are as follows,

$\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }$

Where, is the intensity.

From the graph, at ${\theta }_2={60}^0$ and ${140}^0$ the intensity is zero. From this, at these angles, the second sheet is perpendicular to either the first or third sheet.

$$\begin{gathered} {\theta }_1={90}^0+{60}^0={150}^0 \\ {\theta }_2={90}^0 \\ {\theta }_3={140}^0-{90}^0={50}^0 \end{gathered}$$

The intensity from the first sheet is as,

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

Theintensity fromthesecond sheet as,

$$\begin{gathered} I_2=I_1{\cos }^2({\theta }_1-{\theta }_2) \\ \begin{array}{c}{I}_2=\frac{1}{2}{I}_0{\cos }^2({150}^0-{90}^0)\\ I_2=\frac{1}{2}{I}_0{\cos }^2\left({60}^0\right)\\ I_2=\frac{1}{8}{I}_0\end{array} \end{gathered}$$

Similarly, for the intensity of light fromthethird sheet:

$$\begin{gathered} I_3=I_2{\cos }^2({\theta }_2-{\theta }_3) \\ \begin{array}{l}{I}_3=I_2{\cos }^2({\theta }_2-{\theta }_3)\\ I_3=\frac{1}{8}{I}_0{\cos }^2({90}^0-{50}^0)\\ I_3=\frac{1}{8}{I}_0{\cos }^2\left({40}^0\right)\\ I_3=0.0734I_0\\ \frac{I_3}{I_0}=\frac{0.0734I_0}{I_0}\\ =0.0734\\ \frac{I_3}{I_0}\approx 7.3\%\end{array} \end{gathered}$$

Therefore, the percentage of light’s initial intensity transmitted by the three-sheet system when ${\theta }_2={90}^0$ is $\frac{I_3}{I_0}=7.3\%$

Use the concept of change in the intensity due to polarization of light.