Question

In Fig. 33-76, unpolarized light is sent into a system of three polarizing sheets with polarizing directions at angles, ${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{20}\mathbf{°}$, ${\mathit{\theta }}_{\mathbf{2}}\mathbf{=}\mathbf{60}\mathbf{°}$and${\mathit{\theta }}_{\mathbf{3}}\mathbf{=}\mathbf{40}\mathbf{°}$.What fraction of the initial light intensity emerges from the system?

Short answer

The fraction of the initial light intensity that emerges from the systemis

$\frac{I}{I_0}=0.034$

Step-by-step solution

Determine the given quantities 

The values of the angle is:

${\theta }_1=20°$

${\theta }_2=60°$

${\theta }_3=40°$

Determine the formulas for polarization 

If the incident light is un-polarized, then the intensity of the merging light using one-half rule is given by:

$\mathbf{I}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}{\mathbf{I}}_0$ …… (i)

If the incident light is already polarized, then the intensity of the emerging light is cosine –squared of the intensity of incident light is as follows:

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

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

 Step 3: Determine thefraction of the initial light intensity that emerges from the system

If the original light is initially unpolarized,the transmitted intensity is

$I=\frac{I_0}{2}$

As theoriginal light is initially unpolarizedhere. Therefore,

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

If the original light is initially polarized, the transmitted intensity is

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

The intensity from the second sheet is as follows:

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

And the intensity from the third sheet is

$I_3=I_2{\cos }^2{\theta }^{\text{'}\text{'}}$

Thus,

$I_3=I_1{\cos }^2{\theta }^{\text{'}}{\cos }^2{\theta }^{\text{'}\text{'}}$

$I_3=\frac{I_0}{2}{\cos }^2{\theta }^{\text{'}}{\cos }^2{\theta }^{\text{'}\text{'}}$(1)

Here,the${\theta }^{\text{'}}$ is the relative angle between the first and the second polarizing sheet and the ${\theta }^{\text{'}}$is the relative angle between the second and the third polarizing sheet.

Thus, from the figure,

${\theta }^{\text{'}}=({\theta }_2-{\theta }_1)$

$\begin{array}{c}{\theta }^{\text{'}}=(60-20)\\ =40°\end{array}$

Similarly for the angle,${\theta }^{\text{'}\text{'}}$

$\begin{array}{c}{\theta }^{\text{'}\text{'}}=90-{\theta }_2+{\theta }_3\\ =90-60+40\\ =70°\end{array}$

Substitute the value of${\theta }^{\text{'}}$and${\theta }^{\text{'}}$in equation (1), we get,

$\begin{array}{c}{I}_3=\frac{I_0}{2}{\cos }^2(40°){\cos }^2(70°)\\ \frac{I_3}{I_0}=\frac{1}{2}{\cos }^2(40°){\cos }^2(70°)\\ =\frac{1}{2}\left(0.5868\right)\left(0.117\right)\\ =0.034\end{array}$

Therefore, the fraction of the initial light intensity that emerges from the system is$\frac{I}{I_0}=0.034$