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Expert-verified Found in: Page 1009 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # In Fig. 33-76, unpolarized light is sent into a system of three polarizing sheets with polarizing directions at angles , ${{\mathbit{\theta }}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{20}}{\mathbf{°}}$, ${{\mathbit{\theta }}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{60}}{\mathbf{°}}$and${{\mathbit{\theta }}}_{{\mathbf{3}}}{\mathbf{=}}{\mathbf{40}}{\mathbf{°}}$. What fraction of the initial light intensity emerges from the system? The fraction of the initial light intensity that emerges from the system is

$\frac{I}{{I}_{0}}=0.034$

See the step by step solution

## Step 1: Determine the given quantities

The values of the angle is:

${\theta }_{1}=20°$

${\theta }_{2}=60°$

${\theta }_{3}=40°$

## Step 2: 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:

${\mathbit{I}}{\mathbf{=}}{{\mathbit{I}}}_{{\mathbf{0}}}{\mathbit{c}}{\mathbit{o}}{{\mathbit{s}}}^{{\mathbf{2}}}{\mathbit{\theta }}$ ……. (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 the original light is initially unpolarized here. Therefore,

${I}_{1}=\frac{1}{2}{I}_{0}$

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

$I={I}_{0}{\mathrm{cos}}^{2}\theta$

The intensity from the second sheet is as follows:

${I}_{2}={I}_{1}{\mathrm{cos}}^{2}{\theta }^{\text{'}}$

And the intensity from the third sheet is

${I}_{3}={I}_{2}{\mathrm{cos}}^{2}{\theta }^{\text{'}\text{'}}$

Thus,

${I}_{3}={I}_{1}{\mathrm{cos}}^{2}{\theta }^{\text{'}}{\mathrm{cos}}^{2}{\theta }^{\text{'}\text{'}}$

${I}_{3}=\frac{{I}_{0}}{2}{\mathrm{cos}}^{2}{\theta }^{\text{'}}{\mathrm{cos}}^{2}{\theta }^{\text{'}\text{'}}$(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,

${\theta }^{\text{'}}=\left({\theta }_{2}-{\theta }_{1}\right)$

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

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,

$\begin{array}{c}{I}_{3}=\frac{{I}_{0}}{2}{\mathrm{cos}}^{2}\left(40°\right){\mathrm{cos}}^{2}\left(70°\right)\\ \frac{{I}_{3}}{{I}_{0}}=\frac{1}{2}{\mathrm{cos}}^{2}\left(40°\right){\mathrm{cos}}^{2}\left(70°\right)\\ =\frac{1}{2}\left(0.5868\right)\left(0.117\right)\\ =0.034\end{array}$

Therefore, the fraction of the initial light intensity that emerges from the system is$\frac{I}{{I}_{0}}=0.034$ ### Want to see more solutions like these? 