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Q101P

Expert-verifiedFound in: Page 1009

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 ****, ${{\mathit{\theta}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{20}}{\mathbf{\xb0}}$****, ${{\mathit{\theta}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{60}}{\mathbf{\xb0}}$****and${{\mathit{\theta}}}_{{\mathbf{3}}}{\mathbf{=}}{\mathbf{40}}{\mathbf{\xb0}}$.**** What fraction of the initial light intensity emerges from the system?**

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

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

The values of the angle is:

${\theta}_{1}=20\xb0$

${\theta}_{2}=60\xb0$

${\theta}_{3}=40\xb0$

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.

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,

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

$\begin{array}{c}{\theta}^{\text{'}}=(60-20)\\ =40\xb0\end{array}$

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

$\begin{array}{c}{\theta}^{\text{'}\text{'}}=90-{\theta}_{2}+{\theta}_{3}\\ =90-60+40\\ =70\xb0\end{array}$

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

$\begin{array}{c}{I}_{3}=\frac{{I}_{0}}{2}{\mathrm{cos}}^{2}(40\xb0){\mathrm{cos}}^{2}(70\xb0)\\ \frac{{I}_{3}}{{I}_{0}}=\frac{1}{2}{\mathrm{cos}}^{2}(40\xb0){\mathrm{cos}}^{2}(70\xb0)\\ =\frac{1}{2}\left(0.5868\right)\left(0.117\right)\\ =0.034\end{array}$

Therefore, the fraction of the initial light intensity that emerges from the system is$\frac{I}{{I}_{0}}=0.034$

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