Polarization By Reflection And Brewster's Law

Polarization by reflection and Brewster's Law are key concepts in the study of light behaviour. When light reflects off a surface, it can become polarized, meaning the light waves vibrate in a single plane. Brewster's Law defines the specific angle, known as Brewster's angle, at which this polarization is maximized. Understanding these principles is essential in various practical applications, such as reducing glare in photography with polarized lenses and improving visibility in eyewear. In everyday life, polarized sunglasses utilize this phenomenon to block glare from reflective surfaces, enhancing visual comfort and clarity. In this article, we will delve into the mechanics of polarization by reflection and the significance of Brewster's Law in both scientific and practical contexts.

This Story also Contains

1. Brewster’s Law
2. Solved Examples Based on Polarization By Reflection And Brewster's Law
3. Summary

Brewster’s Law

Brewster's Law describes the phenomenon where light becomes perfectly polarized upon reflection at a specific angle, known as Brewster's angle. When light hits a surface at this angle, the reflected light is entirely polarized perpendicular to the plane of incidence. This principle is fundamental in optics and has practical applications in reducing glare in photography, enhancing contrast in LCD screens, and creating anti-glare coatings for glasses

Brewster discovered that when a beam of unpolarized light is reflected from a transparent medium (refractive index = $\mu$), the reflected light is completely plane polarised at a certain angle of incidence (called the angle of polarisation i.e. $\theta_p$ ).

And also $\mu=\tan \theta_p$.

i.e i.e For $i=\theta_P$

reflected rays will be completely polarised.

For $i<\theta_P$ or $i>\theta_P$
reflected rays will be partially polarised.

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Solved Examples Based on Polarization By Reflection And Brewster's Law

Example 1: Unpolarized light of intensity $I_0$ is incident on the surface of a block of glass at Brewster’s angle. In that case, which one of the following statements is true?

1) transmitted light is partially polarized with intensity $I_0 / 2$.

2) transmitted light is completely polarized with an intensity less than $I_0 / 2$.

3) Reflected light is completely polarized with an intensity less than $I_0 / 2$.

4) reflected light is partially polarized with intensity $I_0 / 2$.

Solution:

When unpolarised light is incident at Brewster's angle then reflected light is completely polarised and the intensity of the the reflected light is less than half of the incident light.

Hence the correct option is (3).

Example 2: A glass plate is placed vertically on a horizontal table with a beam of unpolarised light falling on its surface at the polarising angle of $57^{\circ}$ with the normal. The electric vector in the reflected light on screen S will vibrate with respect to the plane of incidence in a ____.

1) Vertical plane

2) Horizontal plane

3) The plane makes an angle of $45^{\circ}$ with the vertical

4) The plane makes an angle of $57^{\circ}$ with the horizontal

Solution:

Whenever unpolarised light is made incident at a polarising angle then the reflected light is plane polarised in a direction perpendicular to the plane of incidence. Therefore $\vec{E}$ in reflected light will vibrate in the vertical plane with respect to the plane of incidence.

Hence, the answer is the option (1).

Example 3: The angle of polarization for any medium is $60^{\circ}$, what will be the critical angle for this?

1) $\sin ^{-1} \sqrt{3}$
2) $\tan ^{-1} \sqrt{3}$
3) $\cos ^{-1} \sqrt{3}$
4) $\sin ^{-1} \frac{1}{\sqrt{3}}$

Solution:

$\begin{aligned} & \text { By using } \mu=\tan \theta_p \\ & \Rightarrow \mu=\tan 60=\sqrt{3} \\ & \text { also, Critical angle }=\sin ^{-1}\left(\frac{1}{\mu}\right) \\ & \Rightarrow \text { Critical angle }=\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)\end{aligned}$

Hence, the answer is the option (4).

Example 4: A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is $\Theta_{i C}$ and the Brewster’s angle of incidence is $\Theta_{i B}$, such that $\sin \Theta_{i C} / \sin \Theta_{i B}=\eta=1.28$. The relative refractive index of the two media is :

1) 0.2

2) 0.4

3) 0.8

4) 0.9

Solution:

$
\sin \theta_{i c}=\frac{\mu_r}{\mu_d}
$
$\mu_r=$ refractive index of the rarer medium.
$\mu_d=$ refractive index of the rarer medium.
In the case of Brewster's angle
$
r=90-\theta_{i B}
$

From Brew's law: $\mu_d \cdot \sin \theta_{i B}=\mu_r \cdot \sin$
$
\begin{aligned}
& \frac{\sin \theta_{i B}}{\cos \theta_{i B}}=\frac{\mu_r}{\mu_d \text { or }} \tan \theta_{i B}=\frac{\mu_r}{\mu_d} \\
& \sin \theta_{i B}=\frac{\mu_r}{\sqrt{\mu_r^2+\mu_d^2}} \ldots(2) \\
& \because \frac{\sin \theta_{i c}}{\sin \theta_{i B}}=1.28 \\
& \mu_1^2+\mu_d^2=1.638 \mu_d^2 \\
& \text { or } 0.638 \mu d^2=\mu_r^2 \\
& \frac{\mu_r}{\mu_d}=\sqrt{0.638}=0.8
\end{aligned}
$

Hence, the answer is the option (3).

Example 5: The angle of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index, n), is

1) $\sin ^{-1}(n)$
2) $\sin ^{-1}(1 / n)$
3) $\tan ^{-1}(1 / n)$
4) $\tan ^{-1}(n)$

Solution:

According to Brewster's law of polarization,

$n=\tan i_p$ where $i_p$ is the angle of incidence
$
i_p=\tan ^{-1}(n)
$

Hence, the answer is the option (4).

Summary

Polarization by reflection and Brewster's Law explain how light can become completely polarized at a specific angle of incidence, known as Brewster's angle. This principle is crucial in optics, with applications in reducing glare in photography, improving the contrast in LCD screens, and enhancing visual clarity in polarized sunglasses. The reflected light at Brewster's angle is entirely polarized perpendicular to the plane of incidence, a phenomenon utilized in various optical technologies to manipulate and improve the quality of light in practical scenarios.