Polarization

Brewster's law: The tangent of the polarizing angle is equal to the refractive index of the reflecting medium:

At this angle, the reflected light is completely plane-polarized and the reflected and refracted rays are perpendicular to each other.

Proof:

          incident   N   reflected (polarized)
               \     |     /
                \ ip | ip /
     air  _______\___|___/_______
     glass        \  |
                   \ r     reflected ⟂ refracted (90°)
                    \ refracted

1. At the polarizing angle, the reflected and refracted rays are at $90^\circ$:

2. By Snell's law:

3. Therefore

(d) $\sqrt{3}$

When the reflected and refracted rays are perpendicular, the angle of incidence is the polarizing angle, and Brewster's law gives $\mu = \tan i_p = \tan 60^\circ = \sqrt{3}$.

(A) reduce intensity of light

A polarizer lets through only the vibrations in one plane, so it produces plane-polarized light and cuts the intensity of unpolarized light to half. This is why polarizers are used in sunglasses and camera filters to reduce glare. It cannot increase intensity or produce unpolarized light.

(b) $56.3^\circ$

By Brewster's law, $\tan\theta_p = n = 1.5$, so

$\theta_p = \tan^{-1}(1.5) = 56.3^\circ$

(a) Sound waves are longitudinal: the particles vibrate along the direction of propagation. Polarization needs vibrations perpendicular to the direction of travel, which is possible only for transverse waves.

(b) Let two coherent slits $S_1$ and $S_2$ be a distance $d$ apart, with a screen at distance $D\gg d$. For a point $P$ at height $y$ from the central point, the path difference is

$\Delta = S_2P - S_1P \approx \frac{yd}{D}$

Bright fringes: $\Delta = n\lambda$, so $y_n = \dfrac{n\lambda D}{d}$. The distance between successive bright fringes is

$\beta = y_{n+1} - y_n = \frac{\lambda D}{d}$

Dark fringes: $\Delta = (2n+1)\dfrac{\lambda}{2}$, so $y_n = \dfrac{(2n+1)\lambda D}{2d}$, and the distance between successive dark fringes is again

$y_{n+1} - y_n = \frac{\lambda D}{d}$

The bright and dark fringes both have the same width $\beta = \lambda D/d$, which does not depend on $n$, so they are equally spaced.

(c) The refractive index of water is

$n = \frac{1}{\sin C} = \frac{1}{\sin 48^\circ} = \frac{1}{0.743} = 1.35$

By Brewster's law, $\tan\theta_p = n = 1.35$, so $\theta_p \approx 53.4^\circ$.

(d) A grating has a very large number of slits (thousands per cm). It gives very sharp, bright and widely separated principal maxima, so the position of a maximum can be measured much more accurately. Also $d\sin\theta = n\lambda$ gives a larger deviation for a small $d$, so $\lambda$ is found with less relative error. A double slit gives broad, faint fringes that are hard to locate precisely.

(a) Brewster's law: When unpolarized light is reflected from a transparent surface at the polarizing angle $\theta_p$, the reflected light is completely plane polarized, and the tangent of the polarizing angle equals the refractive index of the medium:

$n = \tan\theta_p$

Proof: At the polarizing angle the reflected and refracted rays are perpendicular to each other. If $r$ is the angle of refraction, then

$\theta_p + 90^\circ + r = 180^\circ \;\Rightarrow\; r = 90^\circ - \theta_p$

By Snell's law,

$n = \frac{\sin\theta_p}{\sin r} = \frac{\sin\theta_p}{\sin(90^\circ-\theta_p)} = \frac{\sin\theta_p}{\cos\theta_p} = \tan\theta_p$

(b) No. In a longitudinal wave the particles vibrate along the direction of propagation, so the vibrations are already confined to one line and are symmetric about the direction of travel. There is no perpendicular vibration to be restricted, so polarization (which needs transverse vibrations) is not possible.

Hold one sunglass in front of the other and look through both while rotating one of them. If they are polaroid, the light becomes dim and is almost cut off when they are at $90^\circ$ to each other. (Or look through one at an LCD screen or at glare from water and rotate it: with a polaroid the brightness changes; with an ordinary glass it does not.)

Given: $\mu = 1.3$

The reflected light is completely polarized when the angle of incidence is the polarizing angle $i_p$. By Brewster's law:

Angle of incidence $\approx 52.4^\circ$.