## Section49.7Light Wave Bootcamp

### Subsection49.7.6Miscellaneous

Draw the wave fronts for incident and reflected rays from a concave mirror as they pass through a focus and beyond in the following figure. (Hint: To obtain the wavefronts near the focus, imagine putting a light source there.)

Hint

Wavefronts are drawn one wavelength apart and are perpendicular to the rays.

See solution.

Solution

Wavefronts are drawn one wavelength apart and are perpendicular to the rays. When rays are parallel, you get planar wavefronts and when rays are converging or diverging, you get spherical wavefronts. See Figure 49.7.12.

Draw wave fronts for parallel incident rays and the rays after emerging from a diverging lens.

Hint

First draw a ray diagram. then convert that to a diagram with wavefronts.

See solution.

Solution

See Figure 49.7.14.

An astronaut is at a distance of $R/4$ from the Sun, where $R$ is the average Earth-Sun distance. He holds a perfectly reflecting mirror of area $A$ perpendicular to the rays from the Sun. Find his acceleration due to the radiation pressure alone if his mass along with all the gear is $M\text{.}$ Assume that the astronaut is completely behind the mirror so that light strikes the mirror only and the intensity of the sunlight at Earth is $I_E\text{.}$

Hint

Use $F = PA$ where $P$ is the radiaiton pressure for the case of reflection.

$\dfrac{32 I_E A}{Mc}$

Solution

The intensity of the light at the site of the Astronaut will be $4^2\times I_E$ since the distance from the sun is $1/4^\text{th}$ of the distance to the Earth. We assume light being incident perpendicularly to the mirror and reflecting off perfectly. Therefore, the radiation force on the mirror + astronaut will be

\begin{equation*} F = 2\: \dfrac{I}{c}\: A = \dfrac{32 I_E A}{c}. \end{equation*}

Using Newton's second law on the mirror + astronaut we find their acceleration solely due to the radiation pressure to be

\begin{equation*} a = \dfrac{F}{M} = \dfrac{32 I_E A}{Mc}. \end{equation*}

No light passes through two crossed linear polarizers whose axes are $90^\circ$ to each other. Light however can pass if a third linear polarizer is inserted between the two. What should be the angle the polarizer in the middle should make with the first polarizer so that the intensity of the transmitted light is $5\%$ of the intensity of unpolarized light incident on the first polarizer.

Hint

Use Malus's law for the last two polarizers.

$19.6^\circ\text{.}$

Solution

Let us denote intensities as follows.

\begin{align*} \amp I = \text{starting intensity}\\ \amp I_1 = \text{intensity between first and second polarizers.}\\ \amp I_2 = \text{intensity between second and third polarizers.}\\ \amp I_3 = \text{intensity after third polarizer.} \end{align*}

Now, since the incident light on the first polarizer is unpolarized, we get

\begin{equation*} I_1 = \frac{1}{2} I. \end{equation*}

Let $\theta$ be the direction between first and second polarizer axes. Then, we will have

\begin{equation*} I_2 = I_1\: \cos^2\theta = \frac{1}{2} I\: \cos^2\theta. \end{equation*}

Since first and third polarazers are crossed, meaning their angle is $90^\circ\text{,}$ the angle between second and third will be $90^\circ-\theta\text{.}$ Therefore,

\begin{align*} I_3 \amp = I_2\:\cos^2(90^\circ - \theta) = I_2 \sin^2\theta\\ \amp = \frac{1}{2} I\: \cos^2\theta\sin^2\theta = \frac{1}{8} I\: \sin^2(2\theta). \end{align*}

Requiring this be equal to $0.05I$ gives us a condition on $\theta\text{.}$

\begin{equation*} \frac{1}{8}\sin^2(2\theta) = 0.05. \end{equation*}

Therefore, we get

\begin{equation*} \theta = \frac{1}{2}\sin^{-1}( \sqrt{ 8\times 0.05} ) = 19.6{\circ}. \end{equation*}

The reflected light from the surface of water (refractive index $4/3$) is polarized if the angle of reflection is equal to the Brewster's angle $\theta_B\text{.}$ If you are wearing a polaroid glasses whose axis is $80^\circ$ from horizontal, what is the intensity of light at your eyes if the intensity of the light before reflection from the water surface was $1000\text{ W/m}^2\text{?}$

Hint

Use reflectance of S-wave to find $I_{r\perp} = R_\perp I_{i\perp}$ and then use Malus's law.

$1.2\text{ W/m}^2\text{.}$

Solution

The light gets polarized upon reflection at the Brewster's angle since P-wave reflection is zero there. Assuming the light before incident had half energy in S-wave and half in P-wave, we see that we need to find the energy reflected in the S-wave only. We will work with reflectance of the S-wave.

\begin{equation*} R_\perp = \rho_\perp^2 = \left( \frac{n_1\cos\theta_1 - n_2\cos\theta_2 }{n_1\cos\theta_1 + n_2\cos\theta_2}\right)^2, \end{equation*}

where we have

\begin{equation*} n_1 = 1.0,\ n_2 = \frac{4}{3},\ \ \theta_1=\theta_B,\ \ n_2\sin\theta_2=n_1\sin\theta_1. \end{equation*}

Once we calculate $R_\perp\text{,}$ we can get intensity in the reflected S-wave as

\begin{equation*} I_{r\perp} = R_\perp I_{i\perp}, \end{equation*}

where we have assumed that $I_{i\perp} = \frac{1}{2}I_0\text{,}$ where $I_0=1000\text{ W/m}^2\text{.}$ Now, we have all the pieces and we just calculate.

\begin{align*} \amp \theta_B = \tan^{-1}(n_2/n_1) = \tan^{-1}(1.33/1.0) = 53^\circ. \\ \amp \theta_2 = \sin^{-1}(n_1\sin\theta_1/n_2) = 37^\circ.\\ \amp R_\perp = 0.077.\\ \amp I_{r\perp} = 0.077 \times \frac{1000}{2} = 38.5\text{ W/m}^2. \end{align*}

Now the angle between this polarization and the polaroid is $80^\circ\text{.}$ Therefore, by Malus's law, light entering the eye will have intensity

\begin{equation*} I_\text{into eye} = I_{r\perp}\cos^2 80^\circ = 1.2\text{ W/m}^2. \end{equation*}