Light Wave Bootcamp

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.)

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Figure 46.28. Figure for Exercise 46.7.14
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Hint.

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

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Answer.

See solution.

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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 46.29.

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Figure 46.29. Figure for Exercise 46.7.14.
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15. Drawing Wavefronts for Concave Lens.

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

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Hint.

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

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Answer.

See solution.

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Solution.

See Figure 46.30.

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Figure 46.30. Figure for Exercise 46.7.15.
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16. Computing Intensity from Given Electric Field of Electromagnetic Wave.

A plane electromagnetic wave in glass is given by the following traveling wave.

\begin{equation*} \vec E = \hat u_y\ \left( 3\times 10^{-6}\ \textrm{N/C}\right) \ \cos\left[ 10^{15}\ \pi \left( t - 1.6\frac{x}{c}\right) \right]. \end{equation*}

Find the following: (a) the frequency of the light, (b) the wavelength, (c) the speed of the light in the glass, (d) the refractive index of glass, and (e) the intensity.

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Solution.

(a) Frequency, \(f = 0.5\times 10^{15}\:\textrm{Hz}\text{.}\)

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(b) Wavelength of the light in glass, \(\lambda = (c/1.6)/f = 3.75\times 10^{-7}\:\textrm{m}\text{.}\) The same light will have the wavelength \(= 6\times 10^{-7}\:\textrm{m}\) in vacuum.

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(d) The refractive index of glass \(= 1.6\text{.}\)

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(e) The intensity, \(I = \dfrac{1}{2}v\:\epsilon\:E_0^2\text{.}\) Here \(v = 1.88\times 10^{8}\:\textrm{m/s}\text{,}\) \(\epsilon = \epsilon_0\times \epsilon_r = \epsilon_0\times \sqrt{n} = 1.12\times10^{-11}\:\textrm{C}^2\textrm{/N.m}^2\text{,}\) and \(E_0 = 3\times 10^{-6}\ \textrm{N/C}\text{.}\) Therefore, \(I = 9.4752\times 10^{-15}\text{W/m}^2\text{.}\)

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17. Acceleration of an Astronaut by Light Reflected from the Astronaut.

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 \(I_E\text{.}\)

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Hint.

Use \(F = PA\) where \(P\) is the radiaiton pressure for the case of reflection.

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Answer.

\(\dfrac{32 I_E A}{Mc}\)

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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*}

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18. Suspendinc a Pellet in Air by Light.

A perfectly absorbing cylindrical pellet of cross-sectional area \(A\) and mass \(m\) is to be suspended near Earth’s surface. Assume the orientation of the pellet lengthwise vertical and light strikes only one end and not the sides. Find the intensity of light needed.

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Solution.

Balancing the weight with the radiation force we get the following for a perfect absorber.

\begin{equation*} \dfrac{I}{c}\: A = M g,\ \ \Longrightarrow\ \ I = \dfrac{M g c}{A}. \end{equation*}

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19. Placing a Third Polaroid Between two Crossed Polarizers.

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.

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Hint.

Use Malus’s law for the last two polarizers.

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Answer.

\(19.6^\circ\text{.}\)

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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 \(0.05\,I\) 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*}

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20. Intensity of Light Entering Eye when Wearing Polarized Glasses.

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 \(1000\text{ W/m}^2\text{?}\)

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Hint.

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

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Answer.

\(1.2\text{ W/m}^2\text{.}\)

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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*}

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21. Polarized Reflected Light PAssing Through Polarized Sunglasses.

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 \(10^{\circ}\) from vertical, what is the intensity of light at your eyes if the intensity of the light before reflection from the water surface was \(100\, \text{W/m}^2\text{?}\)

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Answer.

\(1.5\, \text{W/m}^2\text{.}\)

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Solution.

The light gets polarized upon reflection at the Brewster’s angle with the reflected light having the polarization parallel to the interface and the refracted light being polarized perpendicular to the interface. Since the original light is unpolarized, the conservation of energy leads to the conclusion that the reflected light would have half the original intensity.

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Now, this polarized light is incident on a polaroid glasses with an angle of \(80^{\circ}\) between the direction of polarization of the incident light and the direction of the axis of the polaroid. (The angle of the polaroid axis is \(10^{\circ}\) with respect to the vertical but the light is polarized horizontally.) Therefore,

\begin{equation*} I_{\textrm{in eye}} = \dfrac{1}{2} I_0 \cos^2\:80^{\circ} = 1.5\:\textrm{W/m}^2. \end{equation*}

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