chapter_25

# Ch.25: Electromagnetic waves (Apr 10-24)

## Constants Ch.25

• Speed of light $c=299792458\,\frac{\text m}{\text s}\approx 3\times 10^8\,\frac{\text m}{\text s}$
• Electric constant $\epsilon_0$ $\approx 8.854\times 10^{-12}\,\frac{\text F}{\text m}$
• Magnetic constant $\mu_0$ $=\frac{4\pi}{10^7}\,\frac{\text N}{ {\text m}^2}\approx 1.257\times 10^{-6}\,\frac{\text N}{ {\text m}^2}$

## Equation Sheet Ch.25

### Wiki version

1. Speed of light in vacuum:
• $c=3\times 10^8\frac{\text m}{\text s}$ $= \frac{1}{\sqrt{\epsilon_0\mu_0}}$ $= f\lambda$
• $\epsilon_0 = 8.85 \times 10^{-12}\frac{\text F}{\text m}\;$ is the electric constant
• $\mu_0 = 4\pi\cdot 10^{-7}\frac{\text N}{\,{\text A}^2}\;$ is the magnetic constant
2. The Doppler effect for electromagnetic waves:
• $f'$ $= f \left( 1 \pm \frac{u}{c}\right)$
• $u$ is the relative velocity between the source and the observer (difference in speed projected on the axis drawn through both of them)
• Use the plus sign if the distance between the source and the observer is decreasing (they are moving towards each other)
• Use the minus sign if the distance between the source and the observer is increasing (they are moving away from each other)
3. Total energy density of an electromagnetic wave = the energy density of the electric field + the energy density of the magnetic field
• $u=u_E+u_B$
• $u_E=u_B\;\;$ (the two densities are equal at every point in space and in any moment in time, far away from the sources)
• $u=2u_E=2u_B$
4. For an electromagnetic wave described by an electric field of magnitude E and a magnetic field of magnitude B:
• Total energy density at any point in space, some moment in time:
• $u\,=\,\frac{1}{2}\epsilon_0E^2\!+\!\frac{1}{2\mu_0}B^2$ $\,=\,\epsilon_0E^2$ $\,=\,\frac{1}{\mu_0}B^2$
• Time-averaged energy density at any point in space:
• $u_\text{av}$ $\,=\,\frac{1}{2}\epsilon_0E_\text{rms}^2\!+\!\frac{1}{2\mu_0}B_\text{rms}^2$ $\,=\,\epsilon_0E_\text{rms}^2$ $\,=\,\frac{1}{\mu_0}B_\text{rms}^2$ $\,=\,\frac{1}{2}u_\text{max}$ $\,=\,\frac{1}{4}\epsilon_0E_\text{max}^2\!+\!\frac{1}{4\mu_0}B_\text{max}^2$ $\,=\,\frac{1}{2}\epsilon_0E_\text{max}^2$ $\,=\,\frac{1}{2\mu_0}B_\text{max}^2$
• To find the average energy density simply plug in the rms value of the fields instead of E and/or B.
• rms value = square root of 1/2 times the max value:
• $E_\text{rms}=\frac{1}{\sqrt{2}}E_\text{max}$ $\approx 0.7071\!\cdot\!E_\text{max}$
• $B_\text{rms}=\frac{1}{\sqrt{2}}B_\text{max}$ $\approx 0.7071\!\cdot\!B_\text{max}$
5. The energy density of the magnetic field and the energy density of the electric field are equal to each other
• $u_E=u_B\;\;$ (at any point in space and time, far away from the sources)
• $\frac{1}{2}\epsilon_0E^2=\frac{1}{2\mu_0}B^2$
• $E=\frac{1}{\sqrt{\epsilon_0\mu_0}}B=cB\;\;$ (at any point in space and time, in vacuum)
• $E_\text{max}=cB_\text{max}\;\;$ (at any point in space, in vacuum)
• $E_\text{rms}=cB_\text{rms}\;\;$ (at any point in space, in vacuum)
6. Intensity is the energy of the waves, per unit area of the receiver, per unit time:
• $I = \frac{U}{A\cdot\Delta t}$ $=uc\;\;$ (at any moment in time, $\Delta t$ is much shorter than the period $\frac{1}{f}$)
• $I_\text{av} = \frac{U}{A\cdot\Delta t}$ $=u_\text{av}c\;\;$ (average over time, $\Delta t$ is much longer than the period $\frac{1}{f}$)
7. Momentum = total energy absorbed divided by the speed of light:
• $p=\frac{U}{c}$
8. Pressure of the completely absorbed radiation is equal to the intensity divided by the speed of light
• $P=\frac{I}{c}\;\;$ (at any given moment of time)
• $P_\text{av}=\frac{I_\text{av}}{c}\;\;$ (average over time)
9. Pressure of the completely reflected radiation is equal to twice the intensity divided by the speed of light
• $P=2\frac{I}{c}\;\;$ (at any given moment of time)
• $P_\text{av}=2\frac{I_\text{av}}{c}\;\;$ (average over time)

#### Equations Ch.25.5

1. Law of Malus (for a linearly polarized light incident normally (perpendicular to) an ideal polarizer):
• $I = I_0\cos^2\theta$ $=I_0(\cos\theta)^2$
• $I_0$ is the average incident intensity
• $I$ is the average transmitted intensity
• $\theta$ is the angle between the $\vec{\mathbf E}$ of the incident light and the polarization axis of the polarizer
• SI units: $\frac{\text W}{\,{\text m}^2}$
• The more polarizers that are used, hence the more smoothly the direction of polarization changes-the greater the final intensity
2. Transmitted intensity for an unpolarized beam incident normally at an ideal polarizer
• $I = \frac{1}{2}I_0$
• $I_0$ is the average incident intensity
• $I$ is the average transmitted intensity
• SI units: $\frac{\text W}{\,{\text m}^2}$

## Examples Ch.25

### Problem 3.1

Find frequencies of $\lambda=800\,$nm (deep red) and $\lambda=400\,$nm (violet) light. These roughly correspond to the limits of what our eyes can see.

Steps:

1. Use the relationship between the wavelength, speed of light, and frequency:
• $f=\frac{c}{\lambda}$
• For deep red, $\;\;f=\frac{3\times 10^8\,\frac{\text m}{\text s}}{800\times 10^{-9}\,{\text m}}$ $=3.75\times 10^{14}\,$Hz
• For violet, $\;\;f=\frac{3\times 10^8\,\frac{\text m}{\text s}}{400\times 10^{-9}\,{\text m}}$ $=7.5\times 10^{14}\,$Hz

### Problem 3.2

Find the maximum E and B fields of a green light source with the average intensity corresponding to the “threshold of pain” ($I=1000\,\frac{\text W}{\,{\text m}^2}$).

• $c=299792458\,\frac{\text m}{\text s}$ $\approx 3\times 10^8\,\frac{\text m}{\text s}$
• $\epsilon_0$ $\approx 8.854\times 10^{-12}\,\frac{\text F}{\text m}$
• $\mu_0$ $=\frac{4\pi}{10^7}\,\frac{\text N}{ {\text m}^2}$ $\approx 1.257\times 10^{-6}\,\frac{\text N}{ {\text m}^2}$

Steps:

1. Find $E_\text{rms}$ and $B_\text{rms}$ from the average intensity:
• $I_\text{av}=u_\text{av}c$ $=\epsilon_0cE_\text{rms}^2$ $=\frac{1}{\mu_0}cB_\text{rms}^2$
• $E_\text{rms}=\sqrt{\frac{I_\text{av}}{c\epsilon_0}}$
• $B_\text{rms}=\sqrt{\frac{\mu_0 I_\text{av}}{c}}$
2. Find $E_\text{max}$ and $B_\text{max}$ by multiplying the above values by $\sqrt{2}\approx 1.41$
• $E_\text{max}=\sqrt{2}\sqrt{\frac{I_\text{av}}{c\epsilon_0}}$ $= 868\,\frac{\text V}{\text m}$
• $B_\text{max}=\sqrt{2}\sqrt{\frac{\mu_0I_\text{av}}{c}}$ $=2.9\times 10^{-6}\,$T

1. threshold of pain, isn't it $I=1\,\frac{\text W}{\,{\text m}^2}$?
• Answer: for the sound (audio), it is $1\,\frac{\text W}{\,{\text m}^2}$. But for the visible light, I (unofficially) take it to be the intensity of bright sunlight - it is quite painful to stare directly into the sun. After some absorption in the atmosphere, the intensity of bright sun on Earth is about $1000\,\frac{\text W}{\,{\text m}^2}$. — Nicholas Kuzma 2014/04/19 21:45
• Correction: In the previous version of the notes and this wiki example, the threshold of pain for the sound was used by mistake instead of the threshold of pain for the light ($1000\frac{\text W}{\,{\text m}^2}$). Please take a note of the corrected answers!

### Problem 3.3 Assume the solar sail density is comparable to that of diamond $\big(\rho=3500\,\frac{\text{kg}}{\,{\text m}^3}\!\big)$, and that it completely reflects all incident sunlight. Find the maximum thickness of the sail that can be totally supported by the sunlight against the Sun's gravity

(Hint: both the sunlight intensity and gravity drop off with the square of the distance. Therefore, you can pick any position to do your calculation, such as the Earth orbit.
Hint #2: pick any area of the sail, such as 1 m2. In the end, it should cancel out.)

Equations and values:

• Momentum
• $p=\frac{U}{c}$
• Pressure exerted by the completely absorbed light
• $P_\text{av}({\text{absorption}}) = \frac{I_\text{av}}{c}$
• Pressure exerted by the completely reflected light
• $P_\text{av}({\text{reflection}}) = 2\!\frac{I_\text{av}}{c}$
• Average intensity of sunlight at Earth orbit:
• $I_\text{av}$ $=1.36\times 10^3\frac{\text W}{\,{\text m}^2}$

Steps:

1. Acceleration due to the gravity of the Sun at the location of Earth
• $g_\text{sun}$ $= \frac{v^2}{R}$ $= \omega^2R$ $= \left(\frac{2\pi}{T}\right)^2R$ $= \left(\frac{2\pi}{1\,{\text{year}}}\right)^2R$ $= \left(\frac{2\pi}{365.25\cdot 24\cdot 3600\,{\text s}}\right)^2\!\cdot\!1.5\times 10^{11}\,$m $=0.006\,\frac{\text m}{\,{\text s}^2}$
• Here $\omega= \frac{2\pi}{T}= 2\times 10^{-7}\frac{\text{rad}}{\text s}\;$ is the angular velocity of the Earth around the Sun
• $T$ is the period of the Earth (1 year) in seconds
• $R$ is the radius of Earth's orbit (= distance to the Sun, 93.3 million miles, $1.5\times 10^{11}\,$m)
2. Force due to the pressure of sunlight on the area of the sail of $A\!=\!1\,{\text m}^2$:
• $F=P_\text{av}({\text{reflection}})\!\cdot\!A$ $=2\!\frac{I_\text{av}}{c}\!\cdot\!A$ $=2\!\frac{1.36\times 10^3\frac{\text W}{\,{\text m}^2}}{3\times 10^8\frac{\text m}{\text s}}\!\cdot\!1\,{\text m}^2$ $=9\times 10^{-6}\,$N
3. Balance of forces (gravity vs sunlight):
• $mg_\text{sun}=F$
4. Mass of 1 m2 of the sail:
• $m=\frac{F}{\;g_\text{sun}}=\frac{9\times 10^{-6}\,{\text N}}{0.006\,\frac{\text m}{\,{\text s}^2}}$ = 0.0015 kg
5. Density of diamond
• $\rho= 3.5\,\frac{\text g}{\,{\text{cm}}^3}$ $=3500\,\frac{\text{kg}}{\,{\text m}^3}$
6. Volume of 1 m2 of the sail
• $V=\frac{m}{\rho}=4.3\times 10^{-7}\,{\text m}^3$
7. The thickness:
• $d=\frac{V}{A}=4.3\times 10^{-7}\,{\text m}$ $=0.43\,\mu$m

### Problem 3.4

Unpolarized light of average intensity $I_0$ is passed (normally = perpendicular to) through two polarizers (one after another) with polarization axes at an angle $\alpha$ relative to each other. Find the average intensity $I_2$ of the transmitted light.

Steps:

1. Since the first beam is unpolarized the average intensity after the 1st polarizer is:
• $I_1 = \frac{1}{2}I_0$
2. The general trigonometry formula of $\cos^2\alpha$:
• $\cos^2 \alpha$ $=\frac{1}{2}\big(1+\cos(2\alpha)\big)$
3. After passing through the first polarizer, the light becomes polarized with the electric field along the axis of the 1st polarizer:
• $I_2 = I_1\cos^2 \alpha$ $=\frac{1}{2}I_0\cos^2\alpha$ $=\big(\frac{1}{4}+\frac{1}{4}\cos 2\alpha\big)I_0$
• SI units: $\frac{\text W}{\,{\text m}^2}$

### Problem 3.5

Unpolarized light of average intensity $I_0$ is passed (normally) through polarizers 1 and 3 that have their polarization axes perpendicular $\left(\frac{\pi}{2}\right)$ relative to each other. Find the average intensity $I_3$ of the transmitted light. Then, polarizer 2 is inserted between polarizers 1 and 3. The polarization axis of the inserted polarizer 2 is at an angle $\beta$ relative to the axis of the first polarizer. Find the average intensity $I_3'$ in that case as a function of $\beta$.

Steps:

1. After passing the 1st polarizer:
• $I_1 = \frac{1}{2}I_0$
2. With just polarizers 1 and 3:
• $I_3=I_1\big(\!\cos\frac{\pi}{2}\!\big)^2$ $=0$
3. With the 2nd polarizer inserted between polarizers 1 and 3:
• $I_2 = I_1\cos^2\beta$
• $I_3' = I_2\cos^2\big(\!\frac{\pi}{2}-\beta\big)$ $=I_2\sin^2\beta$
• Put it all together
• $I_3' = I_0\frac{1}{2}\cos^2\beta\sin^2\beta$ $=\frac{1}{8}I_0\sin^22\beta$ $=\frac{1}{16}\!\big(1-\cos4\beta\big)I_0$
• So, for example, $I_3'$ is maximized at $\beta=\frac{\pi}{4}$
• $I_3'=\frac{1}{8}I_0$

## Homework Questions Ch.25

### Problem 50

Electromagnetic wave 1 has a maximum electric field of $E_0=52\,\frac{\text V}{\text m}$, and electromagnetic wave 2 has a maximum magnetic field of $B_0=1.5\,\mu$T… Calculate the intensity of each wave.

• Question: For parts B and C, I am having trouble calculating the average intensity.
• Part B: $\;\;I=c\cdot\epsilon_0\cdot E^2$ $\Rightarrow 3\times 10^8\cdot 8.85\times 10^{-12}\cdot 52^2=7.179\,\frac{\text W}{\,{\text m}^2}$
• Part C: same calculations, but using 600 V/m for $E$, my answer is $956\,\frac{\text W}{\,{\text m}^2}$
• Mastering Physics says this is incorrect, where is my problem?
• Answer: I can spot two potential pitfalls:
• If they are asking to find the average intensity, you need to use rms values for the fields, but the given numbers are maximum values. You need to divide them by $\sqrt{2}\approx 1.41$ first, or divide the intensity by 2.
• I am not sure how you converted $\;B_0\!=\!1.5\,\mu{\text T}\;$ to $\;E_0\!=\!600\,\frac{\text V}{\text m}$. Multiplying it by the speed of light should give $450\,\frac{\text V}{\text m}\;$ (unless you had a different number for the magnetic field on the website). — Nicholas Kuzma 2014/04/18 23:29 