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Ch.25: Electromagnetic waves (Apr 14-16)

Concepts Ch.25

After the 1st exam

Units Ch.25

Non-SI units

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 $\mu_0$ $=\frac{4\pi}{10^7}\,\frac{\text N}{ {\text m}^2}\approx 1.257\times 10^{-6}\,\frac{\text N}{ {\text m}^2}$

Lecture notes Apr 15,24

Equation Sheet Ch.25

  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

Questions about Pr. 3.2

  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

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

Question: This problem is very confusing where is this concept explained in the book?

  • Answer: The part about how to calculate the light pressure is explained on page 888 in my book (See Exercise 25-4 just before the section 25-5 “Polarization”).
    • Once you calculate the light pressure, you can calculate the light force on the sail simply by multiplying the pressure by the area.
    • Finally, use your knowledge of gravity from the Fall term to equate this light force with the gravity force.
    • Since both forces drop off as the square of the distance to the sun, you can use any location in the Solar system to perform your balance of forces. The Earth's location is as good as any.
    • I hope this helps, — Prof. Nicholas Kuzma 2015/04/19 23:21

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: $\frac{\text W}{\,{\text m}^2}$

video 2.3: How do polarized sunglasses work?

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 25.16

What distance, d, must separate Galileo and his assistant in order for the human reaction time, $\Delta t = 0.2$ s, to introduce no more than a $16\%$ error in the speed of light?

  • Question: I am having a very hard time with the math in this problem and I can not solve it. Someone, please explain what equation to use, and how to use it.
  • Answer: A combination of unknown distance and percentage error is a little confusing, I agree. You need to use the error calculation method normally used in the labs part of the class:
    • The idea is, one guy is shining light at the mirror held by the other, and is trying to time the delay at which the reflection arrives.
      • The speed of light they get is $c=\frac{2L}{t}$, where
        • $L$ is the one-way distance between them (multiplied by 2 because the light does the round trip)
        • $t$ is the time delay
      • Assuming the distance is measured without error, the relative error of the speed of light measurement is
        • $\frac{\Delta c}{c}=\frac{\Delta t}{t}$ (because for products and ratios the relative (i.e. percentage) errors add up)
      • We know $\frac{\Delta c}{c}=0.16$ (sixteen percent)
      • We also know $\Delta t=0.2$ s
    • Now it is straightforward to calculate $t$ from these data, and then convert $t$ to $L$ using the round-trip equation
  • I hope this helps! — Prof. Nicholas Kuzma 2015/04/19 23:19

Problem 25.50

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

Problem 25.66

Each pulse produced by an argon-fluoride excimer laser used in PRK and LASIK ophthalmic surgery lasts only 10.0 ns but delivers an energy of 2.50 mJ

Part A

What is the power produced during each pulse?

  • What is your question?.. Also, please mention the problem number next time!
    • If you need a hint, here it is:
      • What is the definition of power in terms of energy and time?Prof. Nicholas Kuzma 2015/04/16 22:55

Problem 25.81

A helium-neon laser emits a beam of unpolarized light that passes through three Polaroid filters, as shown in the figure. (See Figure 1.) Suppose that filter 3 is at a general angle θ with the vertical.

  • Part A: Find an expression for the transmitted intensity as a function of θ.
  • Part B: Plot your result from part (A), and determine the maximum transmitted intensity
  • Question: For this question I am able to find the correct answer for part A: $0.375\cdot I\cos^2(\theta-30.0^\circ)$. However I am getting confused when we begin using graphs. I am able to plot a rough graph by plugging in various random angle values for θ, but is this the correct way to be doing this? Is there an easier way?
  • Answer: There are many different ways, and none of them are that “easy”.
    • Basically, you need to:
      1. Generate a list of values for polarization for some values of θ (usually on a grid of equally-spaced θ values).
      2. Input those values into a plotting software, or use pen and graphing paper
        • some software can accept the function directly, and automatically generate the grid values and the corresponding function values
      3. Evaluate your plot and sometimes change the axis range and/or scale, to make sure all the essential features are captured
    • Examples:
      1. Type (or copy-paste) the following expression into http://wolframalpha.com , or click here:
        • plot 0.375 cos(theta-30 deg)^2 from 0 to 90 deg
      2. in Microsoft Excel (or Apple Numbers),
        • generate a column or row of θ values, e.g. 0, 1, 2, 3, … 90
          • Type 1 in cell A1, 2 in cell A2 or B1, then select both cells and “drag” the corner of that block down or to the right
        • generate a parallel column or row of the polarization values
          • use a formula (in Excel, by typing copy-pasting =0.375*cos((A1-30)*PI()/180)^2
          • drag that formula parallel to the θ column or row
        • Use scatter-plot chart to make the graph

chapter_25.txt · Last modified: 2015/05/26 00:20 by wikimanager