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Please check D2L.Prof. Nicholas Kuzma 2015/06/16 10:23

Exam conflict alternate location

  • The early final exam takers (those who have a conflict with Ochem on the exam day) are meeting at
    • SRTC 139D from 09:30-11:30.
    • Please arrive no later than 9:40. Exam will end 11:30.
    • A proctor will be there to answer your questions.
  • All the rest of the students (those who do not have a conflict with another exam) will meet at 10:10
    • at Hoffmann Hall 109.
    • Exam will be given from 10:15 to 12:05

The CLASS survey

The PSU physics department is committed to collecting real data about how effective our teaching is.

  • Upon completion of this survey, your worst quiz grade will be replaced with 1.
  • The survey is about your personal beliefs, there is no right or wrong answer and neither I nor any other instructor will know how you responded. I will only know whether or not you completed the survey (so you can get credit) and the average results of the entire class. The survey should be filled out between our last class and the final exam.
  • DO NOT fill it out on a smart phone they often fail to submit properly. If you take it twice we will only keep your first response unless it is incomplete. The entire survey should take 20-30 minutes and must be completed in one sitting. Plan well because you have a short window of opportunity to meet this requirement. Please respond quickly (do not ponder the questions) according to your first impressions. Remember to respond how you believe, don't even think about how others might respond
  • Here is the link to the survey:

Ch.13,14,25-27 moved

Previous chapters have been moved to the sidebar:

Ch.28: Interference and Diffraction (May 19-21)

Concepts Ch.28

Units Ch.28

  • Micron ($\mu$m) = $10^{-6}$ m
  • (Arc)minute = (1/60)$^\circ$
  • (Arc)second = (1/3600)$^\circ$

Math Ch.28

$\cos\alpha+\cos\beta=$ $2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)$
$\cos\alpha=\cos(-\alpha)$ $\sqrt{1+\epsilon}\approx$ $1\!+\!\frac{1}{2}\!\epsilon$
$\sqrt{1-\epsilon}\approx$ $1\!-\!\frac{1}{2}\!\epsilon$
For small $|\epsilon|\ll 1$
$\tan[\arcsin(x)] = \frac{x}{\sqrt{1-x^2}}$
$\sin\alpha=-\sin(-\alpha)$ $\tan[\arccos(x)] = \frac{\sqrt{1-x^2}}{x}$

Lecture notes May 19,21

Equation Sheet Ch.28

  1. Conditions for bright fringes (constructive interference) in a two-slit experiment and gratings:
    • $d\sin\theta = m\lambda$
      • $d$ is the slit separation
      • $m=0,\,\pm 1,\,\pm 2,\,\pm 3,\,\ldots$
      • $m\!=\!0$ occurs at $\theta\!=\!0$, this is the central bright fringe
      • Positive values of $m$ are above the central bright fringe, negative values are below
    • Solving for $\theta$:
      • $\theta =$ $\arcsin\left(m\frac{\lambda}{d}\right)$
  2. Conditions for dark fringes (destructive interference) in a two-slit experiment:
    • $d\sin\theta =$ $\left(m-\frac{1}{2}\right)\lambda$
      • $m = 1,\,2,\,3\,\ldots$ (above the central bright fringe)
    • $d\sin\theta =$ $\left(m+\frac{1}{2}\right)\lambda$
      • $m = -1,\,-2,\,-3\,\ldots$ (below the central bright fringe)
    • Solving for $\theta$:
      • $\theta =$ $\arcsin\left[\big(m \pm \frac{1}{2}\big)\!\frac{\lambda}{d}\right]$
        • “$+$” or “$-$” depending on location: above or below the central fringe
  3. Linear distance from the central fringe:
    • $y = L \tan\theta$.
      • L is the distance to the screen.
      • Solving for $\theta$ of a bright fringe:
        • $\theta = \arctan\left(m\frac{y}{L}\right)$
      • Solving for $\lambda$:
        • $\lambda = \frac{d}{m}\sin\theta$
      • Solving for $\theta$ of a dark fringe:
        • $\theta = \arcsin\left[\big(m\pm\frac{1}{2}\big)\!\frac{\lambda}{d}\right]$.
          • “$+$” or “$-$” depending on location: above or below the central fringe
  4. Conditions for dark fringes in single-slit diffraction:
    • $w\sin\theta = m\lambda$.
      • $m = \pm 1,\,\pm 2,\,\pm 3\,\ldots$
      • $w$ is the slit width
      • Solving for $\lambda$:
        • $\lambda = \frac{W\sin\theta}{m}$
      • Solving for $\theta$:
        • $\theta = \sin^{-1}(\frac{m\lambda}{W})$
      • approximate angular width of central bright fringe=$2\frac{\lambda}{W}$
  5. First Dark Fringe for the Diffraction Pattern of a Circular Opening:
    • $\sin\theta = 1.22\frac{\lambda}{D}$
  6. Rayleigh's Criterion:
    • $\theta_\text{min} = 1.22\frac{\lambda}{D}$
    • Note: $\lambda$ is dependent on the diffraction of the material that the light is traveling through. If the diffraction is $n, \lambda$ becomes $\frac{\lambda}{n}$
  7. Constructive Interference in a Diffraction Grating:
    • $d \sin \theta = m\lambda$. Here $m = \pm 1,\pm 2,\pm 3\ldots$
    • Solving for $d$:
      • $d = \frac{m\lambda}{\sin\theta}$

Destructive Interference: Waves cancel

$v = c/n$

Wavelength $\lambda$, of light in a medium of index of refraction $n$ greater than 1: $\lambda_n = \lambda$ (in vacuum)$ / n$

  1. For an air wedge:
    1. Constructive interference:
      • Waves add (in phase)
      • $\frac{1}{2} + \frac{2d}{\lambda} = m$
    2. Destructive interference:
      • Waves cancel (out of phase)
      • $\frac{1}{2} + \frac{2d}{\lambda} = \frac{1}{2} + m$
  • Thin Films:
  • Effective path length for ray 1: Path = $\frac{1}{2}\cdot$wavelength
  • Ray 1 reflects from the air to film interface
  • Ray 2 reflects from the film to air interface.
  • Effective Ray 2 path length / wavelength in film = 2t / wavelength in film = 2nt / wavelength in vacuum
  • t = thickness of film
  • Difference in phase changes = (2nt / wavelength in vacuum) - .5
  • Destructive Interference: 2nt / wavelength in vacuum = m
  • Constructive Interference: (2nt / wavelength in vacuum) - .5 = m

$\;\;m = 0,\, 1,\, 2,\, 3\,\ldots$
Refraction index order
($n_1$ and $n_2$ are on either
side of the film $n_f$)

Dark fringes

Bright fringes
Dissimilar reflections (one extra $\pi$) $n_1\!<\!n_f\!>\!n_2\;$ or $\;n_1\!>\!n_f\!<\!n_2$ $t = \frac{\lambda}{2n_f}m$ $t = \frac{\lambda}{2n_f}\left(m+\frac{1}{2}\right)$
Similar reflections (no extra $\pi$ or 2 extra $\pi$) $n_1\!<\!n_f\!<\!n_2\;$ or $\;n_1\!>\!n_f\!>\!n_2$ $t = \frac{\lambda}{2n_f}\left(m+\frac{1}{2}\right)$ $t = \frac{\lambda}{2n_f}m$

Students' questions Ch.28

I am having some trouble understanding the chart above talking about refraction index order. Can someone explain to me what equation to use for what circumstance? — Joshua Vandehey 2015/05/28 18:29

Yeah - I was really confused about this myself. Pg. 983 of the textbook helped me a little bit. This chart is summarizing what happens when a light wave travels from one medium with a certain refraction index (n) to another refraction index. If the light travels from a region with a higher n to a lower n, then it's reflected back with no phase change. As the chart shows, that wave acts in accordance with bright fringes. If the wave travels to a refraction index with a higher n, then there is a phase change, corresponding to an extra $\frac{1}{2}\lambda$ in path difference, reminiscent of what we see in shift to dark fringes. — AlissaG

Examples Ch.28

Problem 6.1

Two sources emitting the light of equal wavelength, phase, and intensity, are located at positions $s_1$ and $s_2$ as shown, a short distance d apart from each other and a distance L away from the vertical screen. Find the locations of bright and dark spots on the screen in terms of the y coordinate along the screen.


  1. Identify given variables:
    • $d$ is the (vertical) distance between the sources $s_1$ and $s_2$
    • $L$ is the (horizontal) distance from the sources to the screen
  2. Determine relevant equations that relate these variables to the question
    • Bright spots:
      • $d\sin\theta$ $= m\lambda\;$ for $\;m = 0,\, \pm 1,\, \pm 2,\,$ etc
    • Dark spots:
      • $d\sin\theta$ $=\big(m-\frac{1}{2}\big)\lambda\;$ for $\;m = 1,\, 2,\, 3,\,$ etc
      • $d\sin\theta$ $=\big(m+\frac{1}{2}\big)\lambda\;$ for $\;m = -1,\, -2,\, -3,\,$ etc
    • $y = L\tan\theta$
  3. Re-arrange the bright spots equation to solve for $\theta$
    • $d\sin\theta$ $= m\lambda$
    • $\sin\theta$ $=\frac{m\lambda}{d}$
    • $\theta$ = $\arcsin\left(\frac{m\lambda}{d}\right)$
  4. Re-arrange the darks spots equation to solve for $\theta$
    • $d\sin\theta$ $= \left(m\pm\frac{1}{2}\right)\lambda$
    • $\sin\theta$ $=\frac{\left(m\,\pm\,\frac{1}{2}\right)\,\lambda}{d}$
    • $\theta=$ $\arcsin\left(\frac{(m\,\pm\,\frac{1}{2})\,\lambda}{d}\right)$
  5. Insert these expressions for $\theta$ into $\;y = L\tan\theta$
    • Bright spots:
      • $y =$ $L\tan\big[\arcsin\big(\!\frac{m\lambda}{d}\!\big)\big]=$ $\frac{Lm\lambda}{d\sqrt{1-\left(\frac{m\lambda}{d}\right)^2}}=$ $\frac{m\lambda L}{\sqrt{d^2-(m\lambda)^2}}\;$ for $\;m = 0,\, \pm 1,\, \pm 2,\,$ etc
        • Here we used the trig identity
          • $\tan[\arcsin(x)] = \frac{x}{\sqrt{1-x^2}}$
    • Dark Spots
      • $y =$ $L\tan\left[\arcsin\left(\!\frac{(m-\frac{1}{2})\,\lambda}{d}\!\right)\right]=$ $\frac{(m-\frac{1}{2})\,\lambda L}{\sqrt{d^2-\left((m-\frac{1}{2})\,\lambda\right)^2}}\;$ for $\;m = 1,\, 2,\, 3,\,$ etc
      • $y =$ $L\tan\left[\arcsin\left(\!\frac{(m+\frac{1}{2})\,\lambda}{d}\!\right)\right]=$ $\frac{(m+\frac{1}{2})\,\lambda L}{\sqrt{d^2-\left((m+\frac{1}{2})\,\lambda\right)^2}}\;$ for $\;m = -1,\, -2,\, -3,\,$ etc
  6. You can now use these equations to determine the locations (the y values) of any bright/dark spot of order $m$.
    • Note: a given spot exists only if $\big|\sin\theta\,\big|\leq 1$, or, equivalently, the square root in the above expressions takes on a positive argument.

Problem 6.2

Two thin slits, 1 mm apart, are illuminated with $\lambda=750\,$nm light. The screen is $L=15\,$m away. Find the locations of the bright fringes on the screen.


  1. Identify variables and convert to SI units:
    • $d =$ $1\,{\text{mm}} \times \frac{1\,{\text m}}{1000\,{\text{mm}}}$ $= 0.001\,$ m
    • $\lambda=750\,{\text{nm}} \times \frac{1\,{\text m}}{10^9\,{\text{nm}}}$ $= 7.5 \times 10^{-7}\,{\text m}$
    • $L = 15\,$m
  2. Determine relevant equations that relate these variables to the available data
    • $d\sin\theta$ $= m\lambda\;$ for $\;m = 0,\, \pm 1,\, \pm 2,\,$ etc
    • $y = L\tan\theta$
  3. In order to combine the two equations, solve $\,d\sin\theta\!=\!m\lambda\,$ for $\,\theta$
    • $d\sin\theta$ $= m\lambda$
    • $\sin\theta$ $= \frac{m\lambda}{d}$
    • $\theta$ $=\arcsin\big(\!\frac{m\lambda}{d}\!\big)$
  4. Insert this $\theta$ expression into $y = L\tan\theta$
    • $y = L\tan\theta$
    • $y = L\tan\big[\arcsin\big(\!\frac{m\lambda}{d}\!\big)\big]$
  5. Insert the given values and simplify
    • $y = L\tan\big[\arcsin\big(\!\frac{m\lambda}{d}\!\big)\big]$
    • $y = (15\,{\text m})\tan\big[\arcsin\big(m\frac{7.5 \times 10^{-7}\,{\text m}}{0.001\,{\text m}}\big)\big]$
    • $y =$ $(15\,{\text m})\tan\left[\arcsin\left(m\cdot 7.5 \times 10^{-4}\right)\right]$
  6. You can now calculate y for $\;m = 0,\, \pm 1,\, \pm 2,\,$ etc
    • For $m = 0$:
      • $y =$ $(15\,{\text m})\tan\left[\arcsin\left(0\cdot 7.5 \times 10^{-4}\right)\right]$ $= 0$
    • For $m = +1$:
      • $y =$ $(15\,{\text m})\tan\left[\arcsin\left(1\cdot 7.5 \times 10^{-4}\right)\right]$ $= 0.01125\,$m $\approx 1.1\,$cm
    • For $m = -1$:
      • $y =$ $(15\,{\text m})\tan\left[\arcsin\left((-1)\cdot 7.5 \times 10^{-4}\right)\right]$ $= -0.01125\,$m $\approx -1.1\,$cm
    • etc

Problem 6.3

An air wedge is getting wider along $y$, such that $d(y)=\beta\!\cdot\!y$, with $\beta=10^{-5}\,$(rad). Find the spacing of interference produced by yellow ($\lambda=600\,$nm) light along the $y$ axis.


  1. A change in phase difference between the two reflected beams of $2\pi$ radians (360$^\circ$) gives the distance to the next fringe along the y axis.
    • One of the reflections will accrue an extra $\pi$ of phase change (from air into glass and back into air)
    • The bright spots are given by the condition:
      • $\frac{\lambda}{2}+2d$ $=m\lambda\;$ where $\;m=0,\,1,\,2,\,3\ldots$
    • Consider subsequent fringes $m_1$ and $m_2=m_1+1$:
      • $\frac{\lambda}{2}+2d_1$ $=m_1\lambda$
      • $\frac{\lambda}{2}+2d_2$ $=(m_1+1)\lambda$
    • Subtract these two equations:
      • $2(\,d_2-d_1)=$ $(m_1+1)\lambda-m_1\lambda$ $=\lambda$
      • $d_2-d_1=$ $\frac{\lambda}{2}$
  2. Now convert this difference in $d$ to a difference in $y$:
    • $y_2-y_1=$ $\frac{d_2}{\beta}-\frac{d_1}{\beta}=$ $\frac{1}{\beta}(d_2-d_1)=$ $\frac{1}{\beta}\frac{\lambda}{2}=$ $\frac{\lambda}{2\beta}$
  3. Plugging in the numbers to find the y distance between the fringes:
    • $\Delta y=$ $y_2-y_1=$ $\frac{\lambda}{2\beta}=$ $\frac{600\times 10^{-9}\,{\text m}}{2\,\cdot\,10^{-5}}$ $= 0.03\,$ m, or 3 cm.

Problem 6.4

A camera lens ($n_\text{lens}=1.42$) is coated with a thin film ($n_\text{film}=1.55$) to prevent reflections at $\lambda_\text{vac}=600\,$nm. Find the minimum thickness $d$ of the film to achieve this, assuming the (normal) reflections from both surfaces of the film, after emerging into the air, are of equal intensity.


  1. Note that the reflection from the front of the film is of “low n – high n – low n” type, therefore it incurs an extra $\pi$ of phase:
    • air–film–air, $\;n_\text{air}\approx 1$
  2. The reflection from the back of the film is of “high n – low n – high n” type, since $\,n_\text{film}>n_\text{lens}\,$. No phase gain there.
    • film–lens–film
  3. Write the equation for the optical path difference between the two normal reflections for dark-fringe conditions:
    • $\Delta{\text{Path}}=$ $2d+\frac{\lambda_\text{film}}{2}$ $=\big(m+\frac{1}{2}\big)\lambda_\text{film}\;$
      • The term $\frac{\lambda_\text{film}}{2}$ is to account for the extra half-wavelength ($\pi$ of phase difference) due to reflections being of different type
  4. Substitute $m\!=\!1$ for the thinnest film, and express the shortened wavelength in the film in terms of the vacuum value:
    • $2d=$ $1\!\cdot\lambda_\text{film}$ $= \frac{\lambda_\text{vac}}{ n_\text{film} }$
  5. Solve for $d$:
    • $d=$ $\frac{\lambda_\text{vac}}{2\,n_\text{film}}$ $= \frac{600\,{\text{nm}}}{2 \times 1.55 }$ $=194\,{\text{nm}}$

Problem 6.5

Please transcribe the problem from the lecture notes Problem Transcribed!!!Joshua Vandehey 2015/05/28 18:46

Using a caliper with a $\frac{1\ }{1000\ }$ ${\text inch}$ gap, a blue monochromatic light is shown through the gap onto a screen L$=$15m away. Find the linear distance from the central fringe to the first dark fringe on the screen.


  1. Convert slit width to S.I. units:
    • $W$ $=\frac{0.0254\, {\text mm}}{1000\ }$ $= 2.54 \times 10^{-5}$ m
  2. Identify wavelength of the light:
    • $\lambda$ = 450 ${\text nm}$ (blue light)
  3. Find the angle between the central and first dark fringe:
    • $W\sin\theta$ $=\lambda$ (m=1)
    • $\sin\theta$ $= \frac{\lambda}{W}$ $=\frac{4.5\times 10^{-7}}{2.54 \times 10^{-5}}$ $=0.0177$ $=1°$
  4. Find the linear distance from the central fringe to the first dark fringe:
    • $y = L\tan\theta$ $=15 \text m \times \tan 1°$ =0.27m
    • Note that as W goes down, y goes up and vice versa.

Problem 6.6

Find diffraction limit of the angular resolution (that is, the smallest angle between two stars that can be resolved) for a telescope with a 10-cm diameter objective lens. Assume $\lambda$ $= 600\,{\text{nm}}$


  1. Convert units to SI:
    • $\lambda$ $= 600\,{\text{nm}}$ $=6\times 10^{-7}\,$m
    • $D=10\,$cm $=10^{-1}\,$m
  2. Use Rayleigh's criterium to calculate the minimally resolvable angle:
    • $\theta_\text{min}$ $=1.22 \frac{\lambda}{D}$ $=1.22 \times \frac{6 \times 10^{-7} \text m}{ 10^{-1} {\text m}}$ $= 7.3 \times 10^{-6}$ rad $\approx 1.5\,$arcsec

This will resolve a 1.7-mile object on the Moon.

Problem 6.7

A laser of $\lambda=633\,$nm wavelength shines normally through a grating with thin slits every 2$\,\mu$m. Find the angle (from the original beam direction) of the 1st and the 2nd peaks.


  1. Convert to SI units:
    • $633\,{\text{nm}}= 6.33 \times 10^{-7}\,{\text m}$
    • $2\,\mu{\text m}=2\times 10^{-6}\,{\text m}$
  2. 1st peak:
    • $d\sin\theta$ $=\lambda$
    • $\sin\theta$ $= \frac{\lambda}{d}$ $= \frac{6.33 \times 10^{-7}\,{\text m}}{2 \times 10^{-6}\,{\text m}}$ $=0.3165$
    • $\theta$ $=\arcsin(0.3165)$ $=0.322\;{\text{rad}}=$ $18.5^\circ$
  3. 2nd peak:
    • $d\sin\theta$ $=2 \lambda$
    • $\sin\theta$ $= \frac{2\,\cdot\,6.33 \times 10^{-7}\,{\text m}}{2 \times 10^{-6}\,{\text m}}$ $=0.633$
    • $\theta$ $=\arcsin(0.633)$ $=0.685\;{\text{rad}}=$ $39.3^\circ$

HW Questions Ch.28

In mastering physics problem 28.74 the maximum spacing between the lines in this grating using m=2. Why was 2 used for m instead of 1? When blue light with a wavelength of 456nm illuminates a diffraction grating, it produces a first-order principal maximum but no second-order maximum. What is the maximum spacing between lines on this grating? used equation d=m(lambda)/sin(theta)

Problem 28.59

The Hubble Space Telescope (HST) orbits Earth at an altitude of 613 km. It has an objective mirror that is 2.4 m in diameter.If the HST were to look down on Earth's surface (rather than up at the stars), what is the minimum separation of two objects that could be resolved using 560nm light? [Note: The HST is used only for astronomical work, but a (classified) number of similar telescopes are in orbit for spy purposes.]

  • Q: I don't understand why the First Dark Fringe for the Diffraction Pattern of a Circular Opening:
    • $\sin\theta=1.22\,\lambda D$ was used to calculate the minimum separation of the two objects.
  • A: The Raileigh citerion states that when the center of the 2nd object's image overlaps the first dark fringe of the 1st object's image, the two images at that point coalesce and are no longer distinguishable
  • Also, when setting the 2 equations together why does “tangent” drop out of the equation:
    • $y=L\tan\theta_\text{min}=L\,\theta_\text{min}$ $=L(1.22 \lambda/D)$ — MISLAYDI GARCIA OLLET 2015/06/02 20:23
  • A: For such small angles (if measured in radians), a small-angle approximation is valid:
    • $\sin\theta\approx\tan\theta\approx\theta$

An image of the Hubble Space Telescop, wanted to share!

Problem 28.36

Problem 7 on mastering physics
Monochromatic light with $\lambda=654$ nm shines down on a plano-convex lens lying on a piece of plate glass, as shown in the figure. When viewed from above, one sees a set of concentric dark and bright fringes, referred to as Newton's rings. If the radius of the twelfth dark ring from the center is measured to be 1.47 cm , what is the radius of curvature, $R$, of the lens?

  • Q: In the homework solutions it says that $R-d$ is the length of the left side of the triangle. Why? I don't understand how they got that.
    Diana King 2015/05/25 22:25
  • A: See the diagram, I added extra $R$ and extra $d$ so it is more clear.
    • Remember, any line connecting the center to any point on the circle or sphere is of lenght $R$.
      Prof. Nicholas Kuzma 2015/06/03 00:00

Problem 28.60

A lens that is “optically perfect” is still limited by diffraction effects. Suppose a lens has a diameter of 120 mm and a focal length of 640 mm. Find the angular width (that is, the angle from the bottom to the top) of the central maximum in the diffraction pattern formed by this lens when illuminated with 540 nm light.

  • Q: I keep getting $5.59\times 10^{-6}$ radians as the angular width, which is not the right answer. Where did I go wrong? — Joshua Vandehey 2015/05/28 18:18
    • A The question is a little bit tricky. It says “Find the angular width (that is, the angle from the bottom to the top) of the central maximum…” I believe you need to multiply what you got by 2 to get the angle from the bottom to the top of the central maximum. That should give you the correct answer. — Angelique Vasquez 2015/06/02 11:01
    • Comment: I guess bottom to top implies that the light travels horizontally — Prof. Nicholas Kuzma 2015/06/03 00:04

Problem 28.65

A diffraction grating has 2200 lines/cm. What is the angle between the first-order maxima for red light (λ=680 nm) and blue light (λ=410 nm)?

  • Q: The equation is $\theta=\arcsin\left(m\frac{\lambda}{d}\right) = \arcsin (mN \lambda)$.
    • Looking at the calculation $\theta_\text{red}=$ $\arcsin(1\cdot 2200/{\text{cm}}\; 680\,{\text{nm}})=$ $8.6^\circ$, I would think $N$ and $\lambda$ would need to be converted to meters before being multiplied but that gives $(8.6\times 10^{-4})^\circ$. Instead the units are left in cm to get 8.6°. Why is this?
  • A: I don't agree with your conclusion. You do need to convert the length units to the same units (e.g. m) in order to get 8.6°:

Problem 28.74

When blue light with a wavelength of 456 nm illuminates a diffraction grating, it produces a first-order principal maximum but no second-order maximum. Explain the absence of higher order principal maxima.

  • Q: I am confused by this question on the assigned homework. Even after reading the explanation on the “Homework Solutions”, it still did not quite make sense to me. Why do we assume a 90 degree angle in order to calculate the first order maxima and how do we determine why a different angle does not work for a second order maxima? Additionally, aren't the maxima also dependent on the slit-width?
    Elin Odegard 2015/06/02 11:23
  • A1: We talked about it in class: the equation gives you $\sin\theta_n$ for various orders. But there is an additional constraint: the sine should be less than 1, that is, $\sin\theta_n \le 1$. In other words, the optical path difference cannot exceed the grating spacing no matter what angle you try. The highest possible $\sin\theta_n$ can be 1, which corresponds to the 90 degrees angle.
  • A2: The width and the brightness of each fringe might depend on the slit widths, but not the fringe positions or spacings
    Prof. Nicholas Kuzma 2015/06/03 00:20

Problem 28.26

When green light (λ = 565nm ) passes through a pair of double slits, the interference pattern (a) shown in the figure(Figure 1) is observed. When light of a different color passes through the same pair of slits, the pattern (b) is observed. Part C: Find the wavelength of the second color. (Assume that the angles involved are small enough to set sinθ≈tanθ.)

I'm having trouble with this part of the question and when I looked up how to do it the answer key said to use the fifth order minima. I didn't follow how they got to this point. Why do you use the fifth order rather than the dsinθ=mλ equation?

I believe you need to use m=5 because there are 5 bright spots (green dots) after the central green dot. Given the interval (the dotted lines), you have to subtract 1/2 from the m value in order to get the correct dsinθ to use the find λ of the second pattern — Ryan Stoner 2015/06/09 09:43 -

Ch.29: Relativity (May 26-28)

Hello Dr. Kuzma, here is a link to that video you discussed in class. Enjoy!

Thomas Ruttger 2015/05/28 12:43

  • Looks like they glossed over the fact that the person on the airplane will be actually older, because of speeding up of his/her time due to the gravity effect.
    • This is sort of the worst of the popular science - nice cinematography but zero physics. And they got the sign of the effect wrong.
      Prof. Nicholas Kuzma 2015/06/06 13:12

Concepts Ch.29

Black holes! This a pretty interesting article about a huge telescope and how black holes are detected. — Ryan Stoner 2015/06/09 09:49

Constants Ch.29

  • Speed of light $c=299792458\frac{\text m}{\text s}$ $\approx 3.00\times 10^8\frac{\text m}{\text s}$
  • Gravitational constant $G=6.67384\times 10^{-11}\frac{ {\text N}\cdot{\text m}^2}{ {\text{kg}^2}}$

Units Ch.29

Math Ch.29

Rotations Relativity
$\cos\left[\arctan\left(\frac{v}{c}\right)\right]$ $=\frac{1}{\sqrt{1+\left(\frac{v}{c}\right)^2}}$ $\cosh\left[{\text{arctanh}}\left(\frac{v}{c}\right)\right]$ $=\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$
$\sin\left[\arctan\left(\frac{v}{c}\right)\right]$ $=\frac{v/c}{\sqrt{1+\left(\frac{v}{c}\right)^2}}$ $\sinh\left[{\text{arctanh}}\left(\frac{v}{c}\right)\right]$ $=\frac{v/c}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$

Lecture notes May 26,28

Equation Sheet Ch.29

Please update if an equation is not included

  1. The laws of physics are the same in all inertial frames of reference
  2. The speed of light in a vacuum is the same in all inertial frames of reference
  3. The time it takes light to travel a distance 2D is simply $\frac{2D}{c}$. This is proper time, or the time elapsed within the same spatial reference.
  4. To the outside observer to a spatial reference system, the time elapsed for them as an object travels 2D at velocity $v$ is
    • $\Delta t$ =$\frac{\Delta t_0}{\sqrt{1-\frac{v^2}{c^2}}}$
      • $\Delta t_0$ is the Proper time
    • time dilation applies to all physical processes (chemical and biological)
  5. Proper length is the length of an object as measured by an observer within its spatial reference system
  6. To the outside observer to a spatial reference system, Contracted length of a object moving at velocity $v$ in the direction of relative motion
    • $L=L_0\times\sqrt{1-\frac{v^2}{c^2}}$
      • $L_0$ is the Proper length
      • $L$ is the Contracted length
  7. Relativistic addition of velocities:
    • $v=\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}$
      • Here $v$ is the velocity of object 2 relative to some frame
      • $v_1$ is the velocity of object 1 relative to the same frame
      • $v_2$ is the velocity of object 2 relative to object 1
  8. Relativistic subtraction of velocities:
    • $v=\frac{v_2-v_1}{1-\frac{v_1v_2}{c^2}}$
      • Here $v$ is the relative velocity of object 2 relative object 1
      • $v_1$ is the velocity of object 1 relative to some frame
      • $v_2$ is the velocity of object 2 relative to the same frame
  9. Relativistic momentum
    • $p=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}$
      • Here $m_0$ is the rest mass
  10. The traditional momentum equation $p = mv$ can be used if we use proper mass $m$:
    • $m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$
  11. Relativistic energy, $E = mc^2=\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}$
  12. Rest energy, $E_0 = m_0c^2$
    • $m_0$ is the rest mass
  13. Relativistic Kinetic energy, $K = \frac{m_0c^2}{\sqrt{1 -\frac{v^2}{c^2}}}-m_0c^2$
  14. Schwarzschild radius or the radius of a black hole, $R$
    • $R = \frac{2GM}{c^2}$
      • $G=$ $6.67\times {10^{-11}}\frac{ {\text N}\cdot{\text m}^2}{ {\text{kg}}^2}\;$ is the gravitational constant
      • $M$ is the mass of an astronomical body
  • $v = c\sqrt{1-(\frac{\Delta t_0}{\Delta t})^2}$

-Professor, regarding the above equation would it be used to calculate the velocity of differing bodies moving in different time frames? If you could clarify its' usefulness and maybe a short example? Thank you -Nichole K

Students' questions Ch.29

  • Hello Dr. Kuzma, I have a question regarding the speed of light in the theory of relativity. We are told that the speed of light does not change in relation to an observer. Does this mean that if a person is moving at the speed of light and shines another light forward, that light will still be propagating forward at the speed of the observer? It seems that the speed of light is a “maximum” speed for all real objects. Can you help me understand this further?
    Thomas Ruttger 2015/05/28 10:25
    • Turns out, only “massless particles” such as photons (i.e. radio waves, light, X-rays, gamma rays), gravitons (gravity waves), possibly neutrinos can propagate exactly with the speed of light. The catch is, they cannot propagate at any other speed. On the other hand, “massive particles”, i.e. electrons, protons, neutrons, atoms, muons, etc., that can move at any (slow) speed, require infinite kinetic energy to reach the speed of light (as seen from the $E_\text{kin}=m_0c^2\left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right)$ when $v=c$). So, essentially, only light and similar radiation alone can move at the speed of light, everything else has to be just a tad slower.
      Prof. Nicholas Kuzma 2015/06/06 23:25
  • Relating to Relativistic Energy and Work: When I was working on the homework problems, I came across one in which we were required to determine the amount of work performed on a system, however the answer was dependent solely on energy. This might be more of a review concept. But is the relativistic energy and the work performed interchangeable in any system? Could you elaborate on this? Thank you!
    Elin Odegard 2015/06/02 11:32
    • In terms of work, I think of it as kinetic energy. In wikipedia page it says “regardless of the choice of reference frame, the work energy theorem remains valid and the work done on the object is equal to the change in kinetic energy.” So the amount of work performed on a system can be calculated by the relativistic kinetic energy formula which says $K=$ Energy $- m_0c^2$, where energy is the relativistic energy formula. That is how I approached the problem. So relativistic energy and work are not directly interchangeable. You need to account for the difference of $m_0c^2$, in order to convert from one form to another.
      Aishwarya Shimpi 6/6/2015
    • Yes, in the more formal treatment of relativistic mechanics, the basic notions of mechanics are converted to 4-quantities, for example space (x,y,z) and time (t) become a 4-coordinate (ct,x,y,z) in the space-time, similarly momentum ($p_x,p_y,p_z$) and kinetic energy ($E$) become “4-momentum” ($p_t,p_x,p_y,p_z$) where $p_t$ is essentially $E/c$. I believe work is still defined as the sum (or integral) of $F\Delta x$, but the definition of force is no longer $ma$, rather it is $\Delta p/\Delta t$, where $p$ is the relativistic momentum. So the only practical way of calculating the work is by using energy conservation and the known changes in kinetic energy
      Prof. Nicholas Kuzma 2015/06/06 23:25
  • Dr. Kuzma, I was reviewing chapter 29 material and was reading about time and time dilation. I guess I am just a bit confused between the actual meaning and differences of $\Delta t_0$ and $\Delta t$. Is $\Delta t_0$ always what is measured/observed on earth while $\Delta t$ is what is observed when the observer is moving at a certain speed? Thank you! — Nichole K.
    • Nichole, I believe that $\Delta t_0$ is the proper time within a spatial reference. So it depends on where you are/what you are measuring. Say someone is in a spaceship traveling to a far away destination, if you want to know the age of the passenger upon reaching his/her destination according to the spaceship's clock, you would be solving for delta t0 and delta t would be equivalent to the time passed on earth. Hope this helps, maybe someone can explain it more clearly.
      Angelique Vasquez 2015/06/05 10:04
    • Yes, $\Delta t_0$ is proper time, defined as the time measured in a reference frame with a clock that is not moving in that frame
      • Examples of the proper time $\Delta t_0$ :
        1. Time on a spaceship measured by the spaceship clock
        2. Time on Earth measured by the Earth clock
      • Examples of $\Delta t$:
        1. Time on a spaceship measured by the Earth clock
        2. Time on Earth measured by a spaceship clock — Prof. Nicholas Kuzma 2015/06/06 23:25
  • I am a bit confused on when we need to use the relativistic addition of velocities vs. the relativistic subtraction of velocities. Could anyone elaborate on the differences between the two and what sorts of circumstances each one is used for. Thanks.
    Angelique Vasquez 2015/06/05 10:04
    • Hi Angelique, I think what you are mentioning here has to do with the velocities being either positive or negative. If an object is traveling forward in respect to the frame of reference, you can consider it to have a positive velocity and therefore the relativistic addition of velocities occurs whereas if the object is travelling the opposite direction in respect to the frame of reference then you would consider it to have a negative velocity and therefore use the relativistic subtraction of velocities. I found a very helpful video that can elaborate more on this type of problem. Here is the link:
      Aishwarya Shimpi
    • Yes, it depends on
      1. sign of velocities (i.e. direction of motion)
      2. whether we are looking for the relative velocity of the two systems or know the relative velocity and looking for the velocity of one of the systems relative to the base frame.
        • Here are all the possibilities:
          • Frames A and B relative to Earth E move in the same direction, A is moving slower than B
            • Knowing A and B relative to E, find B relative to A
              • use subtraction
            • Knowing A relative to E, and B relative to A, find B relative to E
              • use addition
          • Frames A and B relative to Earth E move in opposite directions
            • Knowing A and B relative to E, find A relative to B
              • use addition of the magnitudes of velocities
            • Knowing A relative to E, and B relative to A (in opposite direction), find B relative to E
  • Dr. Kuzma, on the practice exam, Problem 3, part F) is confusing me. The answer key shows that the relative momentum equation is used yet the correct answer is found by using $p=mv$ only. Why don't you divide the $mv$ by the $sqrt{1-(v^2/c^2)}$? Thanks,
    Thomas Ruttger 2015/06/06 15:14
  • Dr. Kuzma, can you explain the reasons why each assumption is valid in problem 3 part H) of the practice exam. I understand the math but I am not sure under what conditions it is okay to use classical conditions over relativist conditions. Thanks,
    Thomas Ruttger 2015/06/06 15:22

We were a bit rushed when reviewing the last quiz question in class, would you be able to post it on online? I think I remember you mentioning that the answer was 1 sec was that because it was speed of light or perpendicular vs parallel? If that variable had not been the case would it have been the 0.866 answer?

Dr. Kuzma, I was reviewing the relativistic addition of velocities and I am still having trouble understanding choosing the appropriate velocities for the variables in the relativistic velocities equation. Could you give me some tips? Thank you! — Dang Nguyen 2015/06/08 18:27

Think of it this way: There are 3 objects. Object 1 is behind object 2 which is behind object 3. Object 1 and 2 are spaceships, while object 3 is the earth. Both object 1 and 2 are moving in the same direction reaching the earth. V23 is the velocity of object 2 in object 3 frame (earth frame). V13 is the velocity of object 1 in object 3 frame (earth frame). V21 is the velocity of object 2 in object's 1 frame. Now just plug and chug in the relativistic addition formula. This also applies to the relativistic subtraction. Be sure to change the signs when they move in different directions however. — Aish Shimpi 2015/06/08 20:48

Here is how I think about this: velocities with numbered subscripts are those with the same frame of reference, the one denoted as simply v is the one with a frame of reference different from the numbered one. All you have left to do, is to decide whether objects in the same frame of reference are moving towards or away from each other (to use plus or minus sign in the equation). Hope this makes sense — Yulia Kotlyar 2015/06/08 19:50

Examples Ch.29

Problem 7.1

How far does the light travel in 1 year? 1 minute? 1 s? 1 ms? 1 $\mu$s? 1 ns?


  1. Convert the given times to SI units:
    • $3.154\times 10^7\,$s, $60\,$s, $1\,$s, $10^{-3}\,$s, $10^{-6}\,$s, $10^{-9}\,$s.
  2. Use the simple formula for distance in terms of speed and time:
    • $x=c\,t$ $=2.998\times 10^8\,\frac{\text m}{\text s}\cdot t$
  3. These distances are SI equivalents of the “light” units for distance:
Distance unit SI equivalent (m) Example
1 light-year $9.5\times 10^{15}$ $\sim\frac{1}{4}$ of the distance to the star nearest to the Sun
1 light-minute $1.8\times 10^{10}$ $\sim\frac{1}{8}$ of the Earth–Sun distance
1 light-second $2.998\times 10^{8}$ $\sim 75\%$ of the Earth–Moon distance
1 light ms $2.998\times 10^{5}$ $186\,$miles: just beyond Seattle (from Portland)
1 light $\mu$s $2.998\times 10^{2}$ $\sim 1000\,$ft from Cramer Hall to Phys. department and back
1 light ns $2.998\times 10^{-1}$ $\sim 1\,$ft about the length of a page of paper

Problem 7.2

At a speed $\;v=0.99\,c\;$ one of the two twins travels to Alpha Centauri, the nearest star (system) to our Sun, which is “only” 4.3 light-years away from us. How much older or younger is the space-traveling twin compared to the earth-bound one upon the return of the spaceship?


  1. In problems like this it is convenient to measure distances in light-years, time in years, and velocities in the units of c (the speed of light). That way all the equations work just fine, and there is no need to convert the units.
  2. The time it takes for the ship to fly out to $\alpha$-Centauri and back, measured from Earth, is
    • $t=\frac{2d}{v}=$ $\frac{2\,\cdot\,4.3\,c\,(1\,{\text{year}})}{0.990\,c}$ $=4.34\,{\text{years}}\cdot 2=8.68\,$years
    • Here we do not take into account the time spent on the distant star, because during that time, both twins age at the same rate
  3. In the ship frame, the time of travel $\Delta t_0$ is…
    • Use the time dilation equation
      • $\Delta t$ =$\frac{\Delta t_0}{\sqrt{1-\frac{v^2}{c^2}}}$
    • Substitute the known $\Delta t$ and solve for $\Delta t_0$:
      • $\Delta t_0= \Delta t\sqrt{1-\frac{v^2}{c^2}}=$ $8.68\,{\text{years}} \sqrt{1-\frac{(0.990\,c)^2}{c^2}}$ $=8.68\,{\text{years}} \sqrt{1-0.990^2}$ $= 1.22\,$years
  4. The difference in the twins' biological age upon return is
    • $8.688\,{\text{years}} - 1.228\,{\text{years}} = 7.46\,$years difference in age

Problem 7.3

A meter-stick is moving at $\;v=0.7\,c\;$ parallel to its own long dimension. How long does this stick appear from stationary Earth frame?


  1. Identify given variables
    • $L_0 = 1.0\,{\text m}$
    • $v=0.7\,c\;$
    • Need to find $L$
  2. Identify relevant equation(s)
    • $L = L_0\sqrt{1-\frac{v^2}{c^2}}$
  3. Plug the variables into the equation and solve for $L$. Note: It is not necessary to multiply $v$ by $c$ because the $c$'s will cancel.
    • $L = L_0\sqrt{1-\frac{v^2}{c^2}}$
    • $L = (1\,{\text m})\,\sqrt{1-\frac{(0.7c)^2}{c^2}}$
    • $L = (1\,{\text m})\,\sqrt{1-\frac{0.49c^2}{c^2}}$
    • $L = (1\,{\text m})\,\sqrt{1-0.49}$
    • $L = (1\,{\text m})\,\sqrt{0.51}$
    • $L = 1\,{\text m}\cdot 0.71$
    • $L = 0.71\,{\text m}$
  4. Note, that if the stick were moving perpendicular to its long dimension, its length would appear unchanged in both frames.

Problem 7.4

A bus, moving at a speed v, turns on its headlights, emitting light at velocity c relative to the bus' frame. How fast does this light appear to travel relative to Earth?


  1. Visualize or draw the problem and, if needed, re-word it to simplify what is being asked.
    • In this case, it is asking: what is the speed of light relative to Earth when the light is emitted from a moving bus?
    • Or, more fundamentally: is the speed of light dependent on the motion of the source?
  2. According to Einstein's postulate of special relativity:
    • “The speed of light in vacuum is the same in all inertial frames of reference, independent of the motion of the source or the receiver.”
  3. Therefore, the answer to this question is: The light appears to travel at velocity $c = 3.00 \times\ 10^8\frac{\text m}{\text s}$ relative to the Earth.
  4. Or, alternatively, use the addition of velocities:
    • $v=$ $\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}$
      • where $v_1=v_\text{bus}$ is the velocity of the bus
      • $v_2=c$ is the speed of emitted light (in bus' frame)
      • $v$ is the speed of light (in the Earth frame) that we need to find
    • $v=$ $\frac{v_\text{bus}+c}{1+\frac{v_\text{bus}c}{c^2}}=$ $\frac{v_\text{bus}+c}{1+\frac{v_\text{bus}c}{c^2}}\times\frac{c}{c}$ $=\frac{v_\text{bus}+c}{c+v_\text{bus}}\!\cdot\!c$ $=c$

Problem 7.5

A spaceship, moving at a speed $\;v=0.5\,c\;$, launches a rocket in the forward direction, also at a speed $\;v=0.5\,c\;$ relative to the spaceship. How fast does the rocket appear to travel relative to our planet? How would this answer change if the rocket were fired in the backward direction, opposite to the spaceship's speed vector?


Part 1: Rocket fired in forward direction
  1. Identify relevant equation(s):
    • $v_{23} = \frac{v_{21}+v_{13}}{1+\frac{v_{21}v_{13}}{c^2}}$
  2. Identify and label variables:
    • Object 1 = spaceship
    • Object 2 = rocket
    • Object 3 = planet
    • $v_{13}= 0.5\,c\;$ is the velocity of the spaceship relative to the planet
    • $v_{21}= 0.5\,c\;$ is the velocity of the rocket relative to the spaceship
    • $v_{23}\;$ is the velocity of the rocket relative to the planet – the quantity we need to find
  3. Plug in the variables and find $v_{23}$.
    • Note: It is not necessary to multiply the velocities by c because the c's will cancel and the answer can be expressed in terms of c.
    • $v_{23} = \frac{v_{21}+v_{13}}{1+\frac{v_{21}v_{13}}{c^2}}$
    • $v_{23} = \frac{0.5c+0.5c}{1+\frac{(0.5c)(0.5c)}{c^2}}$
    • $v_{23} = \frac{1.0c}{1+\frac{0.25c^2}{c^2}}$
    • $v_{23} = \frac{1.0c}{1+0.25}$
    • $v_{23} = 0.8\,c$
  4. Answer: The rocket appears to be traveling at $\;v=0.8\,c\;$ relative to the planet.
Part 2: Rocket fired in backward direction
  1. The equation and the set-up are the same, except now the velocity of the rocket is negative: $v_{21} = -0.5\,c$
  2. Find $v_{23}$ using this negative value of $v_{21}$:
    • $v_{23} = \frac{v_{21}+v_{13}}{1+\frac{v_{21}v_{13}}{c^2}}$
    • $v_{23} = \frac{(-0.5c)+0.5c}{1+\frac{(-0.5c)(0.5c)}{c^2}}$
    • $v_{23} = \frac{0\,c}{1+\frac{-0.25c^2}{c^2}}=0$
  3. Answer: The rocket appears to be motionless relative to the planet.

Problem 7.6

A 50-kg atomic bomb “yields” $4\times 10^{15}\,$J of energy, assuming all of the uranium has fissioned (and not fizzled!). What percentage of the uranium mass is “converted” to energy?


  1. The data is already in SI units:
    • Initial rest mass of the uranium:
      • $m_0= 50\,$kg
    • Amount of energy converted to kinetic energy, radiation and heat:
      • $E_\text{yield}=4\times 10^{15}\,$J
  2. Find the initial rest energy:
    • $E_\text{rest}=m_0c^2$ $=50\,{\text{kg}}\cdot\left(3\times 10^8\frac{\text m}{\text s}\right)^2$ $=4.5\times 10^{18}\,$J
  3. The fraction of the mass converted is the same as the ratio of the yield energy to the rest energy:
    • $\frac{m_\text{conv}}{m_\text{rest}}$ $=\frac{E_\text{yield}}{E_\text{rest}}=$ $\frac{4\,\times\,10^{15}\,{\text J}}{4.5\,\times\,10^{18}\,{\text J}}$ $=9\times 10^{-4}$
  4. About 0.1% of the initial mass, or about 50 g, is converted to the energy of the blast.
    • This mass doesn't disappear, it will contribute to the mass increase of the material that absorbs the energy of the blast

Problem 7.7

During the Chernobyl nuclear disaster in the spring of 1986, one of the dominant and most lethal radioactive isotopes released into environment was iodine-131, usually denoted 131I. It has a half-life of 8 days, and spontaneously decays into a stable 131Xe and an electron. The kinetic energy released in this decay is 0.971 MeV. Calculate the speed of the released electron (assuming it, being much much lighter than 131Xe, acquires the vast majority of the kinetic energy). Also find the mass of such an electron, and the recoil velocity of the 131Xe atom.


  1. Convert units to SI:
    • Conversion factor:
      • $1\,{\text{eV}}=$ $1.6\times 10^{-19}{\text J}$
    • $0.971\times 10^{6}\,{\text{eV}} \cdot\frac{1.6\,\times\,10^{-19}\,{\text J}}{1\,{\text{eV}}}$ $=1.55\times 10^{-13}\,{\text J}$
  2. Calculate the electron's rest energy
    • $E_0=$ $m_0c^2=9.1\times 10^{-31}\,{\text{kg}}\,\cdot\,3.0\times 10^8\frac{\text m}{\text s}$ $=8.19\times 10^{-14}\,{\text J}$
  3. Use the relativistic kinetic energy equation to calculate electron's velocity from its kinetic energy $K_e$:
    • $K_e=$ $\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}-m_0c^2$ $=1.55\times 10^{-13}\,{\text J}$
    • Here we neglect the kinetic energy of xenon, because it is much heavier than the electron. See the last step below.
  4. Now treat this as an equation on $v$ and rearrange the terms to solve for $v$:
    • $\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}=$ $1.55\times 10^{-13}\,{\text J}+m_0c^2$
    • $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=$ $\frac{1.55\times 10^{-13}\,{\text J}\,+\,m_0c^2}{m_0c^2}=$ $\frac{1.55\times 10^{-13}\,{\text J}\,+\,8.19\times 10^{-14}\,{\text J}}{8.19\times 10^{-14}\,{\text J}}$ $=2.9$
    • $\sqrt{1-\frac{v^2}{c^2}}$ $=\frac{1}{2.9}=0.345$
  5. find the velocity $v$
    • $1-\frac{v^2}{c^2}=$ $0.345^2$
    • $\frac{v^2}{c^2}=$ $1-0.345^2$
    • $\frac{v}{c}=\sqrt{1-0.345^2}$
    • $v=c\sqrt{1-0.345^2}$
    • $v=0.939\,c$
      • the speed of this electron is about 94% of the speed of light
  6. Calculate the mass of the moving electron
    • $m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$ $=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}m_0$ $=2.9\,m_0$
      • the mass of a moving electron is 2.9 times higher than the mass of an electron at rest
  7. Calculate the speed of xenon-131. The heavy atom moves so slowly that relativistic (Einstein) formulas are not needed (see last step).
    • Use the conservation of momentum
      • $\mathbf{p}_\text{fin}=$ $\mathbf{p}_e+\mathbf{p}_{^{\,131}{\text{Xe}}}=$ $\mathbf{p}_\text{initial}=0$ $\frac{|}{\big|}$
      • $\big|\,\mathbf{p}_e\,\big|=\big|\,\mathbf{p}_{^{\,131}{\text{Xe}}\,}\big|$
      • $\big|\,\mathbf{p}_e\,\big|=mv=$ $2.9\,m_0v=$ $2.9\,m_0\times 0.939\,c$ $=7.44\times 10^{-22}\frac{ {\text{kg}}\cdot{\text m}}{\text s}$
      • $(m_{^{\,131}{\text{Xe}}})\,(v_{^{\,131}{\text{Xe}}})=$ $7.44\times 10^{-22}\frac{ {\text{kg}}\cdot{\text m}}{\text s}$
    • Use Avogadro constant to convert the number of atoms in one mol to mass
      • $N_A=$ $6.022 \times 10^{23}\frac{1}{\text{mol}}$
      • $m_{^{\,131}{\text{Xe}}}=\frac{M_{^{\,131}{\text{Xe}}}}{N_A}=$ $\frac{0.131\,\frac{\text{kg}}{\text{mol}}}{6.022 \times 10^{23}\,{\text{mol}}^{-1}}$ $=2.17\times 10^{-25}\,$kg
      • $v_{^{\,131}{\text{Xe}}}=$ $\frac{7.44\times 10^{-22}\frac{ {\text{kg}}\cdot{\text m}}{\text s}}{2.17\times 10^{-25}\,{\text{kg}}}$ $=3420\,\frac{\text m}{\text s}$
  8. Check that the original assumptions make sense:
    • relativistic factor for 131Xe:
      • $\frac{1}{\sqrt{1-\frac{v^2_{^{\,131}{\text{Xe}}}}{c^2}}}$ $=\frac{1}{\sqrt{1-\frac{3420^2}{(3\,\times\,10^8)^2}}}$ $=1.000000000065$ $=1+6.5\times 10^{-11}$
        • this is a very good approximation to 1. We are justified in using non-relativistic equations for the xenon atom
    • the kinetic energy of 131Xe:
      • $K_{^{\,131}{\text{Xe}}}=$ $\frac{(m_{^{\,131}{\text{Xe}}})\,(v_{^{\,131}{\text{Xe}}})^2}{2}=$ $\frac{(2.17\times 10^{-25}\,{\text{kg}})\,(3420\,\frac{\text m}{\text s})^2}{2}=$ $1.27\times 10^{-18}\,$J
        • this is much smaller than $K_e=$ $1.55\times 10^{-13}\,{\text J}$. We are justified in assuming the electron carries most of the decay energy

Problem 7.8

Find the Schwarzschild radius of the Earth


  1. Determine what the question is asking. In this case, the “Schwarzschild radius” is the radius that an astronomical body must have in order to be a black hole. (The radius of the body must be equal to or less than the Schwarzschild radius.)
  2. Identify relevant equation(s):
    • $R=\frac{2GM}{c^2}$
  3. Identify relevant variables:
    • R is Schwarzschild radius of the object in question, i.e. what we need to find for the Earth
    • $G=$ $6.67\times 10^{-11}\frac{ {\text N}\cdot{\text m}^2}{ {\text{kg}}^2}\;$ is the universal gravitational constant
    • $M=$ $5.97\times 10^{24}{\text{ kg}}\;$ is the mass of the object in question, i.e. the mass of the Earth in this case
    • $c=$ $3.00\times 10^8\frac{\text m}{\text s}\;$ is the speed of light in vacuum
  4. Plug in the variables and find R:
    • $R=\frac{2\,\cdot\,6.67\times10^{-11}\,\cdot\,5.97\times10^{24}}{(3.00\times 10^8)^2}\,{\text m}$
    • $R=0.00886\,{\text m}$
  5. Convert to other units if desired:
    • $R=0.00886\,{\text m}\times\frac{1000\,{\text{mm}}}{1\,{\text m}}=8.86\,{\text{mm}}$
  6. Answer: The earth would have to condense down to a radius equal to or less than 8.86 mm in order to become a black hole.

Problem 7.9

Is flying internationally on a passenger airplane going to delay or accelerate an atomic clock, compared to the identical clock left behind at home? Assume the speed of the airliner $\;v\approx 500\,$mph $\approx 800\frac{\text{km}}{\text{hr}}\;$, the cruising altitude 35,000 ft, and the radius of the Earth $6.4\times 10^6\,$m.


  1. Picture the problem:
    • The clock will be delayed slightly due to the speed of the airplane itself, but its interaction with the Earth's frame of reference and gravity on the plane will accelerate the clock to a certain extent.
  2. Convert the units to SI units:
    • $v\approx 800\frac{\text{km}}{\text{hr}}$ $\approx 225\frac{\text{m}}{\text{s}}$
    • $h=35000\,{\text{ft}} =$ $10^4\,{\text m}$ is the altitude (height) of the flight relative to Earth's surface
    • $r=6.4\times 10^6\,{\text m}$
    • $r_0=\frac{2GM_\oplus}{c^2}=$ $0.00887\,{\text m}$ is the Schwarzschild radius of the Earth (mass of the Earth is $M_\oplus=5.97219\times 10^{24}\,$kg)
  3. Find the relevant equations:
    • Dilation of the clock's time due to speed: $t_0 =$ $t\sqrt{1-\frac{v^2}{c^2}}$
    • Dilation of time on Earth from Earth gravity, relative to a distant point: $t =$ $t_\infty\sqrt{1-\frac{r_0}{r}}$
    • Dilation of time on the plane from gravity, relative to a distant point: $t_h= $ $t_\infty\sqrt{1-\frac{r_0}{r+h}}$
  4. Solve each equation using the converted units:
    • Dilation due to speed: $t_0=$ $t\sqrt{1-\frac{v^2}{c^2}} =$ $\left(1- 2.8\times 10^{-13}\right)\,t$
      • The relative effect is $\frac{t_0-t}{t}=-0.28$ parts per trillion (the minus indicates slowing down of the traveling clock).
    • Dilation on Earth from gravity: $t =$ $t_\infty\sqrt{1-\frac{r_0}{r}} =$ $t_\infty\sqrt{1-\frac{0.00887\,{\text m}}{6.4\times 10^6\,{\text m}}} =$ $\left(1- 6.9297\times 10^{-10}\right)\,t_\infty$
    • Dilation on the plane: $t_h =$ $t_\infty\sqrt{1-\frac{r_0}{r+h}} =$ $t_\infty\sqrt{1-\frac{0.00887\,{\text m}}{6.4\times 10^6\,{\text m}\,+\,10^4\,{\text m}}} =$ $\left(1- 6.919\times 10^{-10}\right)\,t_\infty$
  5. Find the ratio (relative dilation) between $t_h$ (time at altitude) and $t$ (time on Earth surface):
    • $\frac{t_h}{t}$ = $\frac{1- 6.919\times 10^{-10}}{1- 6.9297\times 10^{-10}}\approx$ $1-\left(6.919\times 10^{-10}-6.9297\times 10^{-10}\right) =$ $1+1.0\times 10^{-12}$
      • The plus indicates that a stationary clock at altitude goes faster by about 1 part per trillion, due to weaker gravity force up there
  6. Interpret the results
    • The atomic clock will slow down by 0.28 parts per trillion because of the speed of the airplane, and will accelerate by about 1 part per trillion due to weaker gravity.
    • The net effect is acceleration by 0.72 parts per trillion.

HW Questions Ch.29

Concept. Question 5

in the textbook When we view a distant galaxy, we notice that the light coming from it has a longer wavelength (it is “red shifted”) than the corresponding light here on Earth. Is this consistent with the postulate that all observers measure the same speed of light? Explain.

  • Q: I am having trouble explaining how this statement is consistent with the postulate that all observers measure the same speed of light. Can someone help me answer this? — Joshua Vandehey 2015/05/28 19:05
  • A: This not a clearly defined question. Is 4=2+2 consistent with the fact that the Sun is larger than Earth? Well, both facts can be true (or are true) at the same time, so they are clearly not inconsistent. Can one prove that one implies the other? I am not so sure.
    • For example, we can observe the sound produced by a moving ship to be “red shifted” compared to the sound produced by an identical but stationary source. However, the speed of sound measured by different observers is different, because it is a fixed number relative to the water.
    • In other words, Doppler effect can be “consistent” with the Einstein's relativity, but you can't derive the latter from the former, that's for sure.
      Prof. Nicholas Kuzma 2015/06/03 00:27

Problem 29.4

Part 1: A clock in a moving rocket is observed to run slow. If the rocket reverses direction, does the clock run slow, fast, or at its normal rate?

  • Q: I understand the answer to the original question. However I did not understand the rational behind the answer that is presented in part two of this question. Which is “The clock will run slow, just as before. The rate of the clock depends only on relative speed, not on direction of motion.” If a person were on the moving rocket with the clock, would the clock run any differently relative to the person depending on the speed? Or are we just looking at a clock on a moving rocket from a distance and noting that the clock is running slowly. How would the change in speed change the speed of the clock? Would that just be relative to an observer or is it asking this question from the position of someone on the clock. Just a bit confused by the context and the effect that change in velocity would have on the running speed of the clock. Any insight into this would be very helpful! Thank you.
    Elin Odegard 2015/06/02 11:39
  • A: Again, we are talking about a moving clock observed by a stationary observer. So, the answer is, the moving clock relative to an identical clock that is stationary, both clocks observed and synchronized from a stationary frame, will tick slower than the stationary clock, no matter which direction (away or towards) it is moving. If both clocks are observed and synchronized by a moving observer (somebody who is riding with the moving clock), then the opposite will be true: the stationary clock in that frame will appear to be moving, and thus will tick slower. Again that is independent of the direction of motion. — Prof. Nicholas Kuzma 2015/06/03 00:33

Ch.30: Quantum physics (June 2)

Concepts Ch.30

Constants Ch.30

  • Planck constant $h=$ $6.62606957\times 10^{-34}\,{\text J}\!\cdot\!{\text s};$ $\;\;\;\;\;$ $\hbar=\frac{h}{2\pi}=$ $1.054571726\times 10^{-34}\,{\text J}\!\cdot\!{\text s}$
  • Mass of an electron: M= 9.11*10^-31kg — Stephanie Moore 2015/06/08 13:40
  • 1 electron volt = 1.60217656*10^-19 J — Stephanie Moore 2015/06/08 13:47
  • h=6.626*10^-34 J/s = 4.136*10^-15 electron volts/s — Stephanie Moore 2015/06/08 13:49

Units Ch.30

  • Micron ($\mu$m) = $10^{-6}$ m
  • (Arc)minute = (1/60)$^\circ$
  • (Arc)second = (1/3600)$^\circ$

Math Ch.30

Lecture notes June 2

Equation Sheet Ch.30

An ideal blackbody absorbs all light that's incident on it. Distribution of energy in a blackbody is independent of the material, it depends only upon the temperature.

  1. Wien's Displacement Law:
    • $f_\text{peak}= (5.88\times 10^{10}\,{\text s}^{-1}\!\cdot\!{\text K}^{-1})\,T$
  2. Plank's Quantum Hypothesis:
    • Planck constant $h=6.63\times 10^{-34}\,{\text J}\!\cdot\!{\text s}$
    • Total energy of radiation at frequency $f$ is
    • $E_n=nhf\;$ where $\;\;n=0,\,1,\,2,\,3,\,\ldots$
  3. Energy of a photon of frequency $f$
    • $E=hf$ $=\frac{hc}{\lambda}$ — Stephanie Moore 2015/06/08 13:43
    • SI units: J
  4. Cutoff frequency $f_0$
    • $f_0= \frac{W_0}{h}$
    • $W_0\;$ is work function (minimum energy to eject an electron from a particular metal)
    • SI unit: Hz = ${\text s}^{-1}$
    • If $\,f>f_0\,$ (frequency of light is above the cutoff frequency), then the electron can be ejected with finite kinetic energy
    • If $\,f<f_0\,$, then no electrons are ejected
  5. Maximum kinetic energy (K) a photoelectron can have:
    • $K_\text{max}= E-W_0$ $=hf-W_0$
  6. Rest mass of a photon
    • $m_0= 0$
  7. Momentum of a photon
    • $p= \frac{hf}{c}$ $= \frac{h}{\lambda}$
  8. Compton scattering (electron is initially at rest):
    1. To conserve energy:
      • Energy of incident photon = energy of scattered photon + final kinetic energy of electron
        • $hf= hf' +K$
    2. Compton Shift Formula
      • $\Delta\lambda = \lambda' – \lambda =$ $\frac{h}{m_e c}\left(1-\cos\theta\right)$
        • SI unit: m
        • When $\theta = 180^\circ$, the change of the photon's wavelength is maximum
        • When $\theta = 0^\circ$, no light actually scatters off of electrons, the change in wavelength is zero.
    3. Change in Kinetic energy of an electron that scattered a photon:
      • $\Delta K=$ $hc\left(\frac{1}{\lambda_\text{final}}-\frac{1}{\lambda_\text{initial}}\right)$ — Stephanie Moore 2015/06/08 13:52
  9. de Broglie wavelength (of any particle or object)
    • $\lambda= \frac{h}{p}$
    • SI unit: m
  10. Constructive interference when scattering from a crystal
    • $2d \sin\theta= m\lambda$
    • $m=1,\,2,\,3,\,\ldots$
  11. The Heisenberg uncertainty principle:
    1. Momentum and position
      • $\Delta p_y\Delta y \ge \frac{h}{2\pi}$
    2. Energy and time
      • $\Delta E\Delta t \ge \frac{h}{2\pi}$
  • Total number of Photons= Total energy/ energy of one photon — Aish Shimpi 2015/06/08 20:52
  • Energy (joules)= intensity (watts/m^2)*area (m^2)— Aish Shimpi 2015/06/08 20:52

Additional notes

  1. Quantum tunneling
    • Particles can pass through regions of space that would be otherwise forbidden to classical particles (because of their wavelike behavior)

Examples Ch.30

Problem 8.1

Find the peak emission frequencies of two black-body light sources, one at 4700 K (“cool white”), and another at 2700 K (“warm white”).


  1. Use Wien's Displacement Law to find
    • $f_\text{peak}=$ $(5.88\times 10^{10}\,{\text s}^{-1}\!\cdot\!{\text K}^{-1})\,T$
  2. Substitute the given T values to find peak frequencies:
    • Peak emission frequency for “Cool White”:
      • $f_\text{peak}=$ $(5.88\times 10^{10}\,{\text s}^{-1}\!\cdot\!{\text K}^{-1})\cdot 4700\,{\text K}$ $= 2.76 \times 10^{14}\,$Hz
    • Peak emission frequency for “Warm White”:
      • $f_\text{peak}=$ $(5.88\times 10^{10}\,{\text s}^{-1}\!\cdot\!{\text K}^{-1})\cdot 2700\,{\text K}$ $= 1.59 \times 10^{14}\,$Hz
    • Note, that in these equations s$^{-1}$ becomes Hz, without the need to divide by $2\pi$.

Problem 8.2

The work function of potassium is $W_0=2.29\,$eV. What (if any) electron emission is observed when the blue ($\lambda=450\,$nm) or, alternatively, the red ($\lambda=750\,$nm) light strikes the metal surface?


  • If the frequency of the light is greater than $f_0$ (cut-off frequency), then the electron can leave the metal with finite (and positive) kinetic energy.
    1. Convert the variables to SI units:
      • $W_0=2.29\,{\text{eV}}\cdot 1.60\times 10^{-19}\frac{\text J}{\text{eV}}=$ $3.67\times 10^{-19}\,$J
      • $\lambda_\text{blue}=4.50\times 10^{-7}\,$m
      • $\lambda_\text{red}=7.50\times 10^{-7}\,$m
    2. Calculate cut-off frequency for both red and blue light using the same work function value
      • The cut-off frequency is the same for any wavelength, it only depends on the type of metal
      • Equation: $\;f_0= \frac{W_0}{h}$
      • Both red and blue Light:
        • $f_0=\frac{3.67\times 10^{-19}\,{\text J}}{6.63\times 10^{-34}\,{\text J}\cdot{\text s}}$ $= 5.53\times 10^{14}\,$Hz
    3. Convert the given wavelengths for red and blue light to frequency.
      • Equation: $\;f= \frac{c}{\lambda}$
      • Blue light: $\;f_\text{blue}=\frac{3.00\,\times\,10^8\,\frac{\text m}{\text s}}{4.50\,\times\,10^{-7}\,{\text m}}$ $= 6.66\times 10^{14}\,$Hz
      • Red light: $\;f_\text{red}=\frac{3.00\,\times\,10^8\,\frac{\text m}{\text s}}{7.50\,\times\,10^{-7}\,{\text m}}$ $= 4.0\times 10^{14}\,$Hz
    4. If calculated frequency is greater than cut-off frequency, electron emission will be observed
      • The blue photon will be able to eject an electron because it exceeds the cut off frequency $(f_\text{blue}>f_0)$
      • The red photon will not be able to eject any electrons, because its frequency is below the cutoff $(f_\text{red}<f_0)$
      • The maximum kinetic energy of the electrons ejected by the blue light is
        • $K_\text{max}=hf_\text{blue}\!-hf_0=$ $h(f_\text{blue}\!-f_0)=$ $6.63\times 10^{-34}\,{\text J}\cdot{\text s}\cdot(6.66\times 10^{14}-5.53\times 10^{14})\,{\text{Hz}}=$ $7.5\times 10^{-20}\,$J
        • The same kinetic energy in eV units is about 0.5 eV

Problem 8.3

Find the maximum change in wavelength for blue light ($\lambda_b=450\,$nm) and for X-rays ($\lambda_X=0.1\,$nm) during Compton scattering off of electrons that are initially at rest. What percentage of the incident wavelength is this change in each case?


  1. The maximum change in the photon's wavelength occurs when it scatters in the reverse direction $(\theta=180^\circ)$, in which case the change is twice the Compton wavelength $\frac{h}{m_ec}$:
    • $\Delta\lambda=\frac{h}{m_ec}\big(1-\cos 180^\circ\big)$ $= \frac{h}{m_ec}\big(1-(-1)\big)$ $= \frac{h}{m_ec}\cdot 2$ $=\frac{2h}{m_ec}$
    • This change in $\lambda$ doesn't depend on the $\lambda$ itself, so it is the same for both wavelengths:
    • $\Delta\lambda=$ $\frac{2h}{m_ec}=$ $4.86\times 10^{-12}\,$m
  2. Percent change for blue light
    • $\frac{\Delta\lambda}{\lambda_b}=$ $\frac{4.86\times 10^{-12}\,{\text m}}{4.5\times 10^{-7}\,{\text m}}=$ $1.08\times 10^{-5}$ $\approx 0.001\%$
  3. Percent change for X-rays
    • $\frac{\Delta\lambda}{\lambda_X}=$ $\frac{4.86\times 10^{-12}\,{\text m}}{1\times 10^{-10}\,{\text m}}=$ $4.86\times 10^{-2}$ $\approx 4.9\%$

Problem 8.4

In a classical model, an electron is orbiting a proton (the nucleus) in a hydrogen atom, with velocity $v=\sqrt{\frac{k_cq_e^2}{m_er}}$ $\approx 2.2\times 10^6\frac{\text m}{\text s}$, where $r=5.3\times 10^{-11}\,{\text m}$ is the radius of its orbit. Find the uncertainty of the electron's energy and momentum, and compare them to the electron's kinetic energy and momentum.


  1. Find the momentum from the known velocity and mass (since $v<0.01c$, no need to use Einstein's relativity)
    • $\big|\,p\,\big|=m_ev=$ $9.11\times 10^{-31}\,{\text{kg}}\cdot 2.2\times 10^6\frac{\text m}{\text s}=$ $2.00\times 10^{-24}\frac{ {\text{kg}}\cdot{\text m}}{\text s}$
  2. Find the kinetic energy of the electron
    • $K=\frac{m_ev^2}{2}=$ $0.5\cdot 9.11\times 10^{-31}\,{\text{kg}}\cdot \left(2.2\times 10^6\frac{\text m}{\text s}\right)^2=$ $2.2\times 10^{-18}\,{\text J}$
  3. Use the uncertainty principle to find $\Delta p_y$, assuming the uncertainty $\Delta y$ is the radius of the orbit
    • $\Delta p_y=\frac{h}{2\pi\Delta y}$ $=\frac{6.63\,\times\,10^{-34}\,{\text J}\cdot {\text s}}{2\pi\,\cdot\,5.3\times 10^{-11}\,{\text m}}=$ $2.00\,\times\,10^{-24}\,\frac{ {\text{kg}}\cdot{\text m}}{\text s}$
  4. Use the uncertainty principle to find $\Delta E$, assuming the uncertainty $\Delta t$ is the time it takes to travel 1 radian
    • $\Delta t=\frac{r}{v}$ $=\frac{5.3\times 10^{-11}\,{\text m}}{2.2\times 10^6\frac{\text m}{\text s}} $ $=2.4\times 10^{-17}\,$s
    • $\Delta E=\frac{h}{2\pi\Delta 4}$ $=\frac{6.63\,\times\,10^{-34}\,{\text J}\cdot {\text s}}{2\pi\,\cdot\,2.4\times 10^{-17}\,{\text s}}=$ $4.4\,\times\,10^{-18}\,{\text J}$
  5. In this classical model, the uncertainty in momentum is equal to the momentum itself, and the uncertainty in energy is twice the value of kinetic energy itself

Homework Questions Ch.30

Problem 30.58

An X-ray scattering from a free electron is observed to change its wavelength by 3.59 pm. Find the direction of propagation of the scattered electron, given that the incident X-ray has a wavelength of 0.530 nm and propagates in the positive x direction.

  1. I need help on this homework question
    • Answer: Follow these steps below.
      1. From the Compton scattering formula, find the angle $\theta$ of the scattered photon.
      2. Find the y component of the photon's momentum after scattering ($p_y'=p'\sin\theta$)
      3. Using the momentum conservation along the y axis, find the electron's y-momentum $p_{ey}'$
      4. Find the electron's kinetic energy $K$ from the energy conservation. Is this a relativistic case (compare to $m_ec^2$)?
      5. Find the electron's total momentum $p_e'$ (after scattering) from its energy $K$
      6. Find the electron's scattering angle from its y-component of momentum and the total momentum $\left(\sin\phi=\frac{p_{ey}'}{p_e'}\right)$

Conceptual Question 11 (Chapter 30)

Why can an electron microscope resolve smaller objects than a light microscope? - The book states that the resolution is determined by the wavelength of the imaging radiation-the smaller the wavelength the greater the resolution. I am wondering where they are getting this answer. I'm thinking that it has to do with Rayleigh's criterion, but I'm not sure that it applies because it is a concept from chapter 28. — Joshua Vandehey 2015/06/06 19:39

  • In Chapter 30 on page 1060 the de Broglie Hypothesis is discussed which I think is the relevant concept for this problem. I think you’re on the right track with the Rayleigh criterion. Basically because we can’t see the object with visible light (due to the Rayleigh criterion) we need either radiation with a smaller wavelength, or use particles with smaller de Broglie wavelengths, in this case the electron. Remember, an electron isn't electromagnetic energy like visible light but rather a subatomic particle. And as the de Broglie equation shows, the wavelength referred to in this conceptual question, or more specifically its de Broglie wavelength, is actually smaller than visible light. Another way to look at this is that you could not observe an electron with visible light because it is smaller than the wavelength of visible light. To take this a step further, even smaller wavelength radiation like Gamma rays might be able to resolve smaller objects, but I think those would have problems due to the fact that they’re higher in energy and might destroy what you’re trying to observe, and also of course you couldn’t see the light from such a microscope with your eye. Similarly an electron microscope feeds the data from the sensor to a computer monitor. — Kris Bugas 2015/06/07 15:44

The main aspect of electron microscopes vs. light microscopes is that electron microscopes use energized electrons to examine nano scale objects, whereas light microscopes use photons. It operates in the same way as the photoelectric effect. The specimen that is viewed is coated with metal, so secondary electrons can be ejected. The factor behind light microscopes is that they use a very narrow range of wavelengths typically the wavelengths of white light for illumination (so longer wavelengths), leading to a decreased resolution. Since electron microscopes deflect electrons. Electrons have shorter wavelengths than light used in light microscopes, so the quality of resolution is increased. - Aishwarya Shimpi 2015/06/07

Conceptual Question: Hello Dr. Kuzma, I have a conceptual question that was presented in the textbook that I do not quite understand. In the textbook conceptual checkpoint 30-2 the text says “A beam of light with a frequency greater than the cutoff frequency shines on the emitter. If the frequency of this beam is increased while the intensity is held constant, does the number of ejected electrons per second from the metal surface a)increase b) decrease c) stay the same? The book says the answer is b (decrease), which does not make sense to me as I would expect it to be c (stay the same). This is because intensity is defined as the number of photons that hit the metal, which is the number of electrons ejected from the metal. Increasing intensity this allows for the same amount of energy in each ejected electron, so the kinetic energy stays the same. Increasing frequency however, with intensity the same, keeps the same amount of ejected electrons but each ejected electron has a greater amount of energy, thus increasing kinetic energy. Can you please let me know if my logic is correct? - Aishwarya Shimpi 6/6/2015

  • I think you may have misread the problem. You wrote “Increasing intensity this allows for the same amount of energy in each ejected electron, so the kinetic energy stays the same.” The problem asked about increasing the frequency, not increasing intensity. The following sentences you seem to understand that increasing intensity increases the number of photons ejected but not their KE, and that increasing frequency increases the KE of the individual photons ejected.. — Kris Bugas 2015/06/07 21:54

I am having trouble understanding and setting up this problem oh mastering physics Problem 30.36 Owl Vision Owls have large, sensitive eyes for good night vision. Typically, the pupil of an owl's eye can have a diameter of 8.5 mm (as compared with a maximum diameter of about 7.0 mm for humans). In addition, an owl's eye is about 100 times more sensitive to light of low intensity than a human eye, allowing owls to detect light with an intensity as small as 5.0×10−13W/m2. Find the minimum number of photons per second an owl can detect, assuming a frequency of 7.0×1014Hz for the light.

Here is step by step how you solve this problem: 1. Intensity x area= Power (watts). Find the power by multiplying the intensity by the owls eye area. Area= pir^2. Use the diameter to find the radius of the owls eye giving the area. 2. Multiply the power by the time in seconds. Energy (Joules)= 1 second x power. The power was calculated in step 1. 3. Realize that step 2 gives you the total energy that goes into the owl's eye in one second. To find the number of photons per second that the owl detects, you need to divide the total energy calculated in step 1 by the energy of one photon. The energy of one photon= h x frequency of the light. 4. This will give you the min number of photons per second. -Aishwarya Shimpi 6/8/2015

I was struggling with this problem myself and I found Aishwarya Shimpi’s solution works great, I’m not trying to criticize it, but I found a way that was a little more direct that made more sense to me. E=hf. Just divide the 5×10^-13 by that number. Then multiply by the area of the owl’s eye. That’s it. [1/{(6.63×10^-34Jxs)x(7×10^14s-1)}]x[(5×10^-13J/s)x(1/m^2)x(5.67×10^-5m^2)] = 61 photons per second. I wrote out the solution so you can compare since it’s an even problem. I wrote it this way so you can see the terms cancel out. — Kris Bugas 2015/06/08 21:11

In chapter 30, the textbook discusses NASA studying the feasibility of constructing light sails. In fact on May 20 a non-profit led by Bill Nye launched a LightSail, using the energy from the sun to move it through space. Embedded is a video and an article link discussing this project.

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