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_{PH203KUZMASPRING2015}
Please check D2L. — Prof. Nicholas Kuzma 2015/06/16 10:23
The PSU physics department is committed to collecting real data about how effective our teaching is.
https://docs.google.com/a/pdx.edu/forms/d/1gAy47sXgeSEfJWmcfgLUiq0d9lOgAOFjk9-u7bh_B-I/viewform
Previous chapters have been moved to the sidebar:
$\cos\alpha+\cos\beta=$ $2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)$ | $\cos\alpha=\cos(-\alpha)$ | $\sqrt{1+\epsilon}\approx$ $1\!+\!\frac{1}{2}\!\epsilon$ $\sqrt{1-\epsilon}\approx$ $1\!-\!\frac{1}{2}\!\epsilon$ For small $|\epsilon|\ll 1$ | $\tan[\arcsin(x)] = \frac{x}{\sqrt{1-x^2}}$ |
$\sin\alpha=-\sin(-\alpha)$ | $\tan[\arccos(x)] = \frac{\sqrt{1-x^2}}{x}$ |
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$
$\;\;m = 0,\, 1,\, 2,\, 3\,\ldots$ | Refraction index order ($n_1$ and $n_2$ are on either side of the film $n_f$) | Dark fringes | Bright fringes |
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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 $\;n_1\!>\!n_f\!>\!n_2$ | $t = \frac{\lambda}{2n_f}\left(m+\frac{1}{2}\right)$ | $t = \frac{\lambda}{2n_f}m$ |
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
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.
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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.
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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.
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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.
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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.
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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 $\lambda$ $= 600\,{\text{nm}}$
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This will resolve a 1.7-mile object on the Moon.
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 1^{st} and the 2^{nd} peaks.
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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)
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.]
An image of the Hubble Space Telescop, wanted to share!
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?
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.
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)?
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.
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 -
Hello Dr. Kuzma, here is a link to that video you discussed in class. Enjoy!
— Thomas Ruttger 2015/05/28 12:43
Black holes! This a pretty interesting article about a huge telescope and how black holes are detected. — Ryan Stoner 2015/06/09 09:49
http://www.nytimes.com/2015/06/09/science/black-hole-event-horizon-telescope.html?mwrsm=Email
Rotations | Relativity |
---|---|
$\cos\left[\arctan\left(\frac{v}{c}\right)\right]$ $=\frac{1}{\sqrt{1+\left(\frac{v}{c}\right)^2}}$ | $\cosh\left[{\text{arctanh}}\left(\frac{v}{c}\right)\right]$ $=\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$ |
$\sin\left[\arctan\left(\frac{v}{c}\right)\right]$ $=\frac{v/c}{\sqrt{1+\left(\frac{v}{c}\right)^2}}$ | $\sinh\left[{\text{arctanh}}\left(\frac{v}{c}\right)\right]$ $=\frac{v/c}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$ |
Please update if an equation is not included
-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
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
How far does the light travel in 1 year? 1 minute? 1 s? 1 ms? 1 $\mu$s? 1 ns?
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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 |
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?
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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?
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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?
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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?
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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?
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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 ^{131}I. It has a half-life of 8 days, and spontaneously decays into a stable ^{131}Xe 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 ^{131}Xe, acquires the vast majority of the kinetic energy). Also find the mass of such an electron, and the recoil velocity of the ^{131}Xe atom.
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Find the Schwarzschild radius of the Earth
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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.
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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.
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?
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.
Find the peak emission frequencies of two black-body light sources, one at 4700 K (“cool white”), and another at 2700 K (“warm white”).
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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?
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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?
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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.
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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.
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
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 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. Lma@pdx.edu
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
http://www.popsci.com/bill-nyes-solar-sail-back-online
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.