The flux of cosmic ray muons is about 1 muon per square centimeter per minute. I was trying to see if I can detect these using a glass of water, putting it in a totally dark room and take a picture of that with a camera mounted on a tripod. I used the highest ISO setting available and exposed for about half a minute, and I let the camera perform a dark frame subtraction. Muons moving through the water should produce photons, some of which will be detected by the camera sensor.  I then converted the raw files using the dcraw program to tiff files without doing any demosaicing (so each pixel corresponds to the actual gray value detected by a sensor photosite). But I'm now a bit bogged down in analyzing the raw files (a problem here is that the Sony ARW files are not real raw files, they have build in lossy compression which is not ideal for doing this sort of analysis).   So far I have not seen any interesting, if I look at gray values that are many standard deviations above the noise then I don't see any obvious (broken) straight line patterns. So, I'm wondering if one would actually expect any signal at all. It seems to me that you would because from close up you would detect some photons from the water, but perhaps I'm too optimistic.
121634
I have read this article    and it talks about estimating the muon cosmic ray flux by means of a DSLR camera.  Is it possible to measure the muon's energy with the same apparatus?  If yes, how can I "calibrate" (?) the camera in order to measure the energy?
111294
I've been thinking about getting into film photography for a while, so recently I grabbed the old Olympus OM-1 from work when they were cleaning out the media cupboard, I'm only mentioned in this so you can understand my experience level with film is minimal. I've shot a few rolls of film for testing and in a couple of them have seen some strange light artifacts, but only in a few of the shots. Rather than try and explain them, I'll show some examples (scanned 6x4's on flatbed scanner, so please excuse the quality):    I also have an example of the orange triangular artifact above on another shot that is much stronger and on a dark landscape shot, but thought it would be useful to show 2 shots that I know were taken close in time.  A second shot was taken just after this, but the artifact is different:   The lighter bar down the left hand side of the above shot appears in a few other shots as well, although with much stronger effect. I've also seen a bar down the side of other shots, but orange in colour rather than white.  These artifacts aren't occuring on every shot, I have shot 3 rolls of film and have found these artifacts on 6/24 or so per film.  My suspicion is that there is a light leak, or multiple light leaks. I've changed the seals on the back door already, but the question I am not sure about it ... is there any other way for light leaks to occur in a film camera that I can investigate? Am I on the right path or is this likely something else ... or a combination of things?     Are the images with the orange artifacts usually the first shots taken after the camera has been unused for a while (hours, days,etc.)? In other words, the frame in question sat directly behind the shutter for longer than just the time between frames in the same session?   I just had a look at the shots in order, and yes, it would seem that the shots with the orange marks would have been the first shots for some time (probably at least a day in all cases).  All the the film shot thus far has been processed by a local professional camera store, so I don't suspect processing at this stage.    update: I've shot another roll through the camera. The good news is this time I had no artifacts appear, however I am conscious of the fact that this film was shot in a matter of hours, rather over multiple days that the films that I had issues with.  Now that I have some time I'll try out the light inside the camera idea that Alan Marcus suggested, an LED with a small battery should do the trick. Note I haven't tried this yet, but something like an LED Throwie should do the trick, just don't attach the magnet!.  for anyone not electronics inclined, they are pretty simple to set up
79928
I know there have been a lot of posts on this topic, but I haven't found exactly what I am looking for.   I noticed some dust spots on my sensor for my D750 after I took a picture with some sky in it and then was playing around with "dehaze" in Lightroom. Okay no problem, I have owned the camera for a year probably time to clean the sensor.   So I do a bit of research online and land on the fact that I can do this myself. I also want to learn how so I don't need to spend $50 or $75 each time I need this done. I decide to buy .  What I do is close my aperture down to F22, put my ISO on 100 and then set my shutter speed to 2 seconds or so and take a picture of a well lit white piece of paper.   I then go into Lightroom and jack the dehaze and look for dust spots. I see a ton. Okay, so now let's clean it. I wet the swab with a few drops to get it just saturated on the top of the swab, apply some pressure, wipe left to right, turn it over, wipe right to left and done.   Then I took the same picture and although the dust spots are different, they are still there! What the heck? Okay maybe it just needs a good blow out. I took my  and blow out everything I can inside of there, then I do the same cleaning method as the first time, except this time I start with a dry swab, throw it out, then get a new swab, wet it and clean it again.  I take the picture again... and they are STILL THERE. Okay let's get out the high power flashlight. I wouldn't say I scrubbed it, but I went back and forth with the swab on what seemed to be a stubborn spot (I know, sketchy). Then I clean it again with a normal swab and take another picture.   This went on and on and I have basically ran out of swabs. I still see spots. I used the same swab a few times (I know, also sketchy) because I just couldn't figure out why it wasn't getting clean. I even tried a can of air (although some don't recommend that because they have solvent in them, but I was desperate and some people on youtube did it). That actually seemed to help, but in the end I still have spots.  So obviously a lot of you will tell me to just "take it in! you're going to break your sensor", but I really want to learn how to do this. And now I am just stubborn.  The spots are at the top and bottom of the frame mostly.  Any ideas? I ordered  today, and I was going to try it out very carefully. I also ordered .       EDIT: I tried the gel stick. It did seem to help, but on the second try something else got on the sensor so I am going to wait until tomorrow to get the swabs and then try to do a swab cleaning + the gel stick again... we will see.  EDIT2: Well it looks . I think the gel sticky stick helped but it's still not spotless. I ran over it one more time. I also included a photo above to the people are nervous about cleaning their own sensor. Obviously I can't guarantee you won't screw up, but I ran a swab over this sensor at least 15 times and it still shots great.
115234
Will I receive an interference pattern on D0 detector   if D1, D2, D3, D4 exist in the set up of the experiment, but they are thousands of light years far? Simply, to set up the experiment so that I detect all of the signal photons on D0 before I detect any of idler photons!
80361
I'm not a physicist; I'm just curious about what would happen.  According to this video "" (at 2:50):     "Somebody decided to leave the detectors on but just not take any data. [...] So if it was also going to a magnetic tape - there was no magnetic tape loaded. [...] And what do you think happed? They got this. They got the diffraction pattern."   So the act of measuring (beeing concious of the data) destroys wave-form pattern?  If so lets consider these cases:   What if I'd throw a dice after the experiment to decide if I should look at the data or to destroy it? (e.g. look at the data if the number on dice is even) What if the information about measurement would need to travel longer period of time than the particle would need to hit the wall?    To above cases: Would it mean that information travelled instantly - faster than light or that "the particle can look into the future"?
89382
In the  (semi-classical, but should still apply here I think), the conducting electrons are in a constant electric field, and, in between collisions with the lattice ions (that happen, on average, in a time $\tau$ since their last collision) in which their velocities are reset to an average of 0, they are accelerated and gain a momentum  $$p = qE\tau (=mv)$$, where $q$ is its charge and $E$ is the electric field from the applied voltage. This obviously means in between the collisions, the electron accelerates $a$, from $v_f = v_i + a\tau$.  Accelerating charges give off . So does a DC carrying wire give off radiation one can measure?  I suspect it does, but it's of an insanely small magnitude. A quick order of magnitude calculation with the following values gives me:  $$q \sim 10^{-19} \,{\rm C}, \\ E\sim 1\,{\rm V}/1\,{\rm cm}=100\,{\rm V/meter},\\ m\sim10^{-30}\,{\rm kg} \\ \Rightarrow a = qE/m \sim 10^{13}{\rm m/s}^2 \\ \Rightarrow P \sim q^2a^2/\epsilon_0c^3 \sim 10^{-28}\,{\rm W}$$  Which is obviously pretty small. I'm not sure what's the smallest power that can be measured, but I'm guessing by increasing the applied voltage one could increase the radiated power. But is the idea even fundamentally sound?
55590
My lecturer told me that $\mu$, the , is zero or negative, and in the following example, mathematically it acts as a normalization constant. But is there any physical insight about why boson gas can be zero or negative? I think it is due to the fact the photon gas can pop up from nowhere (i.e. vacuum fluctuation). $$ f_{BE}(\varepsilon)=\dfrac{1}{e^{(\varepsilon-\mu)/(k_B T)}-1}$$
25458
For  scattering  where he fired alpha particles at a thin sheet of metal to determine the size of the nucleus, please can someone give me the direct link between his observations and his conclusions e.g. this observation lead to this conclusions.
23658
Is this true or false: If A and B have clocks and are traveling at relative velocity to each other, then to B it APPEARS that A's clock moving slower, but A sees his own clock moving at normal speed.  Similarly, to A it APPEARS that B's clock is moving slower, but B sees his own clock moving at normal speed.  If the above is true, then both A seeing his own clock moving at normal speed and B seeing his own clock moving at normal speed means that in reality both clocks are moving at normal speed, and neither has slowed down, whereas the other person's clock APPEARING to move slowly is merely an illusion.   Now is this true or false: If A and B have clocks and are traveling at relative velocity to each other, then A sees his own clock move slowly (compared to the speed of the clock when A was at rest with respect to B) and B sees his own clock move slowly, so that in reality both clocks are moving slowly, but they still remain synchronized (since both are slow by the same amount)  If the above statement is not true, then why do muons decay slowly when moving fast? [it could only be possible if the muon saw its own clock as moving slowly. If we saw the muon's clock moving slowly, but the muon saw its own clock moving at the normal rate, then the muon would decay at the normal rate, and not slowly]  Can anyone please explain where I went wrong?
37607
Last night I took some night sky shots, and I've noticed that with long exposure times (15s, 30s) the time it took to store the images in the camera memory card (jpeg) took roughly as long?  The actual file size is fairly small at 1 to 3MB per file.  Any idea why this is?
49647
Consider a Photoelectric effect experiment apparatus as shown in Figure 11.1    The variation of photocurrent with the voltage applied across $\mathrm{A}$ and $\mathrm{C}$ is as shown in Figure 11.3.    My question –     Why is the current non-zero even if the voltage is zero?    Consider a situation in which the voltage across the plates is zero. Photons of sufficient frequency strike the outermost surface of the cathode, emitter plate $\mathrm{C}$. Electrons become free from the attraction of nucleus as their net energy becomes positive.  Why I think that there should be no net current?  As soon as photons strike the outermost surface of the cathode(assumed to be $1$ to $10$ atomic diameters thick), electrons become free. Now, as the electrons have kinetic energy, they'll come out of that outermost surface. Now, the electrons(with kinetic energy) should have an equal probability of going in every direction of the surface. So, around half of the electrons should go in the right direction towards the anode and strike it and half should go to the left, trying to penetrate the inner surfaces of the cathode plate.   Now, the electrons which will go to the right will face some field due to the space charge present in the tube but nevertheless, some electrons will definitely reach the anode plate. After they strike with the anode plate, they'll face some resistance but still, some electrons will reach the low resistance copper wires connected to the anode plate and milliammeter.  Now, consider the electrons which go towards left. They'll definitely face resistance from cathode plate but some will still reach the copper wire.  Now, as there is no potential difference applied between $\mathrm{A}$ and $\mathrm{C}$, there is no need of commutator and voltmeter. So, the ammeter is connected in series with the anode plates and cathode plates with copper wires.  Now, if there is steady state current in the circuit, electrons emitted to any one of the sides should dominate over the other. But the number of electrons emitted on both the sides is same, and all the electrons should face the same resistance as the current(electrons) should flow in a closed loop. But if the number and velocity of electrons coming from both the sides are same, there should be no net current. But there is. Why?  Edit –  From the answers of anna v and mmainville, it can be seen that the emission of electrons is dependent on the angle of emission of light.  But still, the question remains. If the electrons emitted towards anode have enough energy to complete the circuit and again reach the holes in cathode despite the resistance in between, the electrons emitted to the left should have enough energy to come to anode plate via copper wires and again get ejected through anode plate by the virtue of their kinetic energy and reach the cathode plate. In this case too, the current should be zero which is not true. Why?
94504
It is my understanding of the double-slit experiment that when a measuring device is activated, to measure which of the two slits a particle is travelling through, this measuring is responsible for a different outcome than when the device is left off. When the device is left off particles leave an interference pattern, suggesting a single particle (or its wave) interferes with itself; when the device is activated no interference pattern is left, and particles are observed to travel through a single slit.  Now, does all this imply that this experiment is typically carried out in a vacuum? If not, how is it that interaction with other particles in the environment of the experiment's set-up are not already influencing the outcome, regardless of whether the measuring device is on or left off?  This question may very well be a very naive one, due to my very limited understanding of particle interaction, vacuums and/or physics in general. If so, I would appreciate a brief, layman's, overview of where my understanding of the different subjects at play here is likely to be lacking.
94177
I'm thoroughly confused about the nature of electromagnetic radiation. Light is supposed to exhibit both wave and particle characteristics. But does that mean that it is both a wave and a particle or neither a wave nor a particle? If the latter, then it is not a "real" particle at all, but only occasionally behaves as one (hence, a "virtual" particle). And if the former, then it is, depending on the experimental setup being used to detect it, a "real" particle, but one that only exists (as a particle) for a limited time, which again, corresponds to the definition of a virtual particle ("a transient fluctuation that exhibits many of the characteristics of an ordinary particle, but that exists for a limited time" - Wikipedia).  So, if photons invariably appear to exhibit the characteristics of virtual particles, why are they not actually virtual particles (even in a vacuum)?
39248
I want to experiment taking photos in which the sun appears. I'm afraid of what might happen if I take one with a narrower angle (where sun would be bigger). Can the lens act as a magnifying glass and burn the CCD or CMOS sensor?  Under which circumstances (zoom, exposure, aperture, etc...), can the sensor be damaged by the sun?
116428
$E = h\nu$ and $P = h\nu/c$ in vacuum. If a photon enters water, its frequency $\nu$ doesn't change. What are its energy and momentum: $h\nu$ and $h\nu/c$ ? Since part of its energy and momentum have been transferred to water, it should be less.  If water's refractive index is $n$, are the energy and momentum equal to $h\nu/n$ and $h\nu/c/n$ ?
50431
Why the log? Is it there to make the growth of the function slower?  As this is a common experimental observable, it doesn't seem reasonable to take the range from $[0,\infty)$ to $(-\infty,\infty)$ (For a particle emitted along the beam axis after collision $\theta = 0$ wouldn't be better to have a number that says how close it is to zero rather than one that says how large a number it is. I hope that makes the question clear.)
5190
If I change the ISO settings on my camera, obviously the gain of the system is increased, amplifying the signal from the sensor. What's not clear to me is where the amplification takes place. I see several possibilities:   In the sensor, by increasing voltage or some other mechanism Via an analog amplifier outside of the sensor Digitally, after the signal has been digitized, but before storing data in the RAW file As a parameter applied solely in creating an image from RAW   If #4 is true, then you could take a 4-stop overexposed RAW picture at ISO 1600, and then in post processing produce a JPEG at ISO 100 that would be the same as if the original photo had been shot at ISO 100.  If #1 or #2 is true, then a RAW file shot at ISO 1600 would actually contain more information about shadows, and an ISO 100 RAW would contain more information about highlights.