Would passing horizontally polarized light through a varying width vertical slit allow you to measure the positional (x) amplitude of light? I have found closely related questions on StackExchange, but (surprisingly) not this exact question.  Seems some answers say individual photons do not have amplitude, only when traveling with other photons, forming a wave.  But even then, can't the amplitude be measured?  If so, what are the range of amplitudes and how do they vary?  Based on the frequencies and energies of the light?  Does higher frequency correspond to lower amplitude?  Moderators: This question is not a duplicate to this question:   That prior question is more about the formula for calculating electromagnetic energy amplitude of a single photon. My question is about an experimental way to MEASURE the positional (x) amplitude of the wave passing through a specific apparatus.  Two very different things.  Thanks.  I have updated my question to a more specific thought experiment:  If you had a vertical slit with a horizontal width that could be varied, and you passed horizontally polarized light through this slit, wouldn't the slit block any photons from passing through it if the width of the slit was smaller than the amplitude of the light waves?  Wouldn't we at least get a predicted or average amplitude, even with uncertainty principle?  Wouldn't the energy/frequency/wavelength of the wave also have some correlation to the amplitude?
Amplitude of an electromagnetic wave containing a single photon Given a light pulse in vacuum containing a single photon with an energy $E=h\nu$, what is the peak value of the electric / magnetic field?
If $\alpha_{1},\ldots,\alpha_{m}$ are vectors different from zero vector, then there is a linear functional $f$ such that $f(\alpha_{i})\neq 0$ I get this warning:     Java HotSpot(TM) 64-Bit Server VM warning: Using incremental CMS is deprecated and will likely be removed in a future release   What does it mean? How to cheat and avoid cheating in Flash games? Email from IOS with attachments larger than 7.5Mb gets stuck in outbox and constantly tries to resend. Costs us hundreds in data fees I was watching this  on making a custom UI tab using Python, and in the line from bpy.types import Menu, Panel, UIList, I see the names of objects that look like they can all be used to make different UI elements. Does bpy.types contain all the objects that people use when they want to make custom UI elements? If "I'm thinking" is correct, why isn't "I'm knowing" correct in formal English?  think+ing = thinking is correct know+ing  = knowing isn't correct   Why? How can I reduce the ever-increasing time to push a subtree? Can I run OSX applications? Just wondering, are there any projects like WINE, but for Mac software? That is, can we use a program that acts like a compatibility layer to run Mac apps? In which US accent is “again” pronounced /əˈɡeɪn/ with the FACE vowel, not /əˈɡɛn/ with the DRESS vowel? Most dictionaries provide the US pronunciation of “again” as /əˈɡɛn/ (uh-gen) with the DRESS vowel. This is the most common pronunciation in the USA. However, I think I might have also come across /əˈɡeɪn/ (uh-gain) with the FACE vowel. This sounds like the word “gain” with an “a” at the beginning. Is there a US accent where “again” is pronounced this way? How many attacks can a fighter with two-weapon fighting make at 5th level? I have a 5th level fighter with the Two-Weapon Fighting fighting style.   What is the number of attacks I can make?   I say 4 attacks, but others say 3 because they say the rule on page 72 states that I get (only) 1 extra attack. After Upgrade to Ubuntu 20.04: "Oh no! Something went wrong" Limit of $(\cos1/n)^{n^2}$ when $n\to\infty$ I've been trying to solve this limit: $$\lim_{n\to \infty} \left(\cos\left(\frac{1}{n}\right)\right)^{n^2}$$ I solved it using l'hopital but I have to try in another way. I tried to make the expression:  $$ \lim_{n \to \infty}\exp\left(n^2\ln\left(\cos\left(\frac{1}{n}\right)\right)\right)$$  but it didn't work out for me... Is there a standard pronunciation for gender-neutral -@ or -x More and more I see, especially in activist communities, Spanish-derived words ending in x or @ in order to neuter the gendering inherent in the original language. For example: latinx (or latin@) as opposed to latina and latino. How, though, are these endings pronounced? Please provide evidence of how this orthography is most commonly pronounced. Note 1: I recognize this is in an English Language site, rather than a Spanish language site, but this phenomenon is very much happening in English, albeit with Spanish loan words. Therefore the question is relevant to non-Spanish-speaking English speakers. Note 2: A question like this can lead to politically charged or, at least, sarcastic comments. Please refrain from bringing in your biases for or against. Note 3: This question is not seeking opinions as to how some people pronounce the suffixes. Answers should address evidence of a developing orthodoxy in pronunciation. I would like to make a debian system where you can only open one GUI program at boot, no other graphical interface, no minimize, or any X apart from that program, is there a way to do it on debian if possible, or any other custom distro? I just want to boot open the program and allow the user to only see and use that program. Having trouble adding objects to NSMutableArray in Objective C The Next tag badge block overlaps the badge score block. Undergraduate/High-School-Olympiad Level Introductory Number Theory Books For Self-Learning S(x), which stands for “x is a surgeon”  P(x, y), which stands for “x is a patient of y”  R(x), which stands for “x recovers”  H(x), which stands for “x is happy”  1>"A surgeon with no patients is happy" means: for any person xx, if xx is a surgeon and there is no person yy for which yy is a patient of xx, then xx is happy.  2> "A surgeon is happy if all her patients recover" means: for any person xx, if xx is a surgeon and - if for every person yy for which yy is a patient of xx, we have that yy recovers - then xx is happy.  I have these as solution:  1:∀(x) ( S(x)∧ ∀(y)P(y,x) → H(x) ) 2: ∀(x) [ { S(x)∧ (∀(y)P(y,x) ∧ R(y)) } → H(x)]  Are these okay?  And how do prove that 1 follows from 2 by resolution (not by drawing a truth table)? 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