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Posts posted by Klaynos

  1. Thinking of polarisation as just direction is wrong, I would suggest at this point you look at the derivation of Malus's law, when you come across something in the derivation you don't understand, such as elliptical polarisation then you can read there. That's pretty much the reverse of the most efficient way to learn a complex subject and will result in massive holes in understanding but would be better when where we currently are. 

  2. 43 minutes ago, Dalo said:

    Apparently nobody knows why the first polarizing filter absorbs or transmit 50% of the beam.


    This is why the discussions you start don't seem to go anywhere. 

    As a prerequisite to a discussion on entanglement is a knowledge of polarisation. You don't have that knowledge so don't know the bits of information you're missing when reading about the details. 

    Polarisation isn't just the linear orientation of fields. It's more complicated than that. No one wants to spend the time teaching you the basics when you're wandered in out of your depth insisting you're correct. Sorry to be blunt but that's the way it is. 

  3. 7 hours ago, pzkpfw said:

    There is no discussion if you can't explain (or diagram, or whatever) why you think your result would occur.

    Why would you see all five spots?

    (If you think that your step 5 would show 5 spots, given your step 4; it should be possible for you to draw a diagram showing the path you think the light is taking. Does the light do a loop-de-loop? A figure 8? How do you see all five spots when, given your step 4, two of the beams were blocked?)

    I see we've actually gotten somewhere. 

    If dalo would answer this question his misunderstanding would probably become apt clearer to us. 

    Well done. 

  4. 21 hours ago, pzkpfw said:

    This whole 5 laser thing still seems like a pointless diversion, as I think Dalo has misunderstood post #2 (which he seemed to accept).

    Dalo, the diagram showed how light from one point on the source travels multiple paths to the lens. So the diaphram blocks some of the light from points on the source, but not all. Note how in post #2 the diaphram does not cut off the head of the stick figure; just some of the light from the head. I've slightly expanded the diagram, possibly wasting my time.

    Your five lasers do not replicate what happens for general photo taking, as we don't get single sources of light from the source we are photographing. e.g. there won't be a single ray of light from the head of the stick figure to the lens.



    Dalo, what don't you understand about the quoted post and image?

  5. 4 minutes ago, Dalo said:

    when using the filter the beams are not visible anymore, only the light sources. In this sense, it is just like any other scene.

    Your judgment expresses your confidence in the theory, and there is nothing wrong with that. That does not mean that an empirical confirmation would be superfluous.

    Are you suggesting that the filter completely removed the lasers so they are not dominating the response of the sensor and then you are illuminating the scene with some other light (e.g. room lights) and can image the front of the laser sources?

    If so you are mixing your situations and need to review the first few posts of this thread and where it has been explained to you again and again how non laser sources are radiating in all directions so there are rays (in ray optics) hitting all parts of the lens. There are some nice diagrams above showing this. 

  6. 1 hour ago, Strange said:

    To put that in terms of your lasers, if you were to position one at the extreme edge of the field of view and aim it through the edge of the lens then it would be blocked when the diaphragm is shut down. If you leave the diaphragm in that position and, without changing the position of the laser, rotate the laser to go through the centre of the lens then it will again appear in your image. 

    The same thing is true for a laser position in the middle of the field of view. If you aim it at the edge of the lens than it can be blocked by the diaphragm. If you aim it near the middle of the lens, then it won't be blocked by the diaphragm.

    What that shows is that some rays from every part of the scene are blocked by the diaphragm (so it becomes universally darker as you stop down) but some rays from every part of the image will always reach the film/sensor.

    This seems key to the missing understanding. 

    Light from every point you are imaging is radiating in every direction, so in ray optics you can drawn a ray hitting any part of the lens. 

  7. 1 minute ago, Dalo said:


    As long as the experiment has not been done all anybody can do is speculate on the outcome.

    No, we can apply the well understood and incredibly well tested theory's of electromagnetic propagation that we have. This isn't just guesswork or assumption. 

    Scientist don't just run experiments because they seem like a good idea or some guy on the internet has a pile of misconceptions. 

  8. 10 hours ago, swansont said:

    I recall a talk from an AAPT conference I went to in grad school, on teaching methods in optics. Students had been asked what would happen to an image if part of the lens was blocked. The misconception that part of the image would be missing was common. All that does is make the image dimmer.  

    Blocking the rays reduces the amount of light getting to the image. But light from any point on the object will will pass through any point you pick on the lens. The same concept applies to the aperture area — there will always be light going through the center of it, from all points on the object, to form the image.

    If this wasn't the case the diameter of the lens used (both in bench optics and in photography) would matter to the field of view. It doesn't. 

    I am struggling to see how this is such a long thread! The general problem was solved for both a simple case and with a little extrapolation to some camera optics in the first few posts. 


  9. 2 hours ago, swansont said:

    That's not two waves, or a group of waves. That's multiple peaks of one wave. 

    Two pages and a few hours of my sleep later and this (well the op statement that resulted in this response) is still the misconception that I think is fueling the confusion here. 

    Dalo, a sin wave is a representation of a single continuous wave, not a series of waves. 

    Take water, again, each crest (colloquially each wave) is not technically a single wave, bit just one peak on a series of peaks in a single wave. 


  10. 4 minutes ago, Dalo said:

    I must admit that I have never encountered this explanation of wavelength before. Every textbook or video I have consulted/watched, used the water wave model, and my questions were based on that model.

    This opens a new line of questioning. Since they all have the same color (wavelength), and therefore the same diffraction property, shouldn't they be parallel to each other?

    Also, I find the image of wavelength as the distance between peaks of the same wave very intriguing. In particular, I wonder whether the frequency of a wave also has to be calculated the same way?

    Water waves are used as people are familiar with them. That causes issues as the language people use about them is very loose. What is one water wave? What we see hitting a beach is complicated. A wave comprising a single amplitude peak in the time domain is not a single wavelength wave. 

    Look at a single frequency continuous sound wave, that's a simpler case to start with. 

    To add, each crest isn't a new wave. 

  11. 1 hour ago, Dalo said:

    We are apparently not communicating very well. I want to know how the wavelength is calculated. I have read the information in textbooks and watched the videos, and I still have questions. My first question was answered, but it created its own questions.

    The difference in distances between two lines, from the diffraction grating to the screen, is used to calculate the wavelength. I find it strange because both lines come from the same wave, while the wavelength is the distance between two waves.

    How is that possible?

    The wavelength is the distance between two peaks of the same wave. Not two different waves.

    This is confused by water waves which are complex, and confused by language where every peak is often called an individual wave. 

  12. 10 hours ago, Endy0816 said:

    This might help:


    Like Swansont is suggesting need to consider the convex lens and cornea of the eye.

    If you have a magnifying glass or a farsighted friend, you can see the image flip for yourself.


    Good diagram. Demonstrates exactly why drawing a ray diagram as the first step would have helped op. 

  13. We don't answer questions directly here, but give help. Let me ask you two questions...

    How have you tried to answer this so far?

    Have you drawn a ray diagram of the situation?

  14. 39 minutes ago, Carrock said:

    As the OP appears to want a short range illegal jammer, with a 3m near field for 3cm waves no less, enough leakage for his jammer to be detected by Homeland Security or whatever would not be acceptable.

    I had assumed the shield was instead of a jammer, not to limit its range. The jammer would still be illegal in many jurisdictions. 

  15. 35 minutes ago, John Cuthber said:

    Hold a sheet of foil up to the light from the sun.

    Can you see through it?

    The eye will detect light over a range of something like 12 orders of magnitude.

    If you can't see through the foil it's attenuation must be (at least) something like 1,000,000,000,000 fold.

    Do you agree with my assessment that it is "for all practical purposes" opaque?

    It's much more difficult to do the experiment at other wavelengths because we can't sense them directly.

    You need to be careful using the eye as a detector. It's nonlinear and will self calibrate to the ambient light level. Need to completely cover the eye (e.g. goggles). 

  16. 1 hour ago, Moreno said:

    But far from completely?

    Define completely. The probability of an individual photon getting through will always be non-zero. It depends on your use case. For many an anti-static bag would suffice, others 3m of lead. 

  17. 27 minutes ago, Yaniv said:

    My theory predicts W should decrease at increasing T in vacuum and can be found here <link removed by moderator>

    If you've got a "theory" then use its mathematical framework to give precise numeric predictions. 

  18. 2 hours ago, Moreno said:

    What is the Faraday cage reliable frequency range of shielding? Can it shield from GHz range?

    A microwave shields GHz waves. So yes is the answer to your second question. The answer to your first depends on the configuration. 

    We'd need to know your use case. 

  19. 12 hours ago, koti said:

    I’d say „Ubuntu” and „Suse” but there so many distributions and the whole scene evolves constantly.

    Interesting, I'd say Debian and rhel. 

    In terms of devices, I wouldn't be surprised if raspbian was one of he most used now. 

    On 13/10/2017 at 2:37 AM, Moreno said:

    Is Unix still more secure and reliable than Linux? What about speed anf functionality?


    That depends on which Unix, which Linux and your application. 

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