Klaynos

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.

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.

In the last few years we've quietly disproved most of these things. Almost everyone in the western world and many people in the world in general now all carry acceptable quality cameras with them all the time. Yet good photos just hasn't appeared.

- The Guide Mark II.

It is a simplification of E^2=p^2c^2 +m^2c^4 Which is derived in special relativity. Also, E=mc must be wrong as it fails dimensional analysis.

I see we've actually gotten somewhere. If dalo would answer this question his misunderstanding would probably become apt clearer to us. Well done.

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

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.

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.

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.

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.

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.

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.

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.

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

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?

Pages never seem to finish loading for me at the moment.

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

I agree entirely. And agree with the sentiment of your previous post. Eyes are just annoying optical instruments...

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).

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.

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

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.

Many years ago now people in my research group did florescence experiments. We used IPA all the time for stuff. If there was an issue I'm sure they would have noticed. Edit:crosspost with John.

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