Really stupid physics question

[amolitor]

It doesn't matter what you say, or what pictures you post, or what references you cite, guys. Tris is going to either ignore your post, or dismiss it on some trivial quibble (which might be real but will more likely be nonsensical), and then request that you "please just do one more little thing."

It doesn't end. It's about attention, not learning.

 

[unpopular]

this is the only solution.

 

[amolitor]

The question is, really, who's head to we bang here?

For extra fun, google "tris_d inverse square law". Tris has been fighting this one across the internet for a few days now, and the battle still rages. It's about attention, and Tris is winning.

 

[christop]

Here's a version of my previous post with ASCII art illustration!


Papers:
.....A.........................B
.....|................o........|
......<-----.2m.----->.<-.1m.->

Lenses:
.....A.........................B
|....|................o........|....|
......<-----.2m.----->.<-.1m.->


(Sorry, this forum doesn't like to preserve spaces, and it doesn't have a "code" tag for some reason. Just ignore the periods.)

Paper A and lens A receive the same amount of light. Lens A passes that energy onto the surface behind it. Image A is the real image of the spherical light that lens A projects. Image A consists of 1/4 as much light energy as image B, but image A also has 1/4 the area as image B.

 

[tris_d]

It is mentioned in many textbooks, and in many places on the internet, but I guess that you can't recognize it for what it is. Here's one version from Optics in Photography by Rudolf Kingslake, who used to be the Director of Optical Design at Eastman Kodak, and Emeritus Professor at the University of Rochester. He also received the Gold Medal from The International Society for Optical Engineering. I don't think he has a blog, or even a Facebook page, so I'm not sure if he can really be trusted, to be honest.

E = t &#960; B / (4 N^2)

Where
E is the image illumination,
t is the lens transmittance (dimensionless),
&#960; is Pi (dimensionless),
B is the object illumination, and
N is the f-number (dimensionless).

Note the absence of any term relating to the distance to the object.

Great, thank you. It's just that astronomers use brightness to calculate distance to stars, and if distance is supposed to be irrelevant then you should see where my questions are coming from and why I can not settle with such answer. Ok, so the answer to my confusion is that stars are point light sources and thus their brightness is actually related to distance, right?

 

[fjrabon]

It is mentioned in many textbooks, and in many places on the internet, but I guess that you can't recognize it for what it is. Here's one version from Optics in Photography by Rudolf Kingslake, who used to be the Director of Optical Design at Eastman Kodak, and Emeritus Professor at the University of Rochester. He also received the Gold Medal from The International Society for Optical Engineering. I don't think he has a blog, or even a Facebook page, so I'm not sure if he can really be trusted, to be honest.

E = t &#960; B / (4 N^2)

Where
E is the image illumination,
t is the lens transmittance (dimensionless),
&#960; is Pi (dimensionless),
B is the object illumination, and
N is the f-number (dimensionless).

Note the absence of any term relating to the distance to the object.

Great, thank you. It's just that astronomers use brightness to calculate distance to stars, and if distance is supposed to be irrelevant then you should see where my questions are coming from and why I can not settle with such answer. Ok, so the answer to my confusion is that stars are point light sources and thus their brightness is actually related to distance, right?

The answer is that stars are indiscernibly small from the distance they are, and thus we only perceive the drop in brightness, and cannot perceive their decrease in size. However, they still emit the same amount of light no matter how far they are, until they are no longer in our hubble sphere, and thus, are invisible to us, because the distance between us and them is expanding faster than the speed of light, and thus their light never reaches us.

 

[tris_d]

It doesn't matter what you say, or what pictures you post, or what references you cite, guys. Tris is going to either ignore your post, or dismiss it on some trivial quibble (which might be real but will more likely be nonsensical), and then request that you "please just do one more little thing."

It doesn't end. It's about attention, not learning.

Stop trolling me, please. You are not contributing and your comments are unnecessary. If you don't care just mind your own business. I think we are now close to wrap this thing up, so again, please don't pull my tail. We are so close, don't get me banned now.

 

[fjrabon]

It doesn't matter what you say, or what pictures you post, or what references you cite, guys. Tris is going to either ignore your post, or dismiss it on some trivial quibble (which might be real but will more likely be nonsensical), and then request that you "please just do one more little thing."

It doesn't end. It's about attention, not learning.

Stop trolling me, please. You are not contributing and your comments are unnecessary. If you don't care just mind your own business. I think we are now close to wrap this thing up, so again, please don't pull my tail. We are so close, don't get me banned now.

You've been provided with pictures disproving your theory. How are we 'close to wrapping this up'? It's over, just like we all told you from the beginning.

 

[unpopular]

I don't even know why you even thought this would be a good forum to discuss this. Helen is the only one qualified to concisely answer your question, and she has. If she weren't here, then what could you expect?

Either post some photos, or GTFO.

 

[Helen B]

It is mentioned in many textbooks, and in many places on the internet, but I guess that you can't recognize it for what it is. Here's one version from Optics in Photography by Rudolf Kingslake, who used to be the Director of Optical Design at Eastman Kodak, and Emeritus Professor at the University of Rochester. He also received the Gold Medal from The International Society for Optical Engineering. I don't think he has a blog, or even a Facebook page, so I'm not sure if he can really be trusted, to be honest.

E = t &#960; B / (4 N^2)

Where
E is the image illumination,
t is the lens transmittance (dimensionless),
&#960; is Pi (dimensionless),
B is the object illumination, and
N is the f-number (dimensionless).

Note the absence of any term relating to the distance to the object.

Great, thank you. It's just that astronomers use brightness to calculate distance to stars, and if distance is supposed to be irrelevant then you should see where my questions are coming from and why I can not settle with such answer. Ok, so the answer to my confusion is that stars are point light sources and thus their brightness is actually related to distance, right?

Aaagh. We're going round in circles. The above equation does not contradict the inverse square law at all.

 

[tirediron]

I think we've pretty much done this to death... perhaps the OP could start a new thread to explain why it's possible to divide by zero?