P3D Re: projection

[jacob@xxxxxxxxxxxxxx (Gabriel Jacob)]

Sorry in the delay in replying (will get to other off-list 
email soon also)!. Note: I replied to this off-list to Tom
a few weeks ago but elaborating on some points with this post.

Tom Hubin writes (or wrote!):

>How old is the Catalog? 

(This was with regards to an old Schott catalog that stated among
other things that the heat absorbing glasses of the KG-range contain
a high percentage of phosphoric acid and would have to be toughened
to withstand weathering.)

The catalog is pretty old! Catalog No.365e XII/62
I think it was printed in 1962, although the date stamped (when
it was received, before my time!) is 1966. 

>Do you know if this still applies to the KG1 and KG3 glasses that they
>sell now? The factory rep did not mention any of this. 

The notes apply to the KG1 and KG3 at that time, and I can only guess
that it still applies. Assuming there hasn't been any manufactering
change in the last 30 years. 

>By toughening do they mean tempering?

I don't know. It could very well be. There is an additional
note in the catalog:

  Roughly polished glasses of the KG-series are:
  1) tempered, to increase their thermal resistance, and
  2) provided with a protective film, to substantially 
     decrease the effects of weathering.

I wrote:
>> Actually, not all the energy will be converted to heat. Varying
>> degrees of light energy will be transmitted and illuminate the
>> screen. If all the energy was absorbed and converted to heat the
>> screen would be black! 

Tom writes:
>Please read carefully. I stated that all of the energy that is
>"absorbed" is converted to heat. 
>The energy that is not absorbed is passed on to the screen to form the
>projected image. 

I understand, but alot of people reading this might come to a
wrong conclusion. Your last line above is not in the report. 

I wrote:
>> Also strictly speaking, infrared and ultraviolet energy is not
>> light since it is invisible. Therefore, the term "visible light"
>>is redundant.

Tom writes:
>I was aware of the distinction as I wrote the report. The term light is
>often used to describe that part of the spectrum that can be used to see
>something. Cameras use IR to produce an image that we see. Some animals
>probably use IR or UV to see. Anyway, it is in common usage since most
>of the equations used for optics and Lasers apply for UV and IR as well
>as the visible spectrum. 

Yes it's true that it is common to refer to certain invisible portions
of the spectrum as light. This is often done informally when person A
and B know in what context they are using these terms. But in more
formal presentations (to technical and lay people) using the correct
terminology is essential in understanding what factors contribute to
film deterioration in a projector for example.   

As to the reference about cameras using IR to produce an image that
we can see, the camera is converting an invisible portion of the
spectrum into something we can see (i.e. light). This doesn't make
the IR portion light. As to animals, it's true they can see some
IR and UV which is invisible to humans but again, this doesn't
redefine that portion of the spectrum to "light". 

As for equations applying to UV and IR as well as the visible 
spectrum this is correct, but since they have different physical
effects on matter (or 3-D slides!) the classifications (IR, light,
UV, etc.) are important to keep distinct.

>I think it is comparable to the question about falling trees making
>sound where there is nobody to hear the sound. Just because I cannot
>hear the tree falling does not mean that it did not make a sound. That
>is, if you use the mechanical engineers definition of sound wave.
>Scientists talk about sound waves in crystals at 50MHz. Can you hear
>50MHz?

No I can't. On the same token, according to some defintions I can
see steam and according to others, I can't! Seriously speaking, the
mechanicals engineers defintion of a sound wave is just a short cut
to decribing the nature of the wave (mechanical in nature). This
doesn't translate well to electro-magnetic waves. Would you call a
"50MHz electro-magnetic wave", light?

>I am a scientist and an optical engineer and certainly want to be as
>correct as possible. But I also want to be understood. So when I write
>for scientists I am usually as wordy as necessary to be technically
>correct. When writing for non-scientists I try to be informative without
>being too technical. 

This is an art! ;-) I agree but you make an important distinction here,
"too technical" and "technically correct". The art is keeping it simple
and technically correct, without being too technical! :-) 

>I know that I am inconsistant in this. I chose to use the term "IR
>absorbing glass" rather than the commonly used "heat absorbing glass". I
>chose to do this since the report is about that subject. So extra detail
>seemed like a fair thing to do for educational purposes.

That's what I noticed (regarding consistency) and is the only
reason I made the additional comments. Difficult to keep true
to oneself! I'm guilty of that often myself!
 
Tom wrote:
>>>The glass absorbs infrared energy and radiates heat.

I wrote:
>> The glass absorbs infrared energy (as you mention), but it
>> doesn't radiate heat but rather infrared energy. Something that
>> "radiates" is of electromagnetic nature. As you mentioned earlier,
>> infrared energy is not heat. The main reason the surrounding air
>> temperature increases is because of the conduction path of the
>> heat from the glass to the air.

Tom replies:
>I would debate this. The glass absorbs the IR energy and as a
>consequence gets very hot. In fact, the entrance surface is much hotter
>than the exit surface. If it operated in a vacuum the glass would still
>be very hot. As air comes in contact with the glass the heat is passed
>from the glass to the air by conduction. Then from air to other air by
>convection as the air travels.

I think you missed my point. We're in agreement as to the heat
propagation mechanism (conduction, convection) that contributes
to the high temperatures. The point I'm making is that the
the heat is not "radiated", only infrared energy can radiate. See
my post above.

P.S. Thanks for shining light and not IR on this subject! ;-)

Gabriel


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End of PHOTO-3D Digest 3215
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