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^ we're in the midst of a bad spell of weatherwether forecast suggest you will get soggy tissue
^ we're in the midst of a bad spell of weatherwether forecast suggest you will get soggy tissue
During the eclipse last Friday I saw a guy with a eos 5D/70-200mm f4 combo photograph the eclipse with no protective filters on his camera or no protective eyewear. There was no damage to his eyes or to his 5D.
It was probably this thread. Though the actual cause never seems to have been completely resolved (though the camera was repaired), it looks as though it was shutter damage rather than sensor.I read a post within the last few months from someone who presented prima facie evidence of a "hot spot" of increased sensitivity that was apparently created by photographing the sun. It might have been on this forum. That's the only case I can recall of sun damage to a sensor, assuming that the hot spot was caused in this way. I can't recall what the final verdict was.
The problem is that "the sun" starts out as a small point of light with a wide angle lens...it contains less energy (proportional to the scene). I doubt it will be enough to burn even tissue paper, but maybe.My point precisely. I can use a wide and a long lens to focus the sun at the film plane. The wide angle lens will focus the sun to a much smaller area than the long lens will.
My expectation is that it will be possible to burn a hole in some paper with the wide angle lens but not the long lens because the long lens does not focus to a point of light.
I thought I was agreeing with your response...Regardless how you meter you'll either blow out the sun or block out the rest of the image, I'm not sure what your point is.
Total energy remains the same, area of effect is the issue, silicone and the components can deal with with X heat per sq mm after that things burn out.
I'm quite sure that during an eclipse, with the sun being blocked by the moon, that there is a whole lot less light energy getting to the sensor than in my example image with a bright midday sun in it.During the eclipse last Friday I saw a guy with a eos 5D/70-200mm f4 combo photograph the eclipse with no protective filters on his camera or no protective eyewear. There was no damage to his eyes or to his 5D.
A magnifying glass will do that, but a camera lens will not.
I thought I was agreeing with your response...
The size of the sun in the image changes the exposure due to the "average gray" metering and not due to the total energy collected. It's like cropping an image... if you crop into just the sun the histogram will shift but the exposure (energy per area) does not.
The light doesn't... it's the "optical design" of the lens system that makes the difference. And a camera lens will not focus it's entire FOV into a single point... it's not designed to do that. It's designed to project an image circle.Please explain how the light knows the difference between a camera lens and a magnifying glass.
Steve.
But with a camera lens if you change the (flange) distance you loose focus and have to refocus the internal elements
I checked one of mine... focusing on clouds (close to infinity) it's not f/1 either.I missed post 96.
I just happen to have a magnifying glass on my desk, and I just measured its focal length as about 7". The diameter is 2.5". It isn't f/1...
That was an easy lens - it was circular. I also have a rectangular magnifying glass.
I think the reason it can/could burn is due to a lack of UV filtration (as in the days of film/cloth shutters)... you can't burn something by focusing a "full spectrum visible light bulb" or other "visible light" source (that I am aware of anyways). I believe the reason it is only ever "pinholes" that are burnt is because it is the focus of high energy (UV) light to a point that causes the damage and not the visible light we can identify/see. Meaning both a long FL and a wide FL will burn "pinholes" if focused to do so.The reason I think that it's much more likely with a wide angle lens is that pinholes burned into cloth shutter curtains are well documented. If longer lenses could do the same damage, there would be reports of large holes in shutters - but it's only ever pinholes.
All we need to prove the point is a bit of tissue paper, a couple of lenses and the sun. Not sure when the sun will be willing to take part!
Steve.
Meaning both a long FL and a wide FL will burn "pinholes" if focused to do so.
Steve - I suspect that this "focus to a point" issue is down to the image of the sun produced by a short focal length lens being smaller than that produced by a long focal length one and hence tending more to a point image of the sun. Your earlier point - that it's the energy per unit area that matters - still seems to be the actual factor involved.
I was curious about this question of what kind of lens made the best burning glass more than sixty years ago. So I carried out the appropriate experiments.
A source that starts as a point of light and is "in focus" on the sensor is again a point of light. But maybe I shouldn't have said "pinholes" but rather a longer lens will burn a larger "pinhole."How can a long lens focus UV to a point?
(I'm not arguing that it can't but would like to know how it can).
Steve.
Simple experiment, no camera involved, just a wide-angle and a telephoto, focusing the image of bright sun on a piece of black paper. With the longer lens, 200mm f/4, the paper begins to smoulder after a couple of seconds. With the wide-angle 17mm f/4, I couldn't get it to burn at all.
Pretty obvious then that longer lenses are much more of a problem, but in practise much less likely to be an issue just by leaving the camera sitting in the sun. The longer focal length ensures that the sun is much less likely to be in the right position (field of view) and less likely for the image to be in focus, too.
Yes...Ive been thinking about this some more, and I have an argument as to why there will be no difference between a wide lens and a long lens of equivalent aperture (f ratio, not absolute).
Lets take a 14mm f2.8 and a 200mm f2.8, both at f2.8, both pointing at the sun. If both lenses take a shot of 1/8,000s, they will both be equally exposed, so that means an equivalent number of photons land on each of the photosensors the sun is focused on (there will be more energy overall with the longer lens, but I think what is imporatant is the energy per area). I think this means that the energy / heat will also be equal.