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As you say, focussing the laser/zooming is more powerful & raising the laser away from the subject is like shooting `wider`?
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This Burning Out Of Sensors Through Pointing A Camera At The Sun
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As you say, focussing the laser/zooming is more powerful & raising the laser away from the subject is like shooting `wider`?
Steve Smith
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No. Raising it makes the dot of light larger - just as using a longer focal length lens would with the image of the sun on the sensor.
The longer the focal length, the weaker the Dioptre value of the lens so it's ability to concentrate the light reduces.
Steve.
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Steve Smith
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Think of it another way:
Without a lens, you get the direct light radiated from the sun.
If you added a lens with a focal length of 31.5 million miles and placed it 31.5 million miles away from the earth (a third of the earth to sun distance) the sun's image would be projected onto the earth at half its full size. The image would therefore have a quarter of the area. The light density would therefore be four times as much as a full size sun image.
Keep increasing the power/reducing the focal length of the lens and the image of the sun gets smaller and smaller whilst the light density concentration gets higher and higher.
When you get to a short focal length, you have so much energy concentrated that you can start to do damage to sensors - or pieces of wood or insects with a magnifying glass!
Steve.
Without a lens, you get the direct light radiated from the sun.
If you added a lens with a focal length of 31.5 million miles and placed it 31.5 million miles away from the earth (a third of the earth to sun distance) the sun's image would be projected onto the earth at half its full size. The image would therefore have a quarter of the area. The light density would therefore be four times as much as a full size sun image.
Keep increasing the power/reducing the focal length of the lens and the image of the sun gets smaller and smaller whilst the light density concentration gets higher and higher.
When you get to a short focal length, you have so much energy concentrated that you can start to do damage to sensors - or pieces of wood or insects with a magnifying glass!
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The question of possible risk and the severity of the consequences of being wrong is being confused with real risk. There are lots of variables which need to be quantified before it is known for a fact that looking through a particular telescope will be more damaging than looking through a cardboard tube. But you require a certain amount of scientific understanding to be able to do that, and of course there's always the risk that you might make a careless mistake in your assumptions or arithmetic. Given that the consequences of being wrong could be dire, the safest option is simply to say DON'T TAKE THE RISK -- don't even consider looking through a telescope.
Which is why it's "well known" that looking through a telescope at the sun is very very dangerous. Except that isn't quite correct. It COULD be very very dangerous. But it's not an inevitable consequence of a long focal length. There are circumstances in which looking at the sun through a telescope would be perfectly safe. There are lots more circumstances in which focusing the sun through a long focal length lens on a sensor will not damage the sensor. Because sensors are tougher than retinas. Plus they have IR & UV filters in front of them. Plus all the glass in a camera lens (usually a lot more than in a telescope) shuts out more of the IR and UV on its own. Of course you CAN damage a sensor by focusing the sun on it through a lens. But less easily than you can damage your eye. Also a lot less expensively. So it makes sense, if you're one of those people capable of understanding the science, & willing to accept the consequences of being wrong, to think about the details and evaluate the boundary conditions of the real risk.
Steve Smith
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At the weekend I am going to put some tissue paper in place of the film in an old Zenit body and focus the sun using a 28mm lens and a 300mm lens and see what happens.
Assuming the sun wants to come out to play!
Steve.
Assuming the sun wants to come out to play!
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wether forecast suggest you will get soggy tissue
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Remember that a digital sensor is essentially a piece of silicon and it is soldered in place at about 300 degrees C. So whilst in some respects it is delicate in others it is pretty damn tough. It is also much more capable of conducting heat away from a single spot than film or cloth or tissue. I'm not saying damage won't occur but it is an entirely different can or worms.
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Er, no it doesn't.
With your laser experiment you are taking the same amount of energy and spreading it over a larger area. Camera lenses of different focal lengths capture different amounts of sunlight.
A wide angle lens captures sunlight falling on a teeny-tiny portion of the lens. It is this teeny-tiny amount of energy that is focused on the sensor.
A Telephoto lens captures sunlight falling over a large area of the lens. It is this large amount of energy that is focused on the sensor.
Don't try this at home -
Look at the sun through a carboard tube - (toilet roll centre is about right). This will 'project' a small image of the sun onto your retina. But, if you look for long enough, it'll still be enough to do some damage.
Now look at the sun through a pair of 10x25 binoculars (again, please don't). That sizzling sound you can hear is your retina frying even though the sun's image is larger.
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The amount of light (and therefore energy) captured by a lens is essentially related to the diameter of the front element. Doesn't matter if it is wide angle or telephoto (hence why bigger diameter telescopes are better, they capture more light). It's what happens next that you are thinking of, a wide angle lens focuses a lot of that energy down to a small spot, a telephoto will spread that energy over a bigger portion of the sensor
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No, no, no. Look at the ray diagrams.
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It doesn't, but think about it in terms of exposure. If you were to take a reading using a wide angle lens including the sun it would be different to the reading you would get using a telephoto lens pointed directly at the sun. This must mean that more energy (light) is reaching the sensor (and therefore being collected by the lens) in the case of the telephoto than the wide angle.
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Or you'd meter for the landscape and let the sun blow out, if you exposed for the sun then you'd have big scene of shadows. If you applied the same settings you'd either have a white largely white frame on the tele instead of just a pin point of white.
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OK, here's my experiment to show that the amount of light captured from an object is dependant on the focal length of the lens. For this experiment I used a Tokina 11-16mm (at 11mm) and a 100-400 (at 400mm). Both use a 77mm filter and both have a similarly-sized front element. In each case I used an aperture of f5.6 and manual exposure. The first shot is each pair is with the bare lens. The second is with a mask placed in front of the lens to block all but the central part. Here is the mask -
As you can see most of the lens has been blocked.
As the Sun is a bit dangerous to shoot I used a replacement (a dog's frisbee). Here's the wide-angle image with no mask...
That yellow dot is the Sun. It looks to be around the right size. Now the same image, with the same exposure, but with the mask in place....
Well, there's a surprise. The Sun looks exactly the same. It matters not one jot that most of the lens is masked. I would suggest that is good evidence for my suggestion that it's only sunlight from a very, very small are that is focused to an image in an UWA lens.
Now let's see the Sun at 400mm...
And now with the mask in place -
Hmmmm! What happens if I boost the exposure by 5 stops -
So, there is an image there - just a very faint image. Which is exactly what one would expect to see if the telephoto image was formed from sunlight that hits a large portion of the lens.
My explanation for these results is that the UWA only forms an image of the Sun from light that falls on a tiny part of the lens. So it's only going to focus a small amount of light. The tele lens forms an image of the Sun from light that falls on a large area of the lens - so it focuses a large amount of light.
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Entirely depends on how the metering works, spot, evaluative etc. This whole debate is a bit like saying if I put a lens cap from a 77mm diameter wide angle lens on the ground less sunlight will fall on it than would fall on a 77mm diameter lens cap from a telephoto lens.
The difference is how the energy is focused BY the lens, not the energy incident on the lens.
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The difference, as shown by my experiment above, is how much of the energy falling on the lens is used to form the image. With an UWA lens only a very small portion of that light is used to form the image (which is why masking the rest of the lens has no effect on the image). With a tele lens a lot of the light falling on the lens is used to form the image (which is why masking most of the lens makes the image darker).
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But that small amount of energy is concentrated on a few pixels, less surface area means lower heat dissipation means more likely to damage those pixels.
Honestly I doubt it matters, silicon is very sturdy stuff much more so than anything you'd attempt to brutalise with a magnifying glass.
Honestly I doubt it matters, silicon is very sturdy stuff much more so than anything you'd attempt to brutalise with a magnifying glass.
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But the lens focuses light from a completely different field of view. Light falls on the surface of a lens from many different angles - not a single point - light bounces off all surfaces it hits in all sorts of angles. The telephoto lens collects all the light falling on it from a narrow field of view. The telephoto collects light falling on it from a wide field of view. Each then focuses it onto a sensor (or film, screen etc). For a telephoto lens a lot of the light that is coming in from the side never makes it to the sensor where it would in a wide angle.
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I don't think there is any difference between the telephoto and the WA lens in regards to light energy to damage the sensor. With a wide angle the "sun" starts as a small spot w/in the FOV (less energy) and is focused as a small spot. Where w/ a telephoto the sun starts as a larger spot w/in the FOV and is focused as a larger spot on the sensor. The net effect for "the focused sun" is equivalent energy per area.
This is different from using a simple lens (magnifying glass) to start a fire. In this case you are not focusing "a scene" where all of the light is collected and distributed (focused) proportionally. Instead you are taking all of the light the lens gathers and focusing it into a single point (as much as possible)... it doesn't project "an image."
This crude/simple drawing shows what happens with a point of light. It starts as a focused emission point (reflection), spreads out, gets collected, and is focused back to an equivalent sized point. The distance between F 1 and Pt 1 is the lens' focal length... and in the case of a telephoto Pt 1 may lie some distance beyond the objective element (i.e. the FL is greater than it's physical length). In a wide angle/normal/simple lens design Pt 1 and Pt 2 may be (nearly) coincident (often referred to as the lens' nodal point) and they exist w/in the lens' physical limits.
View attachment 33483
The main risk to a digital camera are heat buildup in the electronics (silicon can take an awful lot) and UV damage (fading) to the color dyes used for the bayer array. I have taken images of the sun using live view...it's the only "safe" way to focus.
Grass by skersting66, on Flickr
The main reasons for using heavy filtration are to block enough light so that you can get an image w/in available settings, and to protect your eye should you be focusing thru the viewfinder.
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Don't go there
sk66
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So should I add a UV filter to my lens?
That's actually a reasonable question... but for it to be of any benefit it would need to block a spectrum that is not already blocked by the built in filtration (i.e. maybe deep UV to which the sensor is not sensitive and therefore isn't already filtered).
In other words, it's probably of no benefit.
And it's a little misleading for me to have said "UV" damage/bleaching... that will happen over time regardless of the spectrum the dyes are exposed to. It's just that the higher energy spectrums (i.e. UV) cause more damage quicker.
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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.
As a very reasonable approximation (the assumptions are ones about optical corrections that should be pretty much true for photographic lenses) the focused image of a subject will be less bright than the subject by an amount that depends on the inverse of the square of the aperture. This is independent of focal length. Given that a long focal length lens will create a larger image of the sun than a wide angle one, it seems to follow that at the same aperture the only difference would be the area of the sensor that is exposed to the heat/light. The total amount of light will be greater with the long focal length lens, but the light per unit area would be the same. On that basis, a wide angle lens would damage fewer pixels than a long focal length lens; but both would cause damage (if damage occurs).
That's as far as I'm prepared to go.
As a very reasonable approximation (the assumptions are ones about optical corrections that should be pretty much true for photographic lenses) the focused image of a subject will be less bright than the subject by an amount that depends on the inverse of the square of the aperture. This is independent of focal length. Given that a long focal length lens will create a larger image of the sun than a wide angle one, it seems to follow that at the same aperture the only difference would be the area of the sensor that is exposed to the heat/light. The total amount of light will be greater with the long focal length lens, but the light per unit area would be the same. On that basis, a wide angle lens would damage fewer pixels than a long focal length lens; but both would cause damage (if damage occurs).
That's as far as I'm prepared to go.
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So, are you willing to look at the sun through a pair of 10x30 binoculars while I look through a cardboard tube? After all, the bins will be doing exactly what a tele lens will be doing
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Are you certain about this? I have to admit that I'm not, as the light from the binoculars will have been concentrated down to a narrow beam that can pass through the eye's aperture, whereas the unfocused light from the tube will not have been. On this assumption, the eye itself will stop more light from the tube from reaching the retina than it will when the beam has been narrowed by binoculars.
Whatever the final opinion - and I can see that this will run and run - the point at issue is whether a sensor can be damaged by (prolonged) exposure to the sun. And if it can't, then the amount of light doesn't matter. If it can, then the amount of light sufficient to cause damage should be quantitified so that we can then discuss what methods would enable this threshold to be reached.
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And a cardboard tube is doing the same as a wa lens? How come they're so expensive then?
Steve Smith
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I still intend to do the test with tissue paper in an old SLR using wide and long lenses. Real life results trump speculation every time.
Steve.
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So are we saying that even with using a solar filter there is a risk of damage to a digital sensor with long exposures or if using live view?
If so, how was it successfully recorded on video?
If so, how was it successfully recorded on video?
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sk66
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I think damage may be possible, but I don't think a single quick exposure will do it. As I demonstrated, I've taken images of the sun directly (more than a couple of times) w/o issue.
I don't think damage is likely regardless unless the sensor's exposure is extreme. The example image I posted is one of five I took in sequence (refining the focus and flare).No. That's what solar filters are for.
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sk66
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You'll find that you can't focus a camera lens to project a single point of light.
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Uhm, no. I'm not willing to look at the sun directly in any manner....
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Or if you spot metered the sun, the exposure settings would remain constant...
It has no relevance to "total energy collected" is has to do with average gray metering.
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No, what you've done is (essentially) created a fixed aperture physical diameter for both lenses... By definition that means the "f/stop" you've created is larger for the wide lens and it is smaller for the telephoto. And that means the telephoto image will be dimmer w/ the same settings.
The reason aperture numbers are constant between lenses, and not aperture diameters, is because longer FL lenses need to collect more light to compensate for light falloff as it travels through the longer FL distance. And it's why some zoom lenses have a variable max aperture... the physical diameter doesn't change, but it's size relationship to the FL does.
Steve Smith
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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.
Going back to the starting fires with magnifying glasses point, magnifiers usually have short focal lengths and they concentrate the sun into a small point of light which has enough concentrated energy to burn. If you had a weaker magnifying glass which focused the sun to a 1" diameter, I doubt that you could start a fire with it.
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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.
Steve Smith
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Hold a piece of paper up so the sun shines on it with no lens - result: the paper doesn't burn.
Hold a weak lens (e.g. 2 Dioptres) in front and focus the sun's image on the paper - result: A large image of the sun on the paper, not enough heat to burn.
Hold a strong lens (e.g. 35 Dioptres) in front and focus the sun's image on the paper - result: A very small image of the sun on the paper with enough energy to start a fire.
Relating that to camera lenses, the weak 20 Dioptre lens has a focal length of 500mm and the stronger 35 Dioptre lens has a focal length of 28mm so equivalent to a long (tele) and short (wide) lens. Replace the paper with a bit of film or a sensor and the only difference remaining is the camera body - which makes no difference.
Note : The first two points also explain why looking at the sun through binoculars is different to using a cardboard tube).
Steve.
Hold a weak lens (e.g. 2 Dioptres) in front and focus the sun's image on the paper - result: A large image of the sun on the paper, not enough heat to burn.
Hold a strong lens (e.g. 35 Dioptres) in front and focus the sun's image on the paper - result: A very small image of the sun on the paper with enough energy to start a fire.
Relating that to camera lenses, the weak 20 Dioptre lens has a focal length of 500mm and the stronger 35 Dioptre lens has a focal length of 28mm so equivalent to a long (tele) and short (wide) lens. Replace the paper with a bit of film or a sensor and the only difference remaining is the camera body - which makes no difference.
Note : The first two points also explain why looking at the sun through binoculars is different to using a cardboard tube).
Steve.
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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.