[paul.fr]

On a sunny day, i can burn paper with my magnifying glass. If i had a really good glass, could i burn through steel? or maybe the earths core!!!

[another_someone]

You can certainly burn through steel (though, it is not just how sunny it is, but how big the magnifying glass).

In theory you could create enough concentration of heat that you could melt the surface of the Earth, but since you are only focusing one one point, you will only melt the absolute surface, and not the layer beneath (at least, not without moving the magnifying glass to shift the point of focus of the light, and continue moving is forward as you melt each level of the surface layer).  You would ofcourse need a very very large magnifying glass to collect enough light to do this with.

[paul.fr]

Excellent. How big would it need to be, to burn a hole throuh steel?

[Seany]

I'm not sure about burning a hole.. But to melt it.. I assume you'll only need about the size of ... 1m squared?

[lightarrow]

On a sunny day, i can burn paper with my magnifying glass. If i had a really good glass, could i burn through steel? or maybe the earths core!!!

The light intensity you can focus on a small area doesn't depend on lens diameter, and is limited from lens aberrations, first chromatic and then spherical aberration; If you take a large lens with high focal distance, aberrations are less, but the effect of chromatic aberration, for that purpose, is actually increased because of the higher distance from the various wavelenghts' focal planes.

So it's not so easy to reach the steel melting point.
However, you could use a mirror, instead; I assume it would be better for that purpose.

[another_someone]

The light intensity you can focus on a small area doesn't depend on lens diameter, and is limited from lens aberrations, first chromatic and then spherical aberration; If you take a large lens with high focal distance, aberrations are less, but the effect of chromatic aberration, for that purpose, is actually increased because of the higher distance from the various wavelenghts' focal planes.

I agree that the aberrations will cause a defocusing of the lens, but I cannot see that it is not dependent on lens diameter, since the larger the lens, the larger amount of light must be being focused (in effect, you have a wider aperture lens).  Also, the larger the lens, the less are likely to be the aberrations, since most of the aberrations tend to be towards the outer parts of the lens, and a larger lens will also have a larger core region (OK, you have mentioned that you would get fewer spherical aberrations, but I would have thought that also you would have less chromatic aberrations - and I am not sure why you think a larger lens would have a longer focal length - that will depend on the curvature, but not the diameter).

[Bored chemist]

You can correct for spherical and chromatic aberation to a fairly good degree. On the other hand the size of the focussed spot depends on the size of the sun and the focal length of the lens. A shorter focal length will give a smaller spot but it's difficult to make big lenses with short focal lengths (there are limits to how big a lens you can make for a given focal length) and the bigger the lens the more difficult it is to correct for spherical aberation.

[lyner]

It's just got to depend upon the actual size of the steel object and the total amount of energy being focussed on 'the spot', more than how small you can make the spot.
"A good big un will beat a good little un" any day.
The steel will conduct away so much heat and reduce the final temperature of the  hot spot so, unless you can produce  a large power flow into your target area, you have no chance of melting a large object.
In the end, you can only rely on  less than 1kW per m squared, whatever you want to do.
Electric arc welders use a few kW of power, in practice.
A huge lens - 1m squared - will direct the best part of 1kW on your spot  on a sunny day and a bit of aberration won't matter much at all. 1kW would melt a small  hole in a thin sheet of steel easily.
It's a different matter if you just want to melt a tiny sample, particularly if you could keep it in an evacuated glass jar (optically flat face, of course). In that case you could melt a small piece of steel with  less than 1W -  so a lens of area  0.001m squared would do - just a few cm across. The basis for this argument is that a small light bulb  filament (1W power) can't be made of steel, but has to be made of tungsten to avoid it melting.

[lightarrow]

The light intensity you can focus on a small area doesn't depend on lens diameter, and is limited from lens aberrations, first chromatic and then spherical aberration; If you take a large lens with high focal distance, aberrations are less, but the effect of chromatic aberration, for that purpose, is actually increased because of the higher distance from the various wavelenghts' focal planes.

I agree that the aberrations will cause a defocusing of the lens, but I cannot see that it is not dependent on lens diameter, since the larger the lens, the larger amount of light must be being focused (in effect, you have a wider aperture lens).  Also, the larger the lens, the less are likely to be the aberrations, since most of the aberrations tend to be towards the outer parts of the lens, and a larger lens will also have a larger core region (OK, you have mentioned that you would get fewer spherical aberrations, but I would have thought that also you would have less chromatic aberrations - and I am not sure why you think a larger lens would have a longer focal length - that will depend on the curvature, but not the diameter).

If you have a bigger lens, you will have a greater amount of heat transferred to the object, so the object won't cool as fast, giving away its heat, we agree on this.

However the light's intensity incident on the object's surface, is the same, if the lens optical quality is the same, because you cannot focus light onto a disk with a diameter less than k*D/f, where D is the lens diameter, f the focal lenght and k a constant, because of aberrations; the theoretical limit is due to the sun's disk, as Bored chemist said. So the rate (lens area)/(focused disk area) = incident light's intensity, is the same. Making an analogy, it's like heating an object with a small or a big bunsen with the same flame's temperature: you cannot reach a much higher temperature of the object, just because the flame's temperature is the same.

To increase the rate (lens area)/(focused disk area) you have to change material (higher refractive index) or decrease the focal lenght increasing the lens curvature (higher aberrations) or increase the optical quality.

[lyner]

Yes - that is reasonable but, practically, the spot size is not the only relevant factor. As soon as the image is focussed, it starts to heat up the material and there is a temperature gradient as around the spot as heat is transferred outward. The actual temperature of the spot, in terms of the spectrum of the radiation arriving at the spot is that of the image of the Sun and that's quite hot enough to melt most things!  That is the limiting temperature of any object you want to heat up.  Isn't it all about Stefan's law?
What limits the actual temperature reached is the rate of transfer of heat away from it.
This is turning into a typical Naked Sci  type discussion - we keep shifting the goal posts.
But, In practical terms, it's just got to work best with a lot of energy from a big lens.

[felixtheferret]

Hi people, bit of fun... Somebody mentioned using a mirror instead.  So the thought occurred to me earlier, just for fun, when reading about the National Ignition Facility project, that maybe the 4.2 billion dollars spent on this could be better used simply by finding a big mountain somewhere, just for fun, and building a humungous 2-kilometer wide parabolic mirror into the side of it and having a moveable target suspended by a crane, following the sun, and generating fusion on the target.  Just for fun, I wondered what the upper limit of temperature was that could be reached, in theory, with such a device ?  Anyone know how many joules per meter squared the sun can normally deliver to the ground on a nice sunny day?  No one has offered any figures, so let's take a guess.  My 10 cm-diameter magnifying glass can easily burn paper with a dot that is, at a guess, about 100 times smaller than the glass itself.  If the temperature generated is say 250 degrees, then we can start having some fun. Let's assume for the sake of argument that a 10cm parabolic mirror would generate the same amount of heat; I suspect it would be more, but let's assume this to make things easier. My glass / mirror is 5 squared times PI  cm squared, which is  78.5 .   So an area of 78.5 cm squared gives you a manageable dot of light at 250 degrees celcius.  So, a mirror of two kilometers wide = 2000 meters = 200000 centimeters.   Square this and times by PI, and you get and area of 125663704000 cm squared.  Now, divide this by the area of my magnifying  glass, 78.5 cm squared, and you learn that the big mirror is 1600811515.9 times bigger in area.  Times this by 250 degrees and you get a temperature of 400,202,878,980   or about 400 thousand million degrees, easily enough to start fusion!  Now, obviously, our great scientists would have tried this, I guess, unless they spotted what would be to them an obvious flaw in the scheme.  Anybody care to put me right?  :-)    thanks :-)

[Bored chemist]

The temperature doesn't scale with the area of the glass.
The limiting temperature happens when the hot thing is losing energy by radiation (etc) as fast as it is gaining it from the light. The radiation rises with the 4th power of the temperature and with the area of the object. The power varies as the area of the magnifying glass. If the object being heated is big then the temperature rises as the square root of the size of the lens (I think) but if you want maximum temperature the object heated must be just bigger than the sizee of the image of the sun. That is roughly proportional to the size of the lens so overall a huge lens doesn't get much benefit in terms of maximum temperature. it does make a difference to the power available
Here's an article about a pretty big mirror (easier to work with than a big lens)
http://www.time.com/time/magazine/article/0,9171,909204,00.html

[another_someone]

I would have though the maximum temperature would be about 6000K, which is the colour temperature of sunlight.

[Batroost]

Surely this would have more to do with the absorbtion/radiation characteristics of the target than of the source of light used?

For example, it's difficult to ascribe a 'colour temperature' to a near-monochromatic light source (e.g. a LASER) but it is certainly possible to use a focussed monochromatic beam to heat a target.

[another_someone]

Surely this would have more to do with the absorbtion/radiation characteristics of the target than of the source of light used?

For example, it's difficult to ascribe a 'colour temperature' to a near-monochromatic light source (e.g. a LASER) but it is certainly possible to use a focussed monochromatic beam to heat a target.

Why would a monochromatic LASER not have a colour temperature (it may not have a black body spectrum, but the light still has an energy associated with its frequency, and that energy equates to a temperature)?

[lightarrow]

Why would a monochromatic LASER not have a colour temperature (it may not have a black body spectrum, but the light still has an energy associated with its frequency, and that energy equates to a temperature)?

Just because it doesn't have a blackbody spectrum. I can find a specific wavelength, let's say 550 nm, in the spectrum of a blackbody of any temperature. I can take a tungsten block at 20°C or at 3000°C, put a very good colour filter in front of it, which let pass only 550 nm light, and how could we establish this light comes from a 20°C or 3000°C block of metal? Certainly at 20°C the intensity of radiation at 550 is less, but this could also be ascribed to source's distance and surphace properties.
It's only the spectrum's shape (specifically: its maximum's position) which is related to temperature.

[Batroost]

It's only the spectrum's shape (specifically: its maximum's position) which is related to temperature

Agreed.

[Bored chemist]

"Surely this would have more to do with the absorbtion/radiation characteristics of the target than of the source of light used?"
All things at all wavelengthts are exactly as good at radiating as they are at absorbing so the effet evens out.

[Batroost]

All things at all wavelengthts are exactly as good at radiating as they are at absorbing so the effect evens out.

A bit of an odd thing to say: What you are describing is an ideal 'black body'. Real objects don't achieve this.

[another_someone]

Just because it doesn't have a blackbody spectrum. I can find a specific wavelength, let's say 550 nm, in the spectrum of a blackbody of any temperature. I can take a tungsten block at 20°C or at 3000°C, put a very good colour filter in front of it, which let pass only 550 nm light, and how could we establish this light comes from a 20°C or 3000°C block of metal? Certainly at 20°C the intensity of radiation at 550 is less, but this could also be ascribed to source's distance and surphace properties.
It's only the spectrum's shape (specifically: its maximum's position) which is related to temperature.

This is certainly true, but I am not sure how it relates to the argument.

We say that the surface of the Sun is approx 6000K, because that is the temperature of the light emanating from the surface.  We know that the temperature beneath the surface can reach millions of K, but that is not the radiation we receive, so we know that the temperature on the surface is not as hot an the depths of the Sun are.

Yes, you can put a filter in-front of an object, and that will cause the system as a whole to cool on its exterior (beyond that filter), but it does not prevent the interior of the system from being hotter.

The question is, should not the radiation emitted from a 6000K body should be able to induce a temperature in a black body that absorbs that radiation that is equal to the colour temperature.  OK, the real world is not composed of ideal black bodies, but it should at least create a benchmark.