Archive for the ‘pic of the moment’ Category

Back to school….   2 comments

Summer’s over, school is back in session.


We had a fairly productive summer: a paper was accepted, we have some encouraging results transfecting our cell line with a GFP-tubulin construct, and started to commercialize our Tissue Interrogator.  This picture has been featured in Physics Today and seems to be getting people’s attention:


Stay tuned for developments on these and other projects.

Summer vacation was also productive- I have been waiting 3 years for skies clear enough to make these images:


These are ‘star trails’: just leave the shutter open for a looooong time, and the stars trace out orbits due to the Earth’s rotation.  See Polaris in the lower left? It’s not located *exactly* on the axis of rotation.  The bright ‘dash’ in the lower right is a Air Force jet hitting it’s afterburner.

Alternatively, by stitching together multiple fields-of-view, we have the entire milky way: (warning, this full size image is *large*: 14k x 4k pixels

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Finally, a smaller region of the milky way, featuring several Messier objects:

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Posted August 28, 2014 by resnicklab in Physics, pic of the moment, Science

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Astrophotography redux times two   Leave a comment

It started as a simple question: “What is the faintest star we can image with our equipment?” Figuring out the answer turned out to be delightfully complicated.

The basics are straightforward: since the apparent magnitude of a star is proportional to the amount of light striking the camera sensor, all we need to do is figure out how much light we need to generate barely-detectable signal.

Sky & Telescope magazine published a few articles in 1989 and 1994 under the title “What’s the faintest star you can see?”, and while much of that discussion is still valid, the question was posed pre-digital imaging, and so the results reflect the natural variability in human vision. It would seem that digital sensors, with their well-defined specifications, would easily provide an unambiguous way to answer that question.

Answering the question seems simple enough- begin by relating the apparent magnitudes of two stars to the ratio of “brightness” of the two stars: m1 – m2 = -2.5 log(b1/b2). So, if you have two stars (or other celestial objects), for example the noonday sun (m1 = -27) and full moon (m2 = -13), the difference in magnitudes tells us that the noonday sun is 400,000 times brighter than the full moon. Going further, the full moon is 3,000,000 times as bright as a magnitude 6 star (the typical limit of unaided human vision), which is 2.5 times as bright as a magnitude 7 star, etc. etc. But that doesn’t really tell us how much light is hitting the sensor, only an amount of light that is relative to some other amount.

Unfortunately, astronomers have their own words for optical things. What astronomers call ‘brightness’, physicists call ‘irradiance’. In radiometry (actually, photometry), ‘brightness’ means something completely different- it is a subjective judgment about photometric (eye-response weighted) qualities of luminous objects.

In any case, we have a way to make the ‘relative brightness’ scale into a real, measureable, quantity. What we do is ‘standardize’ the scale to a convenient standard irradiance, something that stays constant over repeated measurements, is easily replicated, and commonly available- for example, the irradiance of the noonday sun = 1 kW/m2.

So now we have our apparent magnitude scale (say: sun, moon, mag 6, mag 12, mag 16) = (-27, -13, 6, 12, 16) and corresponding relative brightness scale (sun: moon, moon: mag. 6, mag 6: mag 12, mag12: mag 16) that we make absolute by standardizing to the solar irradiance: 1 kW/m2, 3*10-3 W/m2. In table form:

magnitude rel. brightness irradiance [W/m2]
-26.74         1.00E+00           1.00E+03
-12.74         2.51E-06             2.51E-03
6                  8.02E-14             8.02E-11
12                3.19E-16              3.19E-13
16                8.02E-18             8.02E-15
21                8.02E-20            8.02E-17

By specifying the entrance pupil of our telephoto lens (maximum aperture = 140mm; diameter = 1.5 *10-2 m2), for any given star we can calculate how many watts of optical power is incident onto the sensor. But we have to be careful: not all the light emitted by the sun (or any luminous object) is detected by the sensor.

In addition to not being a perfect detector (the ‘efficiency’ or ‘responsivity’ of a detector is always < 1), not all the colors of light can be detected by the sensor: for example, the sun emits radio waves, but our camera sensor is not able to detect those. Of the 1 kW/m2 of light incident on the earth, how much of that light is in the visible region- or equivalently, within the spectral sensitivity of the sensor?

Stars are blackbodies, and the blackbody spectrum depends on temperature- different stars have different temperatures, and so appear differently colored. For ‘typical’ temperatures ranging from 5700K (our sun) to 8000K, the fraction of light in the visible waveband is about 40%. So visible sunlight (V-band light), accessible to the camera sensor, provides an irradiance of about 600 W/m2 on the earth’s surface.

So much for the sources, now the detector- how much light is needed to generate a detectable signal? There are several pieces to this answer.

First is the average ‘quantum efficiency’ of the sensor over the visible waveband. Manufacturers of scientific CCDs and CMOS imagers generally provide this information, but the sensor in our camera (Sony Exmor R) is a consumer product, and technical datasheets aren’t readily available. Basing the responsivity of the Exmor R based on other CMOS imagers on the market, we estimate responsivity as about 0.7 (70%) This means that on average, 1 absorbed photon will produce 0.7 e-. That’s pretty good- our cooled EMCCD camera costs 100 times as much and only has a slightly higher quantum efficiency- 0.9 (90%).

So if we know how many photons are hitting the sensor, we know how many electrons are being generated. And since we know (based on other CMOS sensors) that the Exmor has a ‘full well capacity’ of about 21000 e- and a dark current level of about 0.2 e-/s, if we know the number of photons incident during an exposure, we can calculate how may electrons accumulate in the well, compare that to the noise level and full-well capacity, and determine if we can detect light from the star. How many photons are incident onto the sensor?

We can calculate the incident optical power, in Watts. If we can convert Watts into photons/second, we can connect the magnitude of the star with the number of electrons generated during an exposure. Can we convert Watts into photons/second?

Yes, but we have to follow the rules of blackbody radiation- lots of different colors, lots of different photon energies, blah blah blah. Online calculators came in handy for this. Skipping a few steps, our table looks like this:

magnitude rel. brightness irradiance [W/m2] power incident on sensor [W] 6000K blackbody V-band phot/sec             e-/s
-26.74           1.00E+00         1.00E+03                 1.54E+01                                        6.83E+18                                 4.78E+18
-12.74            2.51E-06           2.51E-03                 3.87E-05                                          1.72E+13                                   1.20E+13
6                     8.02E-14          8.02E-11                  1.23E-12                                           5.48E+05                                  3.83E+05
12                   3.19E-16            3.19E-13                   4.91E-15                                          2.18E+03                                   1.53E+03
16                   8.02E-18           8.02E-15                1.23E-16                                            5.48E+01                                   3.83E+01
21                   8.02E-20          8.02E-17                   1.23E-18                                         5.48E-01                                     3.83E-01
18                    1.27E-18           1.27E-15                   1.96E-17                                          8.68E+00                                  6.08E+00

You may have noticed a slight ‘cheat’- we kept the temperature of the blackbody constant, when in fact different stars are at different temperatures. Fortunately, the narrow waveband of interest means the variability is small enough to get away with 1 or 2 digits of accuracy. As long as we keep that in mind, we can proceed. Now, let’s check our answers:

Moon: we have 1.2E+13 electrons generated every second, which would fill the well (saturated image) after 2 nanoseconds. This agrees poorly with experience- we typically expose for 1/250s. What’s wrong?

We didn’t account for the number of pixels over which the image is spread: the full moon covers 2.1*106 pixels, so we actually generate 5.8E+06 electrons per pixel per second- and the time to saturation is now 1/280 second, much better agreement!

Now, for the stars: a mag. 6 star (covering 20 pixels) saturates the pixels after 1 second, a mag. 12 star after 280 seconds, and a magnitude 16 star requires 10,000 seconds- about 2.5 hours- of observation to reach saturation.

But we don’t need to saturate the pixel in order to claim detection- we just need to be higher than the noise level. A star of magnitude 18.5 produces (roughly) 0.2 e-/s and thus the SNR =1. In practice, we never get to this magnitude limit due to the infinitely-long integration time required, but image stacking does get us closer: our roughly 1-hour image stacks start to reveal stars and deep sky objects as faint as 15 magnitude, according to the database SIMBAD. Calculations show that observing a mag. 15 object requires 4300 seconds of time to saturate the detector, in reasonable agreement with our images.

This is pretty good- our light-polluted night sky is roughly mag. 3!

Why did we even think to ask this question? Wide field astrophotography is becoming more common due to the ready supply of full-frame digital cameras and free/nearly free post-processing software. In fact, the (relatively) unskilled amateur (us, for example) can now generate images on par with many professional observatories, even though we live in a relatively light-polluted area.

What we have recently tried out is panoramic stitching of stacked images to generate an image with a relatively large field of view. This time of year, Cygnus is in a good observing position, so we have had direct views of the galactic plane in all it’s nebulous glory:


This image, and those that follow, are best viewed at full resolution (or even larger- 200% still looks great) on a large monitor to fully appreciate what nighttime at Atacama or Antarctica are probably like.

Some details about how we make these images- we acquire a few hundred images at each field of view, separately stack each field of view with Deep Sky Stacker and then fuse the images with Hugin. It may be surprising to know that each field of view must be slightly ‘tuned’ to compensate for the differences in viewing direction, as opposed to simply stacking all of the images at once and choosing “maximum field of view” in DSS.

Here are two subfields within Cygnus- the edge of the North American Nebula, and the Veil Nebula:



Posted July 31, 2014 by resnicklab in Physics, pic of the moment, Science

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Now… isn’t that special?!?!?!   Leave a comment

We learned something unexpected recently.

Other than microscope objectives, I use three photographic objectives- a wide angle (15/2.8), a ‘normal’ (85/1.4), and a telephoto (400/2.8). For astrophotography, I have always used the 800/5.6 lens combination (400mm + 2x tele) because I wanted to image faint, distant, objects. The large entrance pupil (143mm diameter) of the lens acts as a large ‘light bucket’, and the long focal length provides high magnification:


As I’ve posted previously, using either the 400/2.8 or 800/5.6 configuration, I stack around 100 15-second exposures acquired at ISO2000- and those 100 images are the top 20% in quality of all acquired images. By contrast, using a shorter focal length lens would increase the acceptance rate because the residual alignment error is below the resolution limit of the lens. Using the 85mm lens, I can acquire 30s exposures at ISO100 and use 100% of the images.

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One way to quantify the ‘efficiency’ of a lens is to compare the size of the entrance pupil to the size of an Airy disc- under ideal conditions, the amount of light entering the lens is concentrated into an Airy disc. Analysis shows that the efficiency scales as f^2/N^4, where f is the focal length and N the f-number.

Calculating the efficiency provides the following table:

focal length f/# entrance pupil diameter Airy disc radius irradiance concentration relative gain
400.0 2.8 142.9 1.7E-02 298714.3 1.0E+00
85.0 1.4 60.7 8.5E-03 107910.6 3.6E-01
800.0 5.6 142.9 3.4E-02 149357.2 5.0E-01
15.0 2.8 5.4 1.7E-02 420.1 1.4E-03

What is surprising is the effect of the N^4 dependence: the 85mm is nearly as efficient than the 400mm (72%) in spite of having an entrance pupil substantially smaller.

This chart shows that the 85mm lens is very well suited for astrophotography in terms of viewing faint objects. However, the lower angular magnification results in a loss of spatial detail, meaning the 85mm is not suitable for viewing nebulae, clusters, galaxies, planets, etc. Here are image pairs comparing the 85mm and 400mm lenses:

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The chart also (apparently) shows that the wide-angle lens is very unsuited for astrophotography- as compared to the 400/2.8, the efficiency is 0.1%. So, I was very surprised when I recently used the 15mm lens to image the galactic plane around the constellation Cygnus, with individual exposures of 13s @ ISO 2000:


This was completely unexpected. Even though the 15/2.8 is only 0.1% as efficient as the 400/2.8, the image brightness is as high as what I can generate with the 400/2.8, using essentially identical exposure times. How can this be?

The reason has to do with the density of stars. If I am imaging in a direction pointing out of the galactic plane where the density of stars is low, I would indeed conclude that the 15mm lens is unsuitable. However, each airy disc at the image corresponds to an angular field of view (named the ‘instantaneous field of view’, named back when detectors were single pixels and a scanning system was used to create an image plane), and since the field of view of the wide-angle is considerably more than the telephoto, there is an amplification factor due to the presence of multiple stars ‘sharing’ a single airy disc.



That is to say, when the density of stars is high and the lens magnification low, many stars will be imaged to the same airy disc, resulting in brighter images. How do the lenses compare with this metric?

At this point I have to mention the standard disclaimer about Bayer filters and color cameras. That said, measuring the size of an airy disc from the acquired images results in this table:

f f/# field of view (deg) field per pixel (rad) back projected solid angle for Airy disc (sterad) density gain magnification
400 2.8 5.2 1.50E-05 8.15E-09 1.0
85 1.4 23.9 6.92E-05 7.65E-08 9.4
800 5.6 2.6 7.52E-06 4.09E-09 0.5
15 2.8 100.4 2.91E-04 3.04E-06 372.8

Taking both factors into account results in this metric:

focal length f/# Overall efficiency
400.0 2.8 1.00
85.0 1.4 3.39
800.0 5.6 0.25
15.0 2.8 0.52

When the density of stars is high, I only need to double the ISO when using the 15mm in order to match the performance of the 400mm. Even more surprising, the 85mm lens is more than 3 times as efficient- in practice, this means I can use a lower ISO setting (ISO 100 instead of ISO 400), reducing the gain noise.

The moral of the story is: try new things!

Posted September 16, 2013 by resnicklab in Physics, pic of the moment, Science

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astrophotography, redux   Leave a comment

This time of year, M57 (the ring nebula) is advantageously located in our night sky, so this is a good opportunity to discuss astrophotography.

The basic problem to be solved is that the objects are of very low intensity and are moving. Here’s a sample image:

This is a 20-second exposure using our 800mm f/5.6 lens, ISO 2000. Because we know how fast the stars are (apparently) moving- 360 degrees in 24 hours- we can calculate how long the shutter could be open before this motion blur occurs. For this lens and camera (pixel size = 6 microns, but there is a Bayer filter present), the maximum shutter speed is about 1/4 second: within 1/4 second, the stars move less than a pixel. That’s suboptimal, to say the least. We could try to improve things by using a faster lens and higher sensor gain, but our setup is already close to the limit of what is commercially available.

The solution is to use a ‘tracking mount‘. There are lots of different designs, ours is a ‘German equatorial mount‘. The basic procedure is very simple- align the polar axis of the mount to the North star and turn on the motors. When aligned, The two motors correspond to declination (latitude) and right ascension (longitude). Then, the mount essentially ‘unwraps’ the Earth’s rotation, ensuring the telescope remains pointed at the same part of the night sky. This is also a 20-second exposure, but taken with the tracking mount aligned:


much better! The final step is to take a lot of these images and average them all together (‘image stacking’).

Naturally, life is not as simple as that. Most images look like this:

What’s the deal? There are lots of reasons why this image still has motion blur: vibrations, polar misalignment, gear error, etc., and it’s illuminating to calculate acceptable limits. First, let’s dispense with the pixel size issue- the sensor dimensions result in a single ‘pixel’ as being 6 microns on a side. However, in order to generate color images, a Bayer filter is placed over the pixel array, so that neighboring pixels are assigned different colors (and detect slightly different parts of the object). A detailed analysis is highly complicated- 3 independent non-commensurate samplings of the image plane- but if our image does not have features smaller than say, 12 microns (corresponding to a 2×2 pixel array), software interpolation that generates a color pixel using the neighboring elements will likely give an accurate result, and we can pretend that our sensor has 6-micron color pixels.

And in fact, examining our ‘best’ single image, stars are imaged as bright blobs of radius 3 pixels (and brighter stars appear as even bigger blobs).

Ok, so how much can the sensor move without causing motion blur? The stringent limit is that the sensor (or the image projected onto the sensor) must move less than 0.5 pixel (3 microns) during an exposure. If the lever arm of the lens is 0.5m, the allowed angular displacement is 1.2 arcsec. In terms of vibrations, this is a very stringent requirement! Similarly, we can calculate the maximum allowed polar misalignment: if the telescope pointing is allowed to drift no more than 0.5 pixel during an exposure, since each pixel subtends 1 arcsec (for diffraction-limited performance using this lens), the allowed misalignment is about 6 arcmin ( is a good reference).

Speaking of diffraction-limited, what is the limit of our system? Each star should be imaged as a single pixel! Clearly, there is image degredation not just from movement, but from *seeing*- clear air turbulence appears as blur in long time exposures. How much degredation? Our “best” images correspond to using a lens at f/30, or an entrance pupil diameter of 27mm (instead of f/5.6, 140 mm entrance pupil diameter). The seeing conditions in Cleveland are *awful*!

So why do astrophotography? Our images are not meant to compete with ‘professional’ telescope images. It’s also a nice experience to learn about the night sky and work on our imaging technique. Here’s the result of stacking enough ‘best’ 20-second exposure images to produce a single 29 minute long exposure:

29m Composite crop

Not bad! And we can continue to improve the image- either by ‘dithering’ the individual frames to allow sub-pixel features to emerge:

29m_2x Composite (RGB)

or by deconvolving the final image, using a 3-pixel radius Gaussian blob as the point-spread function:

deconvolved 29m crop

The image improvements may not appear that significant, but as always, the rule of post-processing is *subtle* improvements- no artifacts must be introduced.

Posted July 17, 2013 by resnicklab in Physics, pic of the moment, Science

Chips (the bottom of the bag)   Leave a comment

We’ve gone through most of the early examples of ICs (although we still have quite a few that we haven’t decapsulated yet), so this post is for a few more ‘modern’ ICs. The first is a TI DSP from 1986, and many of the features appear identical to what we have seen on the early ICs- only at a smaller scale. The second chip is a 32 Gb Flash RAM from Samsung manufactured in 2008 using a 32nm process, and this appears completely unrelated to the memory chips we have shown so far- the actual gates are below the resolution limit of optical microscopy, and so identifying specific features is no longer possible (except at more macro-scale items like memory blocks, etc.)

This will likely be the last chipset for a while- now that the weather is nicer, we prefer to image things outside! We have been re-visiting the Virgo supercluster and will post images of that as they become available.

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New-Out99995-_Pyramid Maximum Contrast[1,0,1] (2) New-Out99995-_Pyramid Maximum Contrast[1,0,1]

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Meanwhile….   1 comment

Lately, We have traced back the history of integrated circuits to a few very early devices, primarily manufactured by RCA and examples of ‘MOS’ type ICs. The metal-oxide-semiconductor geometry was the final design revolution, leading directly to modern chips. However, there were (and still are, unlike RCA) other manufacturers. Here’s a Raytheon RC1033 3-input NOR circuit:

The circuit logic is fairly easy to follow. Around the same time (1966), Motorola was producing the MC832P, a 4-input NAND gate:

later (1969!), Motorola was making the 3000L quad 2-input NAND gate:

and 3005L, a triple three-input NAND gate array:

Yet another company, Fairchild Semiconductor, introduced a radically different kind of circuit- the operational amplifier. Here are images of the μA709, Fairchild’s breakthrough product:

the LM710:


and μA741 made by AMD the next year (1972):

Lastly, here’s one we have yet to identify, part code U5B771239:
Fairchild U5B771239

If you know what this is, let us know!

The Fairchild cicuits look radically different than the others- there’s no obvious symmetry to the layout, as opposed to the previous examples.

Solid State Physics: revealed!   Leave a comment

We opened up another RCA COSMOS chip: the 4007, a ‘dual complimentary pair and inverter’, which has a particularly simple layout:
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The layout is simply three pairs of transistors, p-type Field-effect transistors on the left, n-type FETS on the right. We know this because pin 1 is the round contact, and counting counter-clockwise, we identify pin 7 as the common voltage supply for the n-type transistors. On the circuit diagram, p- and n- type transistors are distinguished by the direction of the arrow located on the base terminal, and we also note that the transistors are enhancement mode devices.

Now let’s consider a single transistor. These are field-effect transistors, which mean there are 4 terminals: source, gate, drain, and body/substrate. However, our images seem to have only three metal traces corresponding to source, gate, and drain: where’s the 4th terminal?

We can also see on the epi-illumination image that each transistor has 6 distinct layers: the ‘bottom’ layer where the bonding pads are, and then 4 intervening layers before the metal traces are reached on top. The oblique reflected light image does not show any other reflective metal, and the internal structure is only revealed by epi-illumination. However, one of those interior layers is a conductive layer that links pin 7 to the n-type transistors. A clue is given by the extra feature on the transistor immediately adjacent to pin 7; there is what appears to be a vertical conduction path to an interior layer, and this interior layer spans all three n-type transistors. So, it appears that we can identify the base/bottom layer, an insulating layer, and then a conductive layer followed by another insulating layer. The insulating layers are probably depletion regions.

An even simpler design is the RCA 3018:
RCA CA3018

This IC has 4 transistors, two are isolated and the other two are connected in a Darlington configuration. A more complex IC is the RCA CD4009:

RCA 4009

This IC consists of 6 inverters, and each inverter is constructed out of 5 transistors.

By now we have covered the early history of integrated circuits; these devices, or ones very similar, are the ones fabricated for the Apollo and Minuteman programs. However, so far we have only been looking at RCA chips; it could be interesting examine some examples from Fairchild, Motorola, and Texas Instruments to see what sorts of similarities and differences exist.