How does a lens work to form an image on the camera's sensor...

Started Mar 7, 2014 | Discussions thread
57even Forum Pro • Posts: 12,798
Re: How does a lens work to form an image on the camera's sensor...

GBJ wrote:

Wow, I am very thankful and impressed with the clarity of your explanation! You very quickly have showed me that the lens does not focus an image on the sensor in one point only...which should be obvious, but was not to me.

I can see by your well layed out diagram, that for example, a tree has elements that are closer and further away from the sensor. Each branch of the tree has a varying distance away from the sensor, and must be recorded on the sensor, but only the parts of the tree that are on one parallel plane from the sensor can be in complete focus, and the parts of the tree that are closer, or further away from that parallel plane of focus must be out of focus to the degree to which their distance varies from this plane of focus. Now I can see the development of the idea of depth of field.

Good stuff. Glad it was useful.

From your diagram, I can imagine how each needle of the tree branch reflects it's own light ray towards the lens, and gets focused by the lens onto one pixel, or photon, of the sensor?

Not quite. There is no relationship between a single photon and a pixel.

Light Capture

Imagine photons are rain drops and the photosites/pixels are buckets. You can get more water in the bucket if it rains harder, or if you leave it longer....

There is a constant stream of photons (or a "light wave") coming from each point on the tree. These will strike the light receptive photosite on the sensor which will create an electric charge proportional to the number of photons that hit it (effectively the photon energy will displace an electron from a silicon atom which will then be able to move or create a current). The brighter the light (more intensity) or the longer you expose the pixel, the more photons will hit it and the more electrons will be displaced.

The more raindrops you collect the heavier the bucket.

When the photosite is read out, these electrons create a voltage difference to the baseline which can be detected and stored, one at a time for each photosite. This voltage is therefore proportional to the number of photons captured. The voltage value is stored digitally.

Noise and Signal to Noise Ratio.

Imagine your buckets are not exactly the same size and some were already wet when you started, plus you have some losses when emptying the bucket and some spillage between them because of splashing....

For various reasons (such as heat and the electronics itself) the voltage read out is never zero even if no photons hit it. Similarly, there is statistical variation between photosites caused by manufacturing and the process of reading it out.

This voltage variation between photosites (even with the lens cap on) is called noise. A very low signal is hard to distinguish from noise. This is called the "noise floor" or the point at which the signal cannot be distinguished from other background noise. The noise is actually constant (more or less) but when you amplify the signal (increase ISO) you increase the noise too, which is why cameras get noisier at higher ISO. The signal is smaller (shooting in lower light) so the ratio of the signal intensity to noise intensity is lower (signal to noise ratio).

It is the ratio which dictates how visible the noise is.

Bit Depth

How accurate do the scales need to be to weigh your bucket?

Noise affects how many bits you need to store the voltage values electronically. For instance, a 14bit number can effectively store 2^14 discrete voltage values or about 16,000 intensity values.

However a photosite can record (typically at the larger end) up to 100,000 photons, so why not record 100,000 values? The answer is noise. If the variation (noise) is +/- 50 electrons, then the variation between equal signals is up to 100 electrons, so there is no point in recording more than 1,000 values because you wont be able to distinguish them. If the noise is +/- 5 electrons, then you can distinguish 10,000 values.

This is known as the bit depth of the sensor. Most sensors can just about manage 12bits (4096 values) but the larger the photosite, the bigger the number. A top end MF sensor can manage 14 or even 16 bits. A phone camera will struggle with 8.

Recording more bits than you need is fine (just more storage) but less is obviously a bad idea as you end up with fewer possible tones than you can actually capture.

Well Capacity, Dynamic Range and Exposure Range

How big is your bucket?

Photosites have a limit to their capacity. Any more photons cannot be recorded and that part of the image is "overexposed" or blown out. The limit is set by the physical light capture area of the photosite. The well capacity if defined as the maximum recordable number of electrons per photosite and varies between around 1,000 and 100,000 for phone cameras to MF cameras.

The signal to noise ratio at saturation defines the dynamic range of an individual photosite (number of visible tones or intensity values between noise floor and blow-out which is similar to bit depth above) but exposure range (which most people confuse with DR) is the difference in stops between the brightest and darkest object you can record in one shot.

The two are related, but easily confused. The first is a set property of the sensor, the second is a property of how the signal is processed by the camera (or the RAW processor). Camera makers to all kinds of tricks to make shots "look" nicer by fiddling with the tone curve, but this normally compresses the exposure range to increase contrast. Similarly many fiddle the RAW output by clipping the low end (shadow) bits to reduce apparent noise and smoothing the output. Hence ER seldom reaches the maximum possible DR.

You may think that for maximum exposure range you would want larger photosites. However this is not exactly true, because it is measured for a given print size. If you have more pixels, the difference between pixels is less obvious when you look at the same size print, so the reduced DR at the pixel level is compensated for by having more data to sample (hence reducing the visible variation between pixels hence less noise).

However the larger photosites usually win out at very high ISO when they manage to retain more faithful colour. So whereas base ISO ER may be related to sensor size, high ISO ER will start to take pixel size into account as well.

12 megapixel camera has 12 million pixels, or photo receptors, I think, and I can imagine each part of the tree having a designated pixel on the sensor to capture the image of the tree as a whole.

More or less.

However on a colour sensor each photosite only records one colour (half of them record green and the other half alternate between red and blue). These are combined mathematically to work out what the actual colour of each pixel was (RGB value) by comparing each one with its neighbours. This introduces some statistical uncertainty over the exact extent of that colour value so the solution is to "blur" the edges a bit, which means that the minimum detail level you can record is actually larger than one photosite unless you use a Foveon sensor.

Lenses are not perfect either. None can resolve a point to a point, but will have a "blur radius" even when in perfect focus. If the blur radius is much smaller than the sensor's resolution limit, then the camera is sensor limited. If its much larger it is lens limited.

If the two are comparable, then the overall resolution is actually a combination of the two. There is an approximate formula for working it out which is 1/R = 1/Lr + 1/Sr where R is camera resolution and Lr is lens resolution and Sr is sensor resolution. As you can see from this, if lens and sensor resolution are equal (=x) then the overall resolution is half the resolution of either.

1/R = 1/x + 1/x = 2/x hence R=x/2

This makes more sense if you look at the blur radii instead. Effectively it means you have to add the two blur radii together to get the total amount of blur. B = Lb + Sb.

This is not totally accurate because the blur radius is not uniform, but you get the general idea. However it does mean that as you cram more photosites onto a sensor you reach a point where for most lenses the increase in overall resolution is much less than the increase in sensor resolution. Compare DxO values for the D600 and D800 for some Nikon lenses and the differences are surprisingly small in many cases.

Hope all that is useful.

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