# Comparing Olympus 4/3lenses to FX "Full Frame" offerings

Started Jan 25, 2014 | Discussions thread
Re: Wrong

Tiger1 wrote:

Ian Stuart Forsyth wrote:

Tiger1 wrote:

Ian Stuart Forsyth wrote:

Tiger1 wrote:

Ian Stuart Forsyth wrote:

Tiger1 wrote:

bobn2 wrote:

Tiger1 wrote:

bobn2 wrote:

That is a circular argument, since ISO is defined by exposure. Same ISO, same exposure but four times as much light on the FF sensor, four times as many photons, twice the SNR. It's that simple.

The surface area is four times bigger Bob so what each unit area "sees" is the same amount of photons regardless the format. Hence the SAME SNR.

You've got yourself confused.

I am not confused.

Lets go through it one step at a time.

1. For the same f-number the amount of light per unit area projected on the sensor is the same.

We agree on something.

2. A FF sensor has four times the area as a FT sensor, so four times the 'unit'.

We agree here too.

3. Thus a FF sensor collects four times the light at the same f-number.

But that light is spread over 4 times the area....

Here’s your problem the intensity of the light remainders the same, it’s not spread over 4 times larger surface area but rather that 4 times larger area is being hit with the same light intensity capturing 4 times more light

I know. That is what I am saying! The intensity of light per unit area is the same...... I thought that I have made that obvious.

Then why would you say just above “But that light is spread over 4 times the area “ when responding to uncle Bobs “ thus a ff collect’s 4 times the light at the same F number”

Don’t you see the fallacy of say” but that light is spread over 4 time the area” when the intensity remains the same ?

Uncle Bob is just plain wrong. He claims that each pixel receives 4 times as much light. Pixels don't receive light, photosites do. Even given this error and assuming he means photosites he makes the assumption that each photosite is four times larger in a FF sensor than in a 4/3 sensor. This ignores gaps that may exist between photosites and how these gaps may vary from one sensor to another.

I also am the one that pointed out that the intensity of light per unit area is the same.

What he states if “Thus each pixel of the output image from FF sensor represents four times the number of photons as that made from a FT sensor, assuming the same exposure (f-number, shutter speed and scene luminance)”

Stupefied down, If we view a FF image at 16mp output and compare it to FT image of equal output each pixel will contain 4 time more light information than the FT image

So he knows the photon capture efficiency of each photosite in aFF vs a FT sensor? Wow!!!!! Were did he get that information from? To know the output representation from the pixel you must know the capture efficiency of each photosite. Agree?

With some math and information that is readily available from companies testing the performance of sensors

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