M mode, 1/100 s, f/8, ISO 400, change HTP from disabled to enabled, what happens to the number of photons? Nothing. No difference. Av mode, f/5.6, ISO 200, change HTP from disabled to enabled, what happens to the number of photons? Nothing. No difference.
This is basic physics, assuming of course the situation external to the camera is controlled. Anyone who does not understand this can easily learn it. Admitting one is wrong is another story evidently.
Again, assuming controlled lighting from outside the camera, the number of photons hitting the sensor can only be changed by exposure settings - the f-stop and the amount of time the shutter is open. No artificial computer driven setting in the camera's software can magically create photons or destroy them. That is contrary to the laws of physics.
Hi, How does changing the Exposure increase or decrease the amount of photons hitting the sensor?
It does make sense(brighter/darker image), but i'm just wondering what actually happens.
Photons move from their source over time. They also disperse from their source over time, covering a larger area. We could get into some complicated math, using solid angles, light intensity measurements like cd/m^2, etc. But none of that is really necessary to actually understand the answer to your question.
Lets say you have a 60watt light bulb. That bulb produces light. That means it produces a continuous flow of photons, emitted from the bulb's entire surface area, every moment. If you take a photograph of the bulb with a DSLR camera, there are two things that will affect the "exposure" of the image: aperture and shutter speed. Why these two? Why is ISO not a factor?
If we think of the distance between the bulb and our camera's aperture as a water pipeline, we can come up with an analogy for aperture. For a given quantity of water, a large pipeline can move a large volume of water at a low pressure, and a small pipeline can move a smaller volume of water at a higher pressure. Volume and pressure are both important here, but for now, I'll only cover volume. A large aperture is like a large pipeline...it allows a lot of light to flow through in a given amount of time. Similarly, a small aperture is like a small pipeline, it allows a lesser amount of light to flow through in a given amount of time. Aperture changes the total volume of light passing through the lens.
Shutter speed is a very simple concept. It is simply the time factor for exposure. Since light from the bulb "flows" over time...since a certain amount of photons are emitted in any given unit time, changing shutter speed affects how many photons actually reach the sensor. Shutter changes exposure time at the sensor plane.
Let's say that an exposure of 1/500s f/4 of our light bulb produces a reasonable exposure...the bulb is not blown, but it is very bright. We can do one of two things to reduce the exposure by a factor of two, such that we can start to see some detail in our bulb. Since aperture affects the volume of light passing through the lens, we can reduce the aperture by "one stop" to reduce the volume of light by a factor of two. That gives us 1/500s f/5.6. Similarly, we could double the shutter speed, again a "one stop" change, to reduce the amount of light reaching the sensor by a factor of two. That gives us 1/1000s f/4.
I mentioned that ISO does not actually affect exposure. From a literal standpoint, ISO is not an exposure factor. It is technically a post-exposure factor. From a logical standpoint, ISO as far as the average photographer is concerned, is an "exposure" factor. ISO simply takes the actual exposure on the sensor, and amplifies it. ISO 100 is base ISO for the vast majority of cameras. Increasing ISO from base always has the effect of making an exposure brighter. Changing ISO, however, does not affect the amount of light!!
It uses electronic signal amplification and digital boost to simulate
a brighter exposure, hence the reason noise increases as you increase ISO.
Finally, I compared aperture to a water pipeline. There was a specific reason for this, and it has to do with diffraction. A more realistic analogy would actually have been to compare a lens to a water pipe with a variable exit opening...which is basically the same thing as an aperture. If you force water through a large pipe, lets say a pipe with a 1-foot diameter, at a fixed pressure, it will exit the other end and spread out a little bit. If you double the pressure in the pipe, the amount the water will spread out as it exits the pipe will increase. Double the pressure again, and water will start to spray out at almost a 180 degree angle to the direction of the pipe. Pressure can be changed by adding a cap to the end of the pipe with differing hole sizes...or apertures. We could call the uncapped pipe f/1, so to double the pressure, we add a cap that produces an f/1.4 hole (8.6" diameter), and to double again we add a cap that produces an f/2 hole (6" diameter). Water moving through the pipe is technically a longitudinal wave, and therefor it experiences diffraction when it encounters the cap.
Light also behaves as a wave, so when it passes through a lens and encounters the diaphragm, it must spread out. Reducing aperture therefor must increase the diffraction of light, as it bends more around the edges of the aperture opening by an increasing degree as aperture diameter approaches the wavelength of light. (Similarly, in the case of a telescope, rather than light bending around the edges of the diaphragm blades, it bends around the secondary mirror support in the center.)