dclark
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 Re: Pre-order June 13th: Sony FE 200-600 f5.6-6.3 G OSS ($1,998) | |
molson wrote:
Steve Spencer wrote:
...when the area of a sensor that you are using is twice as big as the area of another sensor you are using you have an additional stop of light.
When the light is spread over a larger sensor area, the intensity of the light reaching the sensor is reduced according to the inverse square law. So, for a sensor twice as big, the light intensity is reduced by four times, and you are in fact losing two stops of light with the larger sensor. Simple physics...
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Steve Spencer wrote:
molson wrote:
Steve Spencer wrote:
...when the area of a sensor that you are using is twice as big as the area of another sensor you are using you have an additional stop of light.
When the light is spread over a larger sensor area, the intensity of the light reaching the sensor is reduced according to the inverse square law. So, for a sensor twice as big, the light intensity is reduced by four times, and you are in fact losing two stops of light with the larger sensor. Simple physics...
No Cliff, you are wrong about that. The brightness is the same at the same aperture. The formula for the amount of light is Brightness X Area X Time. Brightness is controlled by the aperture. Area is controlled by the size of the part of the sensor you are using, and time is controlled by shutter speed. That give you the amount of light the sensor has to work with.
What you said above is only true if you keep the amount of light constant, but if the brightness of the light is controlled by the aperture, which it is, then that formula does not apply.
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codyconway wrote:
Steve Spencer wrote:
molson wrote:
Steve Spencer wrote:
...when the area of a sensor that you are using is twice as big as the area of another sensor you are using you have an additional stop of light.
When the light is spread over a larger sensor area, the intensity of the light reaching the sensor is reduced according to the inverse square law. So, for a sensor twice as big, the light intensity is reduced by four times, and you are in fact losing two stops of light with the larger sensor. Simple physics...
No Cliff, you are wrong about that. The brightness is the same at the same aperture. The formula for the amount of light is Brightness X Area X Time. Brightness is controlled by the aperture. Area is controlled by the size of the part of the sensor you are using, and time is controlled by shutter speed. That give you the amount of light the sensor has to work with.
What you said above is only true if you keep the amount of light constant, but if the brightness of the light is controlled by the aperture, which it is, then that formula does not apply.
kind of - logic says that the area of light that is reflected through an optic is not collected light, but projected. Thus the projection of the light is the area that is applied. In a smaller sensor, you get that "crop" factor because the light is only being captured from the central zone. The excess light is then difussed by the housing / padding, etc. You do not gain or lose light based on sensor size. I wish people would understand that better. Then again, perhaps I'm the one with the twisted view on it?
The number of photons per unit area per unit time at the sensor depends on the f/number. That's the simple physics. Consequently the total number of photons (and photo-electrons) in the image is proportional to the area of the sensor. Larger sensor means more photo-electrons, lower shot noise, and a higher signal to noise ratio.
Dave
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