Understanding light fall off
Hi all,
I have read many times the rule about light fall off.. but I was wondering if light falls with the square of the distance (1/sqr(distance)) why this does not affect us when shooting a portrait (just an example).
Assume that you have a constant source of lighting that does not change for the whole period (that can be a flash firing at exact the same power). Subject also remains the same.
Assume also that I have I am standing at distance x from my subject working with a zoom lens and I am getting a nice exposure (nice exposure is subjective but still this gives me specific settings in the aperture, shutter speed and iso).
Assume now that I am moving back in a bigger distance from my subject.. Lets say that I am doubling or tripling my distance from the subject. I keep though exact the same crop on my subject by zooming my lens as needed (assume that my zoom lens can give me the magnification needed by keeping the same aperture). That means that I get exactly the same two shots/crops (with only the background separation changing as focal length is increasing.. and the compression of my subject).
If I am not wrong, between my initial position at x meters and my future position that is two or three times further I could have used exactly the same settings (aperture, shutter speed and iso) as on my first shot (that I was at x distance).
If this holds true can someone explain me why is this happening? In my first case, when I am at x distance.. the light is generated from my source... reflects on my subject and then travels x meters to reach my sensor.
When I am doubling my distance from the subject.. the same light source produces the same intensity light that goes to my subject gets reflected and travels again 2x meters before hitting my sensor. Would not that mean that the light know coming on my sensor would be degraded by the same law of light fall of? In other words would not I need to compensate for this extra loss by opening my aperture (if I a controlling the flash exposure) or my shutter speed (if my light source was ambient light somehow).
Can someone try to see what really is baffling me here?
I would like to thank you in advance for your reply
Regards
Alex
I have read many times the rule about light fall off.. but I was wondering if light falls with the square of the distance (1/sqr(distance)) why this does not affect us when shooting a portrait (just an example).
Assume that you have a constant source of lighting that does not change for the whole period (that can be a flash firing at exact the same power). Subject also remains the same.
Assume also that I have I am standing at distance x from my subject working with a zoom lens and I am getting a nice exposure (nice exposure is subjective but still this gives me specific settings in the aperture, shutter speed and iso).
Assume now that I am moving back in a bigger distance from my subject.. Lets say that I am doubling or tripling my distance from the subject. I keep though exact the same crop on my subject by zooming my lens as needed (assume that my zoom lens can give me the magnification needed by keeping the same aperture). That means that I get exactly the same two shots/crops (with only the background separation changing as focal length is increasing.. and the compression of my subject).
If I am not wrong, between my initial position at x meters and my future position that is two or three times further I could have used exactly the same settings (aperture, shutter speed and iso) as on my first shot (that I was at x distance).
If this holds true can someone explain me why is this happening? In my first case, when I am at x distance.. the light is generated from my source... reflects on my subject and then travels x meters to reach my sensor.
When I am doubling my distance from the subject.. the same light source produces the same intensity light that goes to my subject gets reflected and travels again 2x meters before hitting my sensor. Would not that mean that the light know coming on my sensor would be degraded by the same law of light fall of? In other words would not I need to compensate for this extra loss by opening my aperture (if I a controlling the flash exposure) or my shutter speed (if my light source was ambient light somehow).
Can someone try to see what really is baffling me here?
I would like to thank you in advance for your reply
Regards
Alex
0
Comments
Otherwise, a simple, non-coupled, and non-diffused electronic flash will indeed spread light at the rate of the square of the distance from the subject. Spreading the light also reduces the illumination falling upon the subject, resulting in a different exposure and requiring either a longer shutter duration, a larger aperture, or a higher ISO (or some appropriate combination of those exposure elements).
Moderator of the Cameras and Accessories forums
We all know the aperture stops f2.0, f2.8, f4, f5.6, f8, f11, f16 each allows 1/2 the previous stops amount of light.
Under constant illumination - say a studio strobe, the light is reduced by half, one stop as you move the strobe from 2 feet to 2.8 feet, and again by half as you move the strobe to 4 feet, and then 5.6 feet, and then 8 feet. See a pattern here? Cool fact that is little known. Try it some time and see if I am not correct.
Merry Christmas!
Moderator of the Technique Forum and Finishing School on Dgrin
I think the answer is that since I am keeping the crop the same as I am movign backwards (my zooming in) the f/4 for example remains the same. If I started at f=50mm at f/4 and then moved to f=100mm (with exact same crop and lighting conditions) that f/4 allows more light to go through the lens.. so the compensation I am looking for is integrated into the way we discuss about apertures.
Regards
A