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|Oct 24, 2002, 6:31 AM||#1|
Join Date: Sep 2002
602 and flash problem
Had to return my 602 I bought the other day.
With the camera set to auto, or any other mode ,it seemed random pictures would come out very dark even though the flash fired. Take the picture next time flash was ready and pic would be OK.
I also thought the flash range was not up to par.
I really liked the feel of this camera and thought the photo quality was very very good for a 3m camera.
Instead of getting a replacment I am now trying out a Nikon 5700.
To be honest, not sure the Nikon is worth the extra 500 bucks.
Anyone else have this problem with the 602?
I may consider returning the Nikon and getting another 602 if this was just an isolated 602 problem with the camera I had
|Oct 24, 2002, 7:44 AM||#2|
Join Date: Aug 2002
You sent the cam back. If you still have all the shots, have a look at the EXIF data, it might give up some clues. If the flash was supposed to be on, the exif says it was, but it didn't fire, I suspect the cam sensed the flash unit was open and ready, but the flash circuit might be faulty.
If as you say, the opposite was happening, what were the aperture/shutter settings for the OK and dark flashshots, does the exif data indicate any warnings for the dark shots you might have missed in the viewfinder?
|Oct 29, 2002, 10:11 PM||#3|
Join Date: Oct 2002
How far were the subjects from the flash?
You have to keep in mind that the flash is very brief, much faster than the shutter speed. So, if you take one picture of a subject at 1/30th of a second with the flash, and another at 1/60th, the main subject will appear exposed exactly the same in both because the same amount of light was reflected by it. (The parts of the scene unaffected by the flash, such as a distant background, will vary accordingly as always.)
Also keep in mind the inverse square law. Light intensity falls off as the inverse square of the distance. That means if the flash is 2 meters from the subject in one picture, and you want the same exposure of that subject with the flash now 5 meters away from it, you have to increase the intensity of the flash intensity by 625% even though you've only increased the distance between the subject and the flash 250%. Why?
In the original image, intensity I incident on the subject is:
I = F/r^2
where F is flash intensity. If you increase the distance to y=2.5*r, look what happens:
I = F/y^2 = F/(2.5*r)^2 = F/(6.25*r^2) = F/6.25*1/r^2
So by moving the subject 250% farther from the flash, you've effectively divided the intensity of the flash by 6.25. That means you have to compensate by making it 625% brighter in order to achieve the same exposure as before.
On the S602, the flash can only be adjusted 3 stops in either direction. So how far can you move the subject and get the same exposure if you bump the flash up +1 eV (3 stops)?
So let G=3*F and the new distance y=a*r:
I = G/(a*r)^2 = 3*F/(a^2*r^2) = 3/a^2*F/r^2
Since the exposure of the main subject is the same in both images, and I=F/r^2 in the first image, we set them equal and solve for a:
I = F/r^2 = 3/a^2*F/r^2
==> 3/a^2 = 1
==> a^2 = 3
==> a = sqrt( 3 ) = 1.732...
So, that means you can get the same exposure of the main subject by juicing the flash up to full power +1 eV and moving the subject back from the flash by about 75%.
So, if the subject was 2 meters away and exposed properly at +0 eV, you'll get that same exposure at +1 eV and 3.5 meters away from the flash.
Incidentally, the inverse square law is why you should never take flash pictures down a table of people gathered. The flash intensity of the subject far down the table will be many, many stops lower than the subjects near the flash, so only the people nearest the camera will be exposed properly.
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