So I was really off on this one. I thought when it says 105mm zoom that means how far the flash is traveling. According to articles I'm reading up on, the higher the number the smaller or tighter the beam of light that's flashed. Would this be accurate?
I have a YN565EX and I was having some practice with it at a graduation for a friend. I must have had it one too high or something because in a couple of shots you can see some "scrunchy" faces in the background of the person I was focusing on (the faces are kind of humorous actually). But based on what I'm reading, in order to reduce the power of the flash you would take it down from 1/4 to 1/8th or even higher..yes?
When you buy a studio flash, they rate the flash power in watt-seconds... really just a representation of the amount of energy that the light might be able to provide. But there's more to it.
Suppose you have a flashlight. You point it at a target such as a white wall and notice how bright the spot is. Now remove the bulb and batteries and direct wire the bulb -- it's just not inside the flashlight anymore. Technically we know that it's the SAME battery and SAME light bulb. It provides the SAME "total" amount of light. But if you look at the wall it will be much less bright.
What's happening here is all about the reflector. When it was inside the flashlight, the reflector was focusing the beam to a concentrated area instead of allowing what was technically the same amount of light travel in every direction rather than all going the same direction. Focusing the beam of light is a really big deal.
This is exactly what the zoom head on your flash is doing.
Suppose your flash emits exactly 1000 photons of light when it fires (yes, clearly I'm making this up to make this easy to understand.)
Let's suppose you have a full-frame camera and a 100mm lens. The angle of view (when measured diagonally) is 24.4º (I looked it up using this website:
http://www.tawbaware.com/maxlyons/calc.htm ).
Let's suppose your subject is 10' away from the camera and flash. The "dimensional" field of view is an area that measures 4' 4" diagonally (4.33')
That means the number of square feet inside that circle (area = pi X radius^2) is half of 4.33 (2.165'). Square that to get 4.69, then multiply by pi (3.14) to get 14.7 square feet. That's the "area" of the circle large enough to cover the field of view we can see in our camera with the 100mm lens. The flash needs to be able to light up this area.
Since we presumed our light emits 1000 photons (a ridiculous number but it makes the math easier) then that means we can divide our 1000 photons across the 14.7 square feet and we discover that assuming even illumination of light, each square foot will be illuminated by 68 photons.
You can think of those 68 photons as a measure of light. The unit does not matter. But lets just presume whatever we named the unit of measure... we have 68 of those "units" of brightness.
Now lets go to a 50mm lens and do the same thing.
The angle of view (diagonally) for a 50mm lens is 46.8º (notice it's not actually double, but it is close.)
The dimensional field of view at 10' is now close to 8' 8" (8.66') diagonally.
The area of an 8.66' circle is now 58.9 square feet (nearly 59). Notice that this is enormously larger than the our original circle when using a 100mm lens.
Now lets divide our 1000 photons to spread them equally amongst each square foot of our image circle. 1000 ÷ 59 = 16.9 (about 17 photons per square foot).
That's actually ONE QUARTER of the light per square foot. Again... same light bulb, but we're going to change the reflector to allow it disperse over a wider area.
As you increase actual distance that the light must travel from the source, the beam spreads out. It spreads out following the "inverse square" law. That laws says that each time the distance increases by a factor equal to the square root of 2 then the amount of light reaching each square foot at that new distance will be HALF. The square root of 2 is very close to 1.4 and for simplicity just use 1.4 as the value. That means if we changed our subject distance to 14' instead of 10' (10 X 1.4) then exactly HALF the number of photons would be illuminating each square foot of our subject. Using the 50mm lens example, it would be 8.5 photons per square foot in stead of 17 per square foot.
Pause for a moment.
You do NOT need to go through all this math to do flash photography, but it DOES help to have a grounding in how it all works because you'll realize that you can actually do flash photography strictly by the numbers rather than by trial and error.
Speedlite flashes simplify a lot of this by using a system called the "guide number". The guide number is a measure of how far the flash can properly illuminate a subject for a good exposure. But this would depend on the camera exposure settings. So they use a baseline exposure value of ISO 100 (that's reasonable) and they use f/1.0. Alarm bells might be sounding because if you check your camera bag, I'll wager you do not actually own an f/1.0 lens. That's ok. The REASON they use f/1.0 is because it makes the math VERY easy. Just divide the "guide number" by the f-stop you ACTUALLY plan to use (yes.. it really is just that) and that's your new distance.
Every time you change the reflector, however, you change the guide number of the flash. That's because the guide number is not a measure of how much light the bulb can emit... it's a measure of how much light will actually be delivered to your subject and THAT depends on how tightly we focus the beam of light. You would have to know the guide number of the flash for each focal length (btw, this is something you can calculate using either an incident meter that can do flash metering OR just by using your camera (but that's another thread.))
Back to your original question... the "zoom" is a measure of an angle and NOT a measure of a distance.
If the flash head is "zoomed" to 100mm it means that assuming you are pointing the flash directly ahead and not using it to bounce off a ceiling, wall, or putting it through type of light modifier (soft boxes, reflectors, etc.) THEN it will illuminate a circularly area large enough that the field of view as seen through the camera using a 100mm would be completely illuminated by the flash.
If, however, you decided to put a 50mm lens on the camera but zoom the flash head to the 100mm position and took a photo you would notice heavy light fall-off around the edges of the frame because the beam is focused so tightly that it's only able to illuminate the center area.
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