happypoppeye: Would probably still buy the 25 f/1.4.
Already have the 1.4. But still itching of getting this one after tried the 42.5/1.7 for 5,000+ shots.
25/1.4 is large and heavy <grin>.
Abu Mahendra: A 50mm f/3.4 eq. lens. Joy. How exciting...
Oh crap! this is 2015 year. Why some T.rex still exists and shows carboniferous math skill.
Abu Mahendra: I agree this whole thing is quite silly. A Canon 50/1.8 will set you back $125. You've made your bed. Now lie in it.
Hmm, seem like toddler who just knows toys wanna express its knowledge as adult does too.
Hope the phone has enough juice left when I badly want to use it as the MOBILE PHONE.
Oh, you talk about spare batteries? OMG .. now the combination is as large and heavy as the old dSLR.
Lensjoy: The human eye is most sensitive to green. When calculating luminance (Y) from the respective RGB values for gamma 2.2, the formula is: Y = 0.2126*R^2.2 + 0.7152*G^2.2 + 0.0724*B^2.2Blue is only 7 percent of perceived luminance! So if I were to design a sensor doing what Sigma is purporting to do, I would put the high resolution four pixels in the center green layer, not the top blue layer.
Perhaps Sigma is doing something more complex than the schematic in the article above implies, but from the above description I expect this sensor design to have faults that we'll see remedied in a future version. I wouldn't buy the camera yet.
We have to go back to Physic 101 classroom <grin>
Photon can be both wave and particle _at the same time_.
The three layers sensor is doped with different elements. The final result is the uppermost P-N junction requires high energy photon to kick the electron (or hole) out of the orbit, generates electrical current.
After photons travel pass through, the high energy photon will be absorbed completely by this high-exitation P-N.
This 'absorption' is occured at the middle GREEN and lower RED layers, too.
Because of the wave/particle property, you may think that the "YELLOW" photon is really composed of RED and GREEN photon pairs.
Green will be absorbed in green. Red will be absorbed in red, and the same vertical axis.
After that - you just calculate what is the true color of that pixel, given signal strength from GREEN and RED layer.
Agree. It's not UW. You've to multiply focal length AND Depth of Field by 2.7.
Best for macrophotography. Shot with 105mm macro @ f/11 will give you 2.7 : 1 photo with DOF of 11 x 2.7 or about f/29-equivalent DOF !
Or low cost super-telephoto? The AF-S 300 mm f/4.0D-ED will give your bird size bigger than that price-of-leg&arm AF-S 600 mm f/4.0G-ED.