On 2006-06-26 12:40, John Francis wrote: > You're both wrong. It's either no darker (just cropped to a > smaller area) -
so you assume that there's no additional magnification within the viewfinder? Personally, I assumed first that the viewfinder image is the same as before. I checked - and I'm wrong. Analog kameras where at abount 0.7 .. 0.8 magnification, covering around 92 % (MX/ZX series). *ist was 0.7 of 90 %. Digital cameras where 0.95 of 95% (*istD or *istDS) and 0.85 of 95% (*istDL and probably K100D). I guess this is somewhere in between: the viewfinder is not half area (as the sensor is, cropped from 24x36 to roughly 18x24). But it's not magnified to full format viewfinder either. > so just as bright in illumination per equal > area (or equal solid angle submitted at the eye), or it's > 1/(1.5)^2 as much total illumination, which is a little over > one stop difference (it's like going from f/1.4 to f/2.1). Hm - I guess you are right. Half sensor area translates directly to half light, which is exactly one aperture step. So what's the total? (1/2 * 0.95) / (1 * 0.7) = 0.67. The K100D has 67% brightness of a ZK-L? Please correct me when I'm wrong... -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
