Nikon S9900 has a
16.0MP 1/2.3' (6.17 x 4.55 mm ) sized CMOS sensor . On the other hand, Fujifilm S8100fd has a
10.0MP 1/2.3-inch (6.17 x 4.55 mm ) sized CCD sensor .
Nikon S9900's sensor provides 6MP more than Fujifilm S8100fd's sensor, which gives a significant advantage in real life. You can print your images larger or crop more freely.
On the other hand, please keep in mind that Max sensor resolution is not the only determinant of resolving power. Factors such as the optical elements, low pass filter, pixel size and sensor technology also affects the final resolution of the captured image.
Below you can see the S9900 and S8100fd sensor size comparison.
Sensor Size and Resolution Comparison image of Nikon S9900 and Fujifilm S8100fd Cameras
Nikon S9900 and Fujifilm S8100fd have the same sensor sizes so they will provide same level of control over the depth of field when used with same focal length and aperture. On the other hand, since Fujifilm S8100fd has 59% larger pixel area (2.81µm 2 vs 1.76µm 2 ) compared to Nikon S9900, it has larger pixel area to collect light hence potential to have less noise in low light / High ISO images.
Sensor size sets the circle of confusion, so it drives how much depth of field each
of these bodies gives you at a given aperture. You can calculate it exactly for the
Nikon S9900
or the
Fujifilm S8100fd ,
including hyperfocal distance.
One thing to watch when comparing these two: they have different sensor sizes
(1/2.3' versus 1/2.3-inch),
so compare them at the same framing rather than the same focal length.
Put the same focal length on the smaller sensor and it will show less depth of
field, not more, because it also crops tighter. The calculator shows the
full frame equivalent focal length for each body so you can match the framing.
