# Tag Info

1

On start up most cameras extend their lenses. They extend even more when zooming. Effectively the camera is increasing its air volume as it zooms and decreases it as it retracts. The air for this increase in volume is drawn in around the telescoping lens sections drawing in dust. Some cameras have better dust seals than others. I was put off from buying a ...

2

As @inkista explains, the "times-zoom" notation is just a relative measure, and doesn't tell you anything absolute. This is different from binoculars, where the number actually means how much bigger the subject appears than with the naked eye — that's because cameras project what's known as a "real image" onto the sensor or film, whereas binoculars form a "...

7

It's impossible to say. This is because a "zoom factor" is a relative measurement, not an absolute one. 10x zoom doesn't mean the picture is enlarged 10x. All that the zoom factor is telling you is how much longer the longest focal length of the lens is from the shortest one. Both a 5-50mm and a 50-500mm lens are 10x zooms. But they have vastly different ...

3

A 10x simply means that the longest (most "zoomed in") focal length is 10x longer than than widest focal length. This has the effect that everything in the photo is 10x bigger. It's a ratio, and doesn't have any meaning in meters or feet. It doesn't mean that it's as if you were 1 foot away instead of 10 feet. In fact, if you took photo with a wide angle ...

-1

A numerical example For years 35mm film cameras dominated. The image size was 24mm by 36mm. let us assume we want a 20mm high image of a 2,0 m high object at a distance u in front of the camera. Provided u is at least 10 times bigger than the focal length f, the ratio of image height to object height is approximately f/u, thus 20mm/2m = f/u, hence ...

0

In theory, it makes no difference how the lens attained focal ratio 4.8 (f/4.8). However, every lens is constructed via a series of compromises. In other words, the perfect lens, able to yield a faithful image has yet to be made. As to the constant zooms design: The problem is; as the focal length is changed, the size of images of objects is changed. Long ...

4

Typically, the constant aperture zoom would be sharper. The vast majority of lenses get sharper when stopped down, this includes constant aperture zooms and prime lenses too. So when you stop down an F/2.8 lens to F/4.5, let's say, you get would get an image which is very sharp. The variable aperture lens though would be wide-open at some focal-length and so ...

3

All other things being equal, yes — mostly. In terms of exposure and depth of field, they would be as close to functionally equivalent as things get in the real world. The shape of the aperture blades will have some impact, so in some cases you could probably tell by looking closely, even if that were the only difference. But in the real world, those lenses ...

4

However, it costs much more than I was planning to pay. As the size of the sensor increases the cost of the lens needed to provide an equivalent picture (same field of view, or depth of field, or brightness, etc.) increases rapidly. Not only must the lens be longer in focal length to provide the same field of view, but the diameter of the entrance pupil (...

2

If you are considering a camera with P pixels, and you have a minimum after-crop megapixel count you are considering (Pmin), then the digital zoom ratio of the camera is ZD = √(P / Pmin). Just multiply this digital zoom ratio by the lens's optical zoom ratio to get the total zoom ratio. Thus for the RX100 cameras, their optical zooms are 100/28 ~= 3.6×. ...

4

Cropping and zooming are basically equivalent, so this is easily calculated. Zooming in 2× — like, focal length 36mm to 72mm — is like cutting the frame in half. Of course, that's each dimension, so you need to square it to calculate megapixels. For example, a 24mpix image might be 6000×4000, and to simulate that 2× zoom, you'd cut it in half each way, to ...

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