41

If you shoot from the same position with both lenses, then taking the 35mm lens and cropping it to the same angle of view of the 50mm lens will give you pretty much the same picture, other than the differences in optical quality between the two lenses and the resolution lost to cropping. But even if you were to shoot with the same lens, shooting from a ...


28

The flattening or compression effect is not caused by a particular kind of lens, it applies to all lens in the same way. Actually, this property of lenses applies to our own eyes as well. The factor that affects flattening is the distance from the camera to the subjects. Consider the following exercise: Place two friends 1 meter away from each other. ...


20

The proof is in the pudding – the focal lengths are not exactly the same as yours, but the differences are obvious...


18

I believe the effect has to do with the RATIO of distances from the camera to various parts of the subject / scene. For example, if you take a wide-angle shot of a person's face, their features are exaggerated because the camera-to-nose distance might be half of the camera-to-ear distance. On the other hand, consider the same shot taken with a telephoto ...


14

Thanks to my schwifty skills in Inkscape, the rotation here is slightly off but the following shows exactly what you're comparing. These are the fields of view of a Nikon 35mm (inner) and a Nikon 50mm (outer). So even when you're getting approximately the same stuff in the frame, the 35mm is much wider, focal distances are slightly different too. If you're ...


11

Many (possibly most) modern SLR lens systems return focus setting data to the camera. Potentially the precision of data returned could be high - something better than 1% of range would be possible and meaningful with modern systems. However, it appears that most if not all systems use a simple gray-coded* system with perhaps 16 steps. Number of steps ...


11

You do get less light as you move further from a object. However, that less light is focused on a smaller area. It so happens that the two effects cancel out, and the focused image of a object will be the same brightness as distance is changed, assuming the f-stop is held the same. For example, moving twice as far away means the lens intercepts ¼ ...


11

The equation assumes a simple single element lens that is bilaterally symmetrical. The camera lens, to mitigate the 7 major aberrations (shortcomings that degrade) is constructed using several individual glass lens elements. Some are positive in power, some with negative power. Some are air-spaced apart and some are cemented together. Because this array ...


9

Perspective is the size relationship of objects. You might say the ratio of size of foreground objects to background objects. Perspective plays a big part when it comes to judging distances to objects. It is a major contributor to our 3D vision. When we change our distance, the size relationships of objects at various distances change. When we change focal ...


8

Why work so hard? I think you are making this much harder than it has to be. Getting something to be exactly 640 pixels in the image will be difficult, require careful measurements and accurate calculations and is very error prone (and for something really close the internal construction of the lens makes a huge difference and this will be impossible to ...


8

The size of the sun or moon in mm in the sensor plane will be approximately f / 110 where f is your focal length. A typical APS-C sensor is 16mm tall (or 15mm for Canon), hence a 1760mm lens would be required to fill the frame (vertically). 800mm would get you about half the frame, 400mm one quarter etc. A "full frame" sensor is 24mm tall, so you'd need ...


7

first of all, kudos on your effort to break a photography problem down to first principles. The discrepancy you've observed stems from a common oversimplification. Your 100mm Lens is actually what optical engineers refer to as a "lens assembly" As you likely know, it is comprised of multiple lens elements in groups working in tandem to form, refine, and ...


5

The relationship is a simple inverse, i.e. object size in image = Object size * focal length / object distance from camera If you keep the same object and the same focal length you get: size = 1/ distance (the =-sign should be proportional-sign).


5

Inversely linear is a good approximation. Imagine a 1,7m tall girl at 1 m distance b. Her head is at point B. How does the size/length of an object vary with distance? Let the girl walk away from you. Her size a stays the same. She appears smaller, because she is appearing under a smaller angle. Her angular size changes. Try to imagine it with the ...


5

According to the FAQ for the particular calculator you're using, the calculations are performed for a thin lens, and hence the front nodal point of the lens should be used. The author recommends using the front surface of the lens, on the assumption that the front nodal point is somewhere inside the lens, and this assumption will yield a conservative ...


5

the only relevant thing is that the same light which falls on your main subject falls on your card. The field of view, distance and so on don't enter in the equation. but you have to be sure that by standing near the subject you don't influence in any way the light, which is not always easy since our eyes are very quick to adapt to variation and they ...


5

It's actually far simpler than any of the answers posted so far! You don't need trigonometry, or field of view calculators at all, all you need is multiplication and division! Firstly (all else being equal) the size of your object in the image is directly proportional to the focal length (if you double the focal length you double the size). So if you know ...


5

TLDR: No because you need an additional variable, either the height of what the subject is that fills the frame or the distance at which you are focused (infinity doesn't work though). Long answer... To do this you can use simple trigonometry, but you would need to know either the current or desired distance to subject or the size of your subject (...


5

You are misunderstanding the compression. The face appears flat because you are observing it from a distance and thus the angles of it ARE flat. If you move in closer, you are viewing it more from the side/off-axis and you get more depth. The focal length is simply used to crop the image for you. You could take a portrait with an 135mm lens at 3 meters ...


5

What lens would work for a canon 6D to take up close photos of birds far away in the trees? It comes down to three questions: How up close do you want to get? How far away are the trees? How big are the birds? In landscape orientation, your 400mm lens takes in 3.4 degrees over the height of the sensor and 5.2 degrees across the width. That means that a ...


5

Sensor height is indeed the physical measurement of the vertical dimension of the sensor's active pixel area. From Wikipedia's Image sensor format article, the phone's sensor format of 1/3" (actually quoted as 1/3.06" in the LG G2 phonearena.com page you linked to) is 4.8 mm wide by 3.6 mm high. I think the problem you encountered is assuming that 1/3" was ...


4

Yes for most camera systems: For Canon EOS, select EF and EF-S lenses transmit distance information through the EF mount. For Nikon, D- and G-type Nikkor lenses transmit distance information through the F mount; this is what the D designation means. G lenses are the same, only without an aperture ring. For Sony, all current lenses transmit distance ...


4

Subject distance is an intrinsic factor in the depth of field formula: DOF = (2Ncf²s²) / (f⁴ - N²c²s²) Where: N = F-Number c = Circle of Confusion f = Focal Length s = Distance to Subject Circle of Confusion (CoC) is a quirky factor here. CoC affects the "depth" of the scene that appears to be in focus. In reality, there is only one infinitely thin ...


4

It's probably not the place to ask, but the answer is fairly easy to arrive at using similar triangles. Given that the feature fills your shot, you have two ratios that should be the same: focal length : sensor height distance to feature : height of feature Since you have a 38mm focal length (at 35mm equivalent and a 35mm frame height is 24mm) then you ...


4

I'm sure this is a duplicate, but I can't find a good answer to the question in the archives so here goes. The relationship between object size and distance is an inverse linear relationship, i.e. size is 1 / distance. This makes sense when you think about it as if you double the distance the size halves. This is why you appear to be observing an ...


4

My answer will be the same as @MattGrum's, but I would like to add a teeny bit of explanation. Say the diameter of the sun/moon is h, the distance is d and your lens' focal length is f. Assume that the image has come to a focus at a distance l behind the lens and has a diameter of i on the sensor. If you quickly sketch out things assuming a thin lens you ...


4

The minimum and maximum focus distances have no direct relation to the focal length of the lens. All lenses have a minimum focus distance. For Macro lenses this is closer than for normal lenses (is the main part of what makes them Macro). The majority of lenses have an infinite maximum focus distance (focus at infinity). Under certain conditions ...


4

Is it possible to calculate the distance of an object to the camera with two pictures which are taken a few cm apart? It depends. You'd need to know something more than just having the two photos. Examples of helpful information include information about the camera and lens such as the sensor size and focal length, and also exactly how the camera's position ...


4

Do you have a selection of lenses (or a zoom lens) now? Shoot a table-top test with different focal lengths, repositining the camera to get tye same view of the foreground object. Then look carefully at the photos to see for yourself. If you're one of those peopke where this doesn't just scream at you, it's good to develop your eye to seeing the ...


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