42

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 ...


21

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


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

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 picture ...


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

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).


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 ...


7

Given the ratio of the Canon APS-C crop factor (1.6) to the MFT (2.0) sensors, the equivalent of a 300mm on your camera is 300*1.6/2=240mm so a bit longer than your intended lens (200mm). The zoom capability you mention is really a zoom range, but since your lens starts at 12mm which is a wide angle, the zoom factor from "normal" is smaller, more ...


7

Assuming all other things are equal: Your Canon 80D has a crop factor of 1.6. This means that your 300mm lens has an effective focal length of 300mm × 1.6 = 480mm. A micro-4/3s camera has a crop factor of 2.0. This means that a 200mm lens has an effective focal length of 200mm × 2.0 = 400mm, a bit less than the 300mm lens on the Canon. Working backwards, ...


5

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 ...


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

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

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 ...


4

That observation is true. The nearer to the lens, the smaller the depth of field. If you use an online DoF calculator like e.g. https://www.dofmaster.com/dofjs.html you can easily see this. Example: full frame, 50mm f1.4 at 1m -> DoF appr 0.02m at 3m -> DoF appr 0.19m at 10m -> DoF appr 2.16m at 50m -> DoF appr 75.40m at 95m -> DoF ...


3

There could be any number of factors coming into play here. Lack of sharpness is many times due to how you captured the image and not what lens you are using. It is not impossible for the lens to be at fault, but if you have a 50mm prime lens already - I would be willing to bet you can achieve very sharp images with it if it is used properly. Some of the ...


3

I think, whenever you reverse a lens by itself, your working distance becomes tiny - of the order of the flange to film distance, since that is how the optics work out. Your best bet for getting good magnification AND good working distance is simply to go for longer and longer focal lengths with short extension tubes. Here are some things you could try re: ...


3

The difference between the compressed look of a telephoto lens like an 85-135mm and the distorted look of a wide angle lens is referred to as perspective. The only factor that determines perspective is distance. Perspective is created by the angles between the camera (more specifically the geometric location of the entrance pupil, which may be in front of ...


3

This is a good question but many of the existing answers seem really convoluted and off the point. All that matters is distance to the subject, as this is what defines the distortions of facial features. Focal length is a secondary issue. If you use a wide angle lens, and you are at an optimal distance for a good portrait, you get a lot of the background ...


3

It really depends on the kind of shot and what conditions allow. That's why I answered in terms of focal lengths.) Shooting around the 65-85mm effective range is generally considered the most natural and most common, but if you want to flatten the image more, you can push it out to the 105-155 range. There are also some shots that work well in the 24-50 ...


3

The only thing that matters is your subject distance. I'll say it again, only. So to answer your question, can you stand 1 meter away and get the same compression? No. Take a look at this for more information: What is background compression?


3

If your lens has a distance scale it will indicate the approximate distance to the point of focus as measured from the film/sensor plane. In the image below, the point of focus is 1.5 meters (5') in front of the film/sensor plane. As the point of focus approaches infinity, the distance measurement becomes less precise. In general, the longer the focal length ...


3

You can go through the lens magnification. M = f / ( f - d) You know the projected size and want the real size, so you divide the projected size by M H_r = H_p / M In this case you get: M = 38 / ( 38-14.000.000), H_r = 24 / -0.0000027143 = -8.818.105mm ~ -8,818km Note the inverted projection by the negative M. The measurement is most accurate if your ...


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