Computes Field of View of camera and lens, both the dimensional size of the Field Of View seen at a specified distance, and also Angle Of View, for any sensor size.
Many other Field of View calculators assume 3:2 DSLR sensors on
Accuracy is best if you can enter the actual sensor dimensions (mm, from camera manual specifications)  Option 1.
Or, if you can determine the crop factor, sensor size can be computed  Option 3, see crop factor notes below.
Or, you may be able to select on
Or, if you can measure the field of view at a specified distance, and know the focal length, sensor size can be calculated  Option 8.
On
Enter Focal Length and Distance, select an Option 17, then click Compute. FOV depends on focal length, distance, and sensor size.
Computing Angle of View (degrees) for a lens focal length and sensor is independent of subject distance, so for that goal, just ignore the Dimension part.
 

Numbers on
Crop factor is the ratio of (Equivalent focal length for 35mm film seeing the same view) divided by the actual focal length on this camera's sensor size. For example, the specification for a compact camera might say:
"Focal Length: 4.5 (W)  81.0 (T) mm (35mm film equivalent: 25450mm)". (the wide angle and telephoto extents)
That case makes crop factor be 25mm/4.5mm or 450mm/81mm, both equal to 5.55 crop factor. However, focal length of compacts is probably not known except at the extremes that the spec mentions. Crop factor is often specified as a slightly rounded number. For example, the 35mm film frame is 36 mm wide, and if the DX sensor is 23.5 mm width, then it is actually 36mm/23.5mm = 1.53 crop. But, the focal length number is also approximated anyway.
Entering exact sensor dimensions above would be the most precise. The camera manufacturers specify the equivalent 35mm crop factor from the diagonal ratio to 35mm film (because many of us are very familiar with 35mm film, and crop factor tells us what to expect now). We may not know sensor size or focal length on compacts, except at either end of the zoom range, but then we can determine crop factor, for example, if they specify their 6.1mm lens is equivalent of 24mm lens on a 35mm camera, then obviously their crop factor is 24/6.1 = 3.93.
The differences in the Aspect menu in Option 3 above is that DSLR and compacts take 3:2 or 4:3 photos, larger than 1920x1080, and their 16:9 movies are constrained within that sensor size, fitting its width. Whereas camcorders typically take 16:9 still photos that are still 1920x1080, with movies fitting the full diagonal, not constrained by any 3:2 or 4:3 sensor width.
Typically photo cameras will use the full sensor width for their HD movie width (D7100, D600, D750) and that is assumed here, but for example, the Nikon D800/D810 use slightly less width (these D8xx manuals specify the sensor image area for HD movies is 32.8 x 18.4 FX, and 23.4 x 13.2 DX).
There are approximations. The math is precise, but the da
The Marked focal length of any lens is a rounded nominal number, like 50 or 60 mm. The actual can be a few percent different. Furthermore, the Marked focal length is on
Actual focal length can be determined by the magnification (Wikipedia). Or, the focal length (f), the distance from the front nodal point to the object to photograph (s1), and the distance from the rear nodal point to the image plane (s2) are related by this equation:
If OK with a little geometry and algebra, you can see the derivation of this classic Thin Lens Equation at the Khan Academy.
In this equation, we can see if the subject at s1 is at infinity, then 1/s1 is zero, so then s2 = f. The marked focal length applies when focused at infinity.
The field of view math is basic trigonometry. The focal length measures from lens node to sensor. We compute the right triangle on center line, of half the sensor dimension, so the half lens angle = arctan (sensor dimension / (2 * focal length)). The Subject distance is in front of lens node, with same opposite angle. Field dimension = 2 * distance * tan (center line half angle). The problem is that focal length f becomes longer when focused at close distances (but the opposite can be true of a few zoom lenses). That becomes an insignificant field of view difference at normal distances, 1 meter or more.
Multielement camera lenses are "thick" and more complex. We are not told where the nodes are designed, normally inside the lens somewhere, but some are outside. For telephoto lenses, the rear node (focal length from sensor plane) is in front of the front lens surface. The designer's term telephoto is about the reposition of the nodal point so that the physical lens is NOT longer than its focal length. Yet, this rear node is generally behind the rear lens surface of a wide angle lens (lens moved well forward to provide room to allow the larger SLR mirror to rise... 12mm lens, 24mm mirror, etc). This nodal difference is on
The Subject distance S is measured to the sensor focal plane (it is the "focus distance"), where we see a symbol like Φ marked on the top of the camera (near rear of top LCD). The line across the circle indicates the location of the sensor plane (for focus measurements). However, the Thin Lens Equation uses the distance d in front of lens. This is why we often see in equations: (S  f) used for d.
For Macro, computing magnification is more convenient than focal length (since we don't really know focal length at macro extension). Focal length and subject distance determine Magnification, which is the ratio of size of image to size of actual subject. Or size of sensor to the size of the remote field. We could compute that here, but magnification has more significance up closer (easier for macro), which is where our knowledge of the actual focal length is weakest. We could measure the field to compute the actual magnification, to then know the actual focal length. However Magnification is simply:
m = s2/s1. Or m = f/d. Or m = f/(Sf).
So from this, we know macro field of view is simply the sensor dimensions, divided by the magnification. Let's say it this way:
1:1 macro (magnification 1), the field of view is exactly the same size as the sensor.
1:2 macro (0.5 magnification), the field of view is twice the size of the sensor.
1:4 macro (0.25 magnification), the field of view is four times the size of the sensor.
This is true of any focal length for any lens (or method) that can achieve the magnification. Focal length and subject distance are obviously the factors determining magnification (it is still about them), but magnification ratio is simply easier work for macro.
The easiest method to determine field of view for macro is to simply put a mm ruler in the field. If a 24mm sensor width sees 32 mm of ruler, then that is the field of view, and the magnification is 24/32 = 0.75 (this scale of magnification is 1 at 1:1, and is 0 at infinity).
The definition of macro 1:1 magnification is that the focal length and subject distance are equal (distances in front of and behind the lens nodes are necessarily equal, creating 1:1 magnification). In this Thin Lens Equation, if s1 and s2 are equal, the formula is then 2/s1 = 1/f, or 2f = s1. So lens extension to 2f gives 1:1. And since f/stop number = f / diameter, then if 2f, then f/stop number is 2x too, which a double f/stop number is 2 stops change, which is the aperture loss at 1:1. We know those things, this is just why.
But the point here, if f is actually 2f at 1:1 macro, the field of view changes with it. None of the FOV calculators are for macro situations (too close, magnification is instead the rule there). Field of View calculators expect subject distance to be at least a meter or two, reducing the focal length error to be insignificant.
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