of a Camera and Lens

Computes Field of View seen by 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 (film or digital). The Angle of View applies to this focal length on this size sensor, and the angle is independent of distance. But the dimensional field of view is calculated at the one specified distance from camera.

There is also a large table of angular Field of View (degrees) for many lens focal lengths and a few popular sensors lower down on this page (or Option 1 or 3 shows more specific cases).

This calculation works when sensor size and focal length are known. These values may be difficult to determine for compact cameras and phones and camcorders, but DSLR should be easy. Some Field of View calculators assume only one default sensor size or ratio, but this one is general purpose, offering 3:2, 4:3, and 16:9 too. Or entering the sensor or film size can be any ratio.

The calculator is NOT for fisheye lenses (their view will be wider), and it is not for macro distances (at end below). Field accuracy will be better at a distance of at least several feet (say 6 feet or 2 meters), because the lens focal length number changes with closer distance. The stated value of focal length applies to Infinity focus.

Enter Focal Length and Distance, select a sensor size in Option 1-4, then click Compute. FOV is computed from focal length, distance, and sensor size. Or Options 5-8 are more special purpose.

**Numbers only** (a NaN result means input is Not A Number). Periods as decimal points are OK.

Please report ( Here ) any problems with the calculator, or with any aspect of this or any page. It will be appreciated, thank you.

Focal length is entered as millimeters, but distance units can be feet or meters, miles or cubits, your choice. Dimension results will be those same units. If the units are to be feet, then clicking the little **Green button** (by Distance, marked I for Inch) will assume feet, and will repeat Dimensions in feet ' and inches " format. You can click the Green button again to toggle this option off.

Angle of View (degrees) for a lens focal length and sensor is independent of subject distance, so if only that Angle goal, you can simply ignore the Distance and field dimension parts. Dimensions are calculated for the Distance, for that Angle.

Sensor size is all important to know. If unknown, it can be computed from Crop Factor if known (see below).

Computing also must know correct focal length used. For fixed lens small cameras that do not zoom, the computed sensor Diagonal shown by the calculator (click it once) might be a crude ballpark guess for focal length? Just saying, a focal length equal the diagonal is considered to be a "normal lens", but many are a little wider (shorter).

Select an Option, and click the Compute button (for all Option numbers).

**Option 1**- Best accuracy is when entering actual exact sensor dimensions (mm, from camera manual specifications).**Option 3**- Or, if you can determine the lens crop factor (from specifications), sensor size can be computed. See crop factor notes and 16:9 notes below.Special Feature: In Option 3, the

*All Focal*button will instead show Angular field of view of many focal lengths, used with this sensor size and aspect ratio. The crop factor (1, 1.5, 1.6, 2.7, etc) shown in Option 3 computes the sensor size. Or for other specific numbers, you can just use Option 1 or 3, etc.**Option 4**- You may be able to select one of the general sensor descriptions. Film sizes should be accurate, and the larger sensors with actual WxH dimensions too, but the 1/x inch sensor numbers are Not meaningful. The 1/x inch description is NOT the sensor dimension, not even related to the sensor. It seems like fraud, it compares the sensor to the picture size of an old glass vidicon tube (1950s, even before CCD). But the 1/inch dimension was that outer glass tube diameter, and there is nothing about the digital sensor that is that dimension (its diagonal is probably no more than 70% of that dimension, if that). It is an inflated false number. Instead, we need to know the sensors actual real W x H size in mm. Or an accurate Crop factor can compute it here. The list here tries to substitute some known sensor sizes for the 1/x numbers, but there are usually a few different sensor sizes claiming the same 1/x number, so it can be wrong. The correct calculation really needs the correct sensor size, WxH in mm.Special Feature: In Option 4, the

*All Sensors*button will instead show summary of all sensors in the Option 4 list.**Option 6**for angular field does not depend on distance, so you can enter a known angular goal, like 0.5 degrees for the Moon, and just ignore entering any distance. Or if there is a distance, the field dimensions will be computed for that distance.**Option 8**- Probably not practical, but technically, if you can measure the field of view dimensions at a specified distance (at several feet), and know the focal length accurately, sensor size can be calculated. (the diagonal will be computed for you, from your horizontal and vertical dimensions).

The Circle of Confusion (CoC) is not used here, but is shown as the standard diagonal/1500 value used in Depth of Field computation, merely mentioned here for the various sensors.

The biggest risk to accuracy is not actually knowing the specific sensor size, and not actually knowing the specific accurate focal length. DSLR cameras do better describing those specifications. Compacts and phones may be difficult to know (and Crop Factor might be your best tool).

Possibly the most usable general understanding to compare focal lengths (for a given sensor, and a medium or longer lens focal length, not too short) is that the resulting image size is the simple ratio of the two focal lengths. A 400 mm lens will show an enlarged view 4x the subject size and 1/4 the field of view of a 100 mm lens (100/400 = 1/4).

One example: Suppose you want a portrait to include a 2x3 foot subject area. You know you need to stand back six or eight feet for proper portrait perspective. What focal length is that field size going to require? And the background may be five feet further back, how large does it have to be? This calculator can plan or verify your choice. If you want to see fractions of inches better, you can enter distance as inches (ten feet = 120 inches), but realistically, the accuracy (of the focal length) may not be up to fractions of an inch.

**Determine Crop Factor**: Field of View requires knowing focal length and sensor size. In some cases, focal length is in the Exif data. Sensor size can be computed from Crop factor.

Crop Factor is the ratio of

(Equivalent focal length with 35 mm film that sees the same view)

divided by the

(actual focal length with this camera's sensor size).

(Equivalent focal length with 35 mm film that sees the same view)

divided by the

(actual focal length with this camera's sensor size).

For example, the specification for a compact camera's zoom lens might say:

Focal Length: 4.5 - 81.0 mm (35 mm film equivalent: 25 - 450 mm)

That case makes crop factor be 25 mm / 4.5 mm or 450 mm / 81 mm, both equal to 5.55 crop factor.

Or, crop factor is also the ratio of the 35 mm film diagonal divided by this sensors diagonal. We know 35 mm size, so knowing crop factor can tell us sensor size. So computing sensor size will be as accurate as your data numbers. However, the focal length actually used in compacts is probably not known except at the extremes that the zoom spec mentions (and default power up zoom is probably some unknown intermediate point).

Smart phones don't zoom, but official specs are rare. Sensor size can be computed from crop factor, if known. Sources differ, and Apple specs don't say, but one estimate is that an iPhone 5 sensor size is 4.54 mm * 3.42 mm and crop factor 7.93x. If you can determine either sensor size or crop factor, the calculator above will compute the other. iPhone 5 Exif reports focal length of 4.2 mm.

Crop factor is often specified as a slightly rounded number. For example, the 35 mm film frame is 36 mm wide, and if the DX sensor is 23.5 mm width, then it is actually 36 mm/23.5 mm = 1.53 crop, but called 1.5x. 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 35 mm crop factor from the diagonal ratio to 35 mm film (because many of us are very familiar with 35 mm film, and crop factor tells us what view 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 6.1 mm lens is equivalent of 24 mm lens on a 35 mm camera, then obviously their crop factor is 24/6.1 = 3.93. And after the fact, the Exif (Manufacturers Data section) often shows the zoom focal length used for the picture (see a viewer that will show this).

**HDTV movies:** Special considerations. The differences in the Aspect menu in Option 3 above is that DSLR and compacts take 3:2 or 4:3 photos, and their diagonal fits in the diameter of the lens circular view. HDTV movies are of course resampled to 1920x1080 or 1280x720 pixels. It possibly might be smaller, but it can't be larger. Because 16:9 movies in still cameras are constrained within that still camera sensor size, *limited to the same width*. Whereas camcorders *fit the full diagonal*, not constrained by any 3:2 or 4:3 sensor width. Generally, if camcorders also take still pictures, those are 16:9 too.

In this image at right, the blue circle represents the diagonal of the image that the lens projects. 16:9 in camcorders is wider, but not if constrained within the more narrow frame sizes. I suppose there could be some exception, but camcorders should have a 16:9 sensor, and DSLR, compacts and phones typically have 4:3 or 2:3 camera sensors. Their 16:9 width will be the same as the 4:3 width, but then the height is less. Another page also describes this effect numerically.

Determine aspect ratio by dividing image width by the height (both in pixels), which is its aspect ratio. 3:2 divides as 1.5, 4:3 divides as 1.33, and 16:9 divides as 1.78. So if your movie mode takes still pictures of aspect 1.78, then it is probably a camcorder with a 16:9 chip.

Many photo cameras will use their full sensor width for their HD movie width (Nikon D7100, D600, D750 specify their movies are full sensor width) and that is assumed here. But for example, the D800/D810 use slightly less width (these D8xx manuals specify the sensor image area for FX HD movies is 32.8 x 18.4 mm FX, and DX HD is 23.4 x 13.2 mm). That is 35.9/32.8 = 91% of full width. An iPhone 5S (on a tripod) measures the movie field of view to be 91% less too. This field of view reduction could be computed by using effective Focal Length = 1/0.91 times longer (10% longer), OR a crop factor 1/0.91 times larger (10% larger).

But actually knowing the actual sensor size is the key.

Here's a chart of Angular Field of View (Width, Height, Diagonal, in degrees) for many lens focal lengths ("Lens mm") on popular sensor sizes and common aspect ratios. One example of the chart use might be to compare the width of view on two cameras.

The calculation is NOT for fisheye lenses (their view will be wider), and it is not for macro distances. Field accuracy will be better at a distance of at least several feet, because the lens focal length number changes with closer distance. The stated value of focal length applies to Infinity focus.

The top row marked WxH shows the sensor dimensions in mm, computed from the crop factor. Also 16:9 image size can be limited to be no wider than camera 3:2 or 4:3 sensors (or left at full size for camcorders). See *HD movies* above.

The 4:3 sensors are missing here, but there are so many sizes, and some web screens are so small. Phones normally don't zoom, and we probably don't know focal length for compacts except at end extremes. The Pentax Q7 with 1/1.7" sensor is mentioned because it is a smaller sensor, but with interchangeable lenses (its normal zoom is 5-15 mm). Tiny sensors require very short focal length to achieve any usable field of view, but the short focal lengths can be included.

**You can easily add another sensor into the chart** (specify crop factor), which will replace the last sensor (that original default 4.65 crop in last column now is for the Q7 mentioned.) Numbers only. See *Determine Crop Factor* above.

FWIW, the size of our full Moon is very near 0.5 degrees (its size varies slightly in its elliptical orbit).

We can realize from this chart that generally, a 2x longer focal length shows a view only half as wide. Or 4x is 1/4, etc. But this is not linear, meaning it is Not true of wide angles. This non-linearity is due to the angle trig, not the focal length. I am arbitrarily suggesting that a horizontal view width of the "normal" lens, or roughly about 40 degrees horizontal reasonably satisfies this, and then 2x focal length will be near half angle, about 20 degrees. This is true of full frame at about 50 mm, and true of DX at 30 mm, and true of the 1/1.7" Q7 at about 10 mm. And really, still "close enough" if a little wider, within a degree or two. It's another detail of crop factor. But it is true of longer lenses, that 2x focal length is half the view width.

There are approximations in calculations. The math is precise, but the data is less so. Focal length is a little vague, as might be precise sensor size. However, the results certainly are close enough to be very useful in any practical case. My experience is that the field is fairly accurate (at distances at least a meter or two), assuming you actually know your parameters. Some problems are:

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 *only applicable to focus at infinity*. Focal length necessarily increases when lens is extended forward to focus closer. Also zoom lenses can do other internal tricks with actual focal length (some zooms can be shorter when up close, instead of longer). Focal length will be less accurate at very close distances, and field of view becomes a little smaller. So macro distances are left out, but any error should be small if focused beyond one or two meters. You also have to measure your distance and field dimensions accurately too. And of course, we are only seeking a ballpark number anyway, we adjust small differences with our subject framing.

And a fisheye lens is a different animal, wider view than this formula predicts. A regular lens is rectilinear, meaning it shows straight lines as straight lines, not curved. A fisheye is rather unconcerned about this distortion, and can show a wider view, poorly purists might say, but very wide, and very possibly interesting.

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 **Thin Lens 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 that if the subject at s1 is at infinity, then 1/s1 is zero, so then s2 = f. This says that the marked focal length applies when focused at infinity.

Also if at 1:1 magnification, then s1 = s2, saying that the working macro distance in front of the lens is equal to the (extended at 1:1) focal length of the lens.

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.

Multi-element 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... 12 mm lens, 24 mm mirror, etc). This nodal difference is only a few inches, but it affects where the focal length is measured. And it shifts a bit as the lens is focused closer. Repeating, the focal length marked on the lens is specified for focus at infinity.

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/(S-f).

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, the focal length f is same as the distance d in front of lens (each with its own node).

1:1 macro (magnification 1), the field of view is exactly the same size as the sensor.

1:2 macro (magnification 0.5), the field of view is twice the size of the sensor.

1:4 macro (magnification 0.25), 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 24 mm 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.