For lenses projecting
rectilinear (non-spatially distorted) images of distant objects, the effective
focal length and the image format dimensions completely define the angle of view. Calculations for lenses producing non-rectilinear images are much more complex and, in the end, not very useful in most practical applications. (In the case of a lens with distortion, e.g., a
fisheye lens, a longer lens with distortion can have a wider angle of view than a shorter lens with low distortion) Angle of view may be measured horizontally (from the left to right edge of the frame), vertically (from the top to bottom of the frame), or diagonally (from one corner of the frame to its opposite corner). For a lens projecting a rectilinear image (focused at infinity, see
derivation), the angle of view (
α) can be calculated from the chosen dimension (
d), and effective focal length (
f) (
f is defined as the distance of the lens with respect to the image plane. For a thick lens, it is the distance of the rear
principal plane of the lens w.r.t the image plane) as follows: \alpha = 2 \arctan \frac {d} {2 f} d represents the size of the film (or sensor) in the direction measured
(see below: sensor effects). For example, for 35 mm film which is 36 mm wide and 24 mm high, d = 36\,\mathrm{mm} would be used to obtain the horizontal angle of view and d = 24\,\mathrm{mm} for the vertical angle. Because this is a trigonometric function, the angle of view does not vary quite linearly with the reciprocal of the focal length. However, except for wide-angle lenses, it is reasonable to approximate \alpha \approx \frac{d}{f} radians or \frac{180d}{\pi f} degrees. The effective focal length is nearly equal to the stated focal length of the lens (
F), except in
macro photography where the lens-to-object distance is comparable to the focal length. In this case, the absolute transverse
magnification factor (
m) (m = S_2/S_1) must be taken into account: f = F \cdot ( 1 + m ) (In photography, the magnification is usually defined to be positive, despite the inverted image.) For example, with a magnification ratio of 1:2, we find f = 1.5 \cdot F and thus the angle of view is reduced by 33% compared to focusing on a distant object with the same lens. Angle of view can also be determined using FOV tables or paper or software lens calculators.
Example Consider a 35 mm camera with a lens having a focal length of . The dimensions of the 35 mm image format are 24 mm (vertically) × 36 mm (horizontal), giving a diagonal of about 43.3 mm. At infinity focus, , the angles of view are: • horizontally, \alpha_h = 2\arctan\frac{h}{2f} = 2\arctan\frac{36}{2 \times 50}\approx 39.6^\circ • vertically, \alpha_v = 2\arctan\frac{v}{2f} = 2\arctan\frac{24}{2 \times 50}\approx 27.0^\circ • diagonally, \alpha_d = 2\arctan\frac{d}{2f} = 2\arctan\frac{43.3}{2 \times 50}\approx 46.8^\circ
Derivation of the angle-of-view formula Consider a rectilinear lens in a camera used to photograph an object at a distance S_1, and forming an image that just barely fits in the dimension, d, of the frame (the
film or
image sensor). Treat the lens as if it were a
pinhole at distance S_2 from the image plane (technically, the center of perspective of a
rectilinear lens is at the center of its
entrance pupil where
chief rays meet): . Now \alpha/2 is the angle between the
optical axis of the lens and the ray joining its optical center to the edge of the film. Here \alpha is defined to be the angle-of-view, since it is the angle enclosing the largest object whose image can fit on the film. We want to find the relationship between: • the angle \alpha • the "opposite" side of the right triangle, d/2 (half the film-format dimension) • the "adjacent" side, S_2 (distance from the lens to the image plane) Using basic trigonometry, we find:\tan ( \alpha / 2 ) = \frac {d/2} {S_2}which we can solve for
α, giving: \alpha = 2 \arctan \frac {d} {2 S_2} To project a sharp image of distant objects, S_2 needs to be equal to the
focal length, F, which is attained by setting the lens for
infinity focus. Then the angle of view is given by: \alpha = 2 \arctan \frac {d} {2 F} Note that the angle of view varies slightly when the focus is not at infinity (See
breathing (lens)), given by S_2 = \frac{S_1 F}{S_1 - F} as a rearrangement of the lens equation.
Macro photography For macro photography, we cannot neglect the difference between S_2 and F. From the
lens formula, \frac{1}{F} = \frac{1}{S_1} + \frac{1}{S_2}. The absolute transverse
magnification (the absolute ratio of the image height to the object height) can be expressed m = S_2/S_1, we can substitute S_1 and with some algebra find: S_2 = F\cdot(1+m) Defining f=S_2 as the "effective focal length", we get the formula presented above: \alpha = 2 \arctan \frac {d} {2 f} where f = F\cdot(1+m). A second effect which comes into play in macro photography is lens asymmetry (an asymmetric lens is a lens where the aperture appears to have different dimensions when viewed from the front and from the back). The lens asymmetry causes an offset between the nodal plane and pupil positions. The effect can be quantified using the ratio (
P) between apparent exit pupil diameter and entrance pupil diameter. The full formula for angle of view now becomes: \alpha = 2 \arctan \frac {d} {2 F\cdot ( 1 + m/P )} == Measuring a camera's field of view ==