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*All numbers used in this formula, and all others listed MUST be in the same units, i.e. millimeters, inches, etc. This excludes f-stop because it has no dimensions, and therefore always stays the same*
This is the most commonly asked for number from clients.
Use when they want to know how much of the foreground will be in focus, when you want a shallow depth of field, and need to know how much your subject can move within that depth of field, or when you want to figure out how much of the foreground will be in focus.
Hyperfocal distance, at its simplest, is the focusing distance that gives your photos the greatest depth of field. For example, consider a landscape where you want everything — foreground and background — to appear sharp. If you focus on the foreground, the background will appear blurry in the image.
Use hyperfocal distance when you want the entire scene to be acceptably sharp. For example, you may be photographing a scene that has elements in the foreground (not very close to the lens) and the background that you want to be in focus. By focusing at hyperfocal distance, you get these objects acceptably sharp in your image.
Simply put, depth of FIELD concerns the image quality of a stationary lens as an object is repositioned, whereas depth of FOCUS concerns a stationary object and a sensor’s ability to maintain focus for different sensor positions, including tilt.
The more distant you are from your subject, the wider or larger the depth of field. As a result, the depth of focus is shorter.
The precise focus distance at which you get the maximum depth of field for a given aperture and focal length combination!
H = (F^(2))/(fCc)
Where
H = hyperfocal distance
f = f-stop of lens
Cc = circle of confusion
In order to calculate hyperfocal distance, you need to know three things:
*More on circle of confusion: For circle of confusion consult the lens and camera manuals. You can also use .0010 inches (1/1000) as a conservative circle of confusion, although modern cine lenses may have an even smaller circle of confusion.*
Example
Question: Imagine you have a 20mm lens at f/11 on a full-frame camera, what is the hyperfocal distance?
Answer: Hyperfocal distance = (20 x 20) / (0.03 x 11) = 400/0.33 = 1212.12mm
So, you get a hyperfocal distance of 1212 mm, or 1.2 meters (almost 4 feet). You should focus on an object that is approximately 1.2 meters away; everything from 0.6 meters (half the hyperfocal distance) away to infinity will be in focus.
*NOTE: There are numerous ways of writing these equations. Below are the equations as given in the ASC Handbook*
The range in front of and behind a designated focusing distance, where an object still appears to be in acceptable focus.
There are two equations for calculating depth of field, one for the far limit, and one for the near limit.
Camera to Near Limit: Dn = (H)(S)/((H=+(S-F))
Camera to Far Limit: Df = (H)(S)/((H=-(S-F))
Total depth of field = Df-Dn
where
Dn = camera to near limit
Df = camera to far limit
H = hyperfocal distance
S = distance from camera to subject
F = focal length of lens
Example
Question: A 35mm camera lens of 50 mm focal length is focused at 20 feet and is set to f/2.8. Over what range of distances will objects be in acceptable focus?
Answer:
50 mm becomes 1.969 inches (rounding up to 2 inches is sufficient)
20 feet becomes 240 inches.
35 mm cameras have a circle of confusion equal to .001 inches.
H = (F^(2))/(fCc)
H = ((2)(2))/((2.8)(0.001))
H = 4/0.0028
H = about 1,429 inches
Dn = ((1429)(240))/((1429+(240-2))
Df = ((1429)(240))/((1429-(240-2))
Dn = 205.7 inches (17.1 feet)
Df = 288 inches = (24 feet)
Therefore, when a 50 mm lens at 4/2.8 is focused at 20 feet, everything from 17.1 feet to 24 feet will be in acceptable focus.
Dn = (HS)/(H+S)
Df = (HS)/(H-S)
where
Dn = camera to near limit
Dƒ = camera to far limit
H = hyperfocal distance
S = distance from camera to subject
F = focal length of lens
Total depth of field = Df-Dn
where
Dn = camera to near limit
Dƒ = camera to far limit
Use this to calculate the lens settings needed to accommodate for a set near and far focal length.
Ls = ((2)(Dn)(Dƒ))/ (Dn+Dƒ)
H = ((2)(Dn)(Dƒ))/ (Dn-Dƒ)
f = (F^2)/((H)(Cc))
Where
Dn = camera to near limit
Dƒ = camera to far limit
H = hyperfocal distance
Ls = Lens focus distance setting
F = focal length of lens
ƒ = f-stop setting of lens
Cc = circle of confusion
Example
Question: a scene is being photographed on a 35mm camera using a 75mm lens. Everything in the scene from 15 to 27 feet must be in acceptable focus. How much the lens f-stop and focus be set?
Answer:
Focal length is 2.95 inches (75/25.40)
Dn is 180 inches (15×12)
Df is 324 inches (27×12)
Ls = focus distance setting = ((2)(180)(324))/(180+324) = 231 inches = 19.3 feet is your lens focus distance!
H = hyperfocal distance = (2.953^2)/(1429+(240-2)) = 810 inches = 67.5 feet is your hyperfocal distance
ƒ = f-stop =(2.953^2)/(0.001×810) = f/10.77 = F/11 is your f-stop/aperture
Te = 1/((Sx360)/a)
a = 360STe
where
a = shutter angle (in degrees)
S = frames per second
Te = exposure time (in seconds)
Example
Question: What is the exposure time when shooting at 24 fps with a 180 degree shutter?
Answer:
Te = 1/((24×360)/180) = 1/48 second
*This equation only applies to spherical lenses*
O/A = D/F
Or
D = distance to subject = (OF)/A
O = size of subject being photographed = (AD)/F
F = focal length of lens = (AD)/O
A = aperture size = (OF)/D
Where
F = focal length of lens
D = distance to object being photographed
O = size of object being photographed
A = aperture size
θ = viewing angle
To solve it simply input any three known variables.
The following equation is a good approximation of the depth of focus of a lens. This is most often used with film, because it refers to distance between the lens and the actual film plane.
Depth of focus approximately = (Fƒ)/1000
Where
F = focal length of lens (in mm)
ƒ = f-stop of lens
Example
A 50 mm f/2.8 lens has the following depth of focus
(50×2.8)/1000 = 0.14mm = 0.0055 inches
So we know the film must stay within plus or minus half that value.
This equation is helpful to know, but not as commonly used.
ƒ = F/Da
Where
ƒ = f-stop value of lens
F = focal length of lens
Da = diameter of aperture
To make action in miniatures look convincing at normal speed, you usually need to shoot in a faster frame rate.
Rf = 24√(1/s)
where
Rf = frame rate
S = scale of miniature
Example
Question: What frame rate should be used to shoot a 1:4 (quarter scale) miniature?
Answer:
Rf = 24√(1/0.25) = 24√4 = 24×2 =48 fps!
Tf = (S)(t)
Where
Tf = total footage
S = speed of film (in ft/min)
t = time (in minutes)
Get the speed of film using the film gage and frames per foot. Usually you can Google it.
Example
Question: How many feet of 35 mm film runs through a sound camera in 4 and a half minutes?
Answer:
Tf = 90×4.5
Tf = 405 feet of film needed to shoot 4 and a half minutes
Ft = ((t)(Fs))/Ff
Where
Ft = total footage
Fs = frames per second
t = time (in seconds)
Ff =frames per foot (look this number up, it depends on the type of film camera and the film gage. Find a chart.)
Example
Question: How much film goes through a 16mm sound camera in 8 seconds?
Answer:
Ft = (8×24)/40 = 4.8 feet
To convert to frames, multiply the decimal part by number of frames per foot.
.8×40 = 32 frames
Therefore the answer is 4 feet 32 frames goes through the camera in 8 seconds.
People often use the terms “aperture” and “f-stop” interchangeably, but they are, in fact, two very different, but related, measures.
The “aperture” is the diameter of the entrance pupil of the lens, and is measures in mm.
The “f-stop” is the ratio of the focal length and the aperture diameter: f-stop = focal length / aperture diameter.
The aperture, in combination with shutter speed, determines how much total light that reaches the sensor. This is of central importance because it is the total light, along with sensor efficiency, that determines the total image noise (not ISO or sensor size, as most people think). The f-stop, on the other hand, determines the intensity of the light falling on the sensor, and, in combination, determines the density of light falling on the sensor. This is important, because the density of the light is the exposure.
Circle of Confusion (Cc) characterizes the degree of acceptable focus. The smaller the circle of confusion is the higher the resulting image sharpness. 
Technically it is an optical spot caused by a cone of light rays from a lens not coming to a perfect focus when imaging a point source. It is also known as disk of confusion, circle of indistinctness, blur circle, or blur spot.
The term 35mm equivalent focal length is a comparison of the field of view seen through a digital camera lens compared to the field of view produced by the older 35mm film cameras.
The light capturing area (imaging area) of most digital camera image sensors is smaller than the imaging area of film.
The widest field of view is produced when 35mm film or a full frame image sensor is used. With each smaller sized image sensor, more of the outer areas of the scene are lost. (cropped)
The field of view from a 35mm film camera and its lens is the standard field of view to which digital camera fields of view are compared. That is why you will see the phrase 35mm equivalent used.
The bottom line is that the standard fields of view from 35mm cameras can be duplicated by digital cameras with their smaller sensors and lenses by using wider focal length lenses.
Calculating the 35mm focal length equivalent of a camera lens is done by using a number that is called the crop factor or the focal length multiplier. The crop factor or focal length multiplier represents the ratio of the size of an image sensor compared to 35mm film.
Simply multiply the focal length by the crop factor. Find the crop factor in your camera’s manual.
35mm equivalent focal length = (Digital Camera Lens Focal Length)(crop factor)
Example
Question: 35mm equivalent focal length = (18mm)(1.5)
Answer: 35mm equivalent focal length = 27mm