How to Calculate Field of View From Focal Length

You cannot determine a camera’s field of view from focal length alone. The sensor or film size matters too: the same lens gives a wider view on a larger sensor and a narrower view on a smaller one.

For a rectilinear camera lens, the field-of-view angle is twice the inverse tangent of half the sensor dimension divided by the focal length. This article shows how to choose the right dimension, calculate the angle, and interpret the result in a camera or 3D scene.

The field-of-view formula

For a sensor dimension $s$ and focal length $f$, both in the same units, the angle of view is

$$\theta = 2\arctan\left(\frac{s}{2f}\right).$$

The result is in radians if your calculator evaluates inverse tangent in radians; convert it to degrees if needed. Most calculators have a degree mode. Use:

• Sensor width for horizontal field of view.
• Sensor height for vertical field of view.
• Sensor diagonal for diagonal field of view.

The formula follows from the camera’s pinhole projection geometry. Half the sensor dimension is opposite the angle’s half-angle, and the focal length is adjacent to it, so $\tan(\theta/2)=(s/2)/f$.

Worked example: 50 mm lens on full frame

A full-frame sensor is approximately $36\text{ mm}$ wide and $24\text{ mm}$ high. Its diagonal is approximately $43.3\text{ mm}$. With a $50\text{ mm}$ lens, calculate each field of view separately.

Horizontal:

$$\theta_h=2\arctan\left(\frac{36}{2(50)}\right) =2\arctan(0.36)\approx 39.6^\circ.$$

Vertical:

$$\theta_v=2\arctan\left(\frac{24}{2(50)}\right) =2\arctan(0.24)\approx 27.0^\circ.$$

Diagonal:

$$\theta_d=2\arctan\left(\frac{43.3}{2(50)}\right) \approx 46.8^\circ.$$

So this setup has about a $39.6^\circ$ horizontal field of view, not one single angle that describes every direction. State which field of view you mean when comparing cameras or entering a value into 3D software.

Applying the result to a 3D scene

If an object or plane is a distance $D$ from the camera, the width covered across a plane perpendicular to the viewing direction is

$$W=2D\tan\left(\frac{\theta_h}{2}\right).$$

For the example above, at $D=10\text{ m}$ the horizontal coverage is

$$W=2(10)\tan(39.6^\circ/2)\approx 7.2\text{ m}.$$

This is the width of the view at that distance, under the simple pinhole-camera model. The same calculation using the vertical angle gives the height covered. In a 3D program, check whether its camera setting is horizontal, vertical, or diagonal: entering the right numerical angle in the wrong direction changes the framing.

What if you know only the image aspect ratio?

The aspect ratio tells you how width compares with height, but it does not give the sensor’s physical size. That means focal length plus aspect ratio still does not determine field of view. You need at least one sensor dimension, or a specified format such as full frame that supplies the dimensions.

If you know the vertical angle and aspect ratio, you can convert between vertical and horizontal angles. For aspect ratio $r=\text{width}/\text{height}$:

$$\theta_h=2\arctan\left(r\tan\left(\frac{\theta_v}{2}\right)\right).$$

This conversion is useful in 3D applications that report vertical field of view while a camera specification gives horizontal field of view.

Common mistakes

• Using focal length alone: A $50\text{ mm}$ lens does not have one fixed field of view across every camera format.
• Mixing units: The sensor dimension and focal length must use the same units. For example, use millimetres for both; the units then cancel in the ratio.
• Confusing horizontal and diagonal angles: A camera maker’s advertised angle may be diagonal. Compare like with like.
• Treating focal length as a zoom setting: A longer focal length narrows the angle of view for a fixed sensor. Cropping an image also narrows its recorded field of view, even though the lens focal length has not changed.
• Expecting exact results for every lens: The formula assumes an ideal rectilinear projection. Real lenses can distort the image, and close focusing or focus breathing can alter the effective angle of view.