# Calculating the field of view for a Nexus 5

The specs on the Nexusw 5 rear camera are: Aperture size: F2.4 Focal length (35mm equivalent): 30.4 mm Camera sensor size: 1/3.2"

I want to calculate the field of view for this camera. Using FOV=2*atan(W/2/f) where: FOV=field of view W=camera sensor size f=focal length From a table of image sensor formats 1/3.2" sensor has a horizontal size of 4.54mm. This gives a FOV=2*atan(4.54/2/30.4)=8.54 degrees.

So if the subject is 1m away from the lens, the calculated field of view at the subject is D*tan(FOV/2)*2=1*0.0747*2=0.149m.

If I position a nexus 5 camera 1m away from a subject I can see much more than 0.149m of the subject.

Why is my calculation wrong?

## 2 Answers

The actual focal length is 4mm. The 35mm equivalent tells you what focal length you'd need to use with a 35mm film/sensor to achieve the same result, but it doesn't have much optical sense.

Assuming the sensor width is 4.54mm:

``````FOV = 2*atan(.5*4.54mm/4mm)
FOV ~ 1.03235913 rad
FOV ~ 59.1498211 deg
``````

That is roughly 40% of a human's field of view.

• And your calculations are backed by comparing to the stated 35mm equivalent of 30.4mm,which also works out to a horizontal field of view of about 60°. – mattdm Sep 6 '14 at 14:55

Camera2 API returns f = 3.97mm and 4.6032 x 3.5168mm sensor size.

so field of view on Nexus 5 is 1.0507905x0.83390445 in radians

60°x48° (rounded)