# 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?

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°. Sep 6, 2014 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)