# Another focal length question

1. May 9, 2009

### questions_uk

Hi. If using:

f (focal length) = a / ( 2 x arctan (theta / 2)) where a = 13 and is the Field of View

the value I get for the focal length is 0.03 and this seems too small. Unless the calculation has not been done right?

Thanks.

2. May 10, 2009

### Redbelly98

Staff Emeritus
Something is weird about your equation. If theta is an angle, it makes no sense to take the arctan of an angle.

That depends. 0.03 mm would be too small. 0.03 m would actually be reasonable. 0.03 miles could be too large.

Moral: units are an important part of the answer.

Last edited: May 10, 2009