The distance between two opposite fiducial marks of a photo is 20cm. What is the field of view if the camera used has a focal length of 115mm.?
\[
\alpha = 2\arctan\frac{d}{2f}
\]
\[
\alpha = 2\arctan\frac{20cm. (\frac{10mm.}{1cm.})}{2(115mm.)}
\]
\[
\alpha = 82^{\circ}01’5.43″ \leftarrow \text{answer}
\]