Question

Calculate the optical invariant for a photographic objective with F = 50 mm. N = 2, field of view +26.6°.

Answer 1

Given,

$F=50mm=50\times 10^{-3}m$

$N=2$

View angle $(\theta) =\pm 26.6^\circ$

Optical invariant $(I)=FN\sin\theta$

Now, substituting the values,

$(I)=50\times 10^{-3}\times 2\times \sin(26.6^\circ)$

$\Rightarrow (I)=0.1\times \sin(26.6^\circ)$

$=0.1\times 0.447$

$=0.0447$