Diffraction limit on resolution

[mrbling]

Q:Spy planes fly at extremely high altitudes (25.3 km) to avoid interception. Their cameras are reportedly able to discern features as small as 5.20 cm. What must be the minimum aperture of the camera lens to afford this resolution? (Use lambda = 550 nm.)

I first found theta.. .052m/2 = .026m (since .026 above and below the center = .052m)
tan theta = .026m/25300m, and theta = 4.578x10^-5

Then plugging into the rayleigh criterion for resolution limit of theta = 1.22lamda/D, where D is the diameter of the objective lens (same as aperture of camera??)... I get
theta = 1.22lamda/D
4.578x10^-5 = 1.22(550x10^-9)/D
D= .0147m, which is incorrect..

anyone know where I went wrong?
Thanks!

 

[gnome]

I get d = .326m using
$$sin \theta = 1.22\frac{\lambda}{d}$$
and letting $$\theta = arctan \left( \frac{.052}{25300}\right)$$
so:
$$d = \frac{1.22x5.5x10^{-7}}{sin({arctan \left( \frac{.052}{25300}\right)})} = .326m$$

Not sure though. Do you know what the answer is supposed to be?

 

[mrbling]

Gnome,
Your answer was the correct one. I tried your process earlier, but I think I was confusing the theta since it was given in radians as opposed to doing sin theta for the degrees..

Thanks!

 

[gnome]

You can get the same answer using the $$\theta=sin\theta$$ small angle approximation, but then you have to be consistent & do the arctan calculation in radians also.

So $$\theta = arctan \left(\frac{.026}{25300}\right) = 2.0553x10^{-6}$$

and $$d = \frac{1.22x5.5x10^{-7}}{\theta} = .326 m$$

 

[krab]

Gnome's answer in slightly more legible LaTeX:

You can get the same answer using the $\theta=\sin\theta$ small angle approximation, but then you have to be consistent & do the arctan calculation in radians also.

So

$$\theta = \arctan \left(\frac{.026}{25300}\right) = 2.0553\times 10^{-6}$$

and

$$d = \frac{1.22\times 5.5\times 10^{-7}}{\theta} = .326\,\mbox{m}$$