What is the Diffraction-Limited Object Size at 25 cm?

[dancergirlie]

Homework Statement

What is the diffraction-limited object size (at 25 cm) imposed by the numerical aperture of the eye (if the eye is a diffraction-limited optic)? Use 4mm for the iris diameter and 550 nm for the wavelength.

Homework Equations

Rayleigh criterion states: for angular resolution, theta:
sin(theta)=1.22(wavelength)/Diameter

The Attempt at a Solution

I can solve for the angular resolution, which I got as 1.6775E-4. However, I don't know how to solve for the object size from that. Any help would be great!

 

[NobodySpecial]

if you have an angle and a distance - you can get the size.
Picture it as a narrow triangle of rays going from your eye to a distance of 25cm.
- draw a diagram.

Hint. if the angle is small and in radians then it's even easier

 

[dancergirlie]

would I have to take the half-angle of that in order to make a right triangle, or is the angle that I solved for before? If I do have to take the half angle, would I have to multiply the object size by 2?

 

[NobodySpecial]

Yes the formula you have is for the half angle.
The diffraction limit (in radians) is approx 2 * wavelength / diameter

You should also learn somethign about the sin() of small angles in radians,

 

[dancergirlie]

wouldn't that have to be tangent and not sin. Yes, I know about paraxial approximation where tantheta and sintheta would be approximately theta. So, I would take 2*25E-2m*(theta)?