Determine the focal length of a Fresnel Zone Plate

[Ratpigeon]

Homework Statement

Derive an expression for the Fresnel Zone Plate's focal length

Homework Equations

The fresnel zone plate has a transmittance function
1/2 (1+cos (ar^2)circ(r/2l)

The Attempt at a Solution

Using Eulers formula. this splits into e^(iar^2)+e^(-iar^2) Since the transmittance funciton of a lens is e^(i$\pi$r^2/$\lambda$f) The fresnel zone transmittance function is pretty much equivalent to a lens which satisfies
a=$\pi$/$\lambda$f
Hence; the focal length is f= $\pi/$/$\lambda$a
Is this right?

 

[mark726]

Yes, your derivation is correct. The focal length of a Fresnel Zone Plate is given by f = \pi/\lambda a, where a is the zone width and \lambda is the wavelength of the incident light. This is because the transmittance function of the Fresnel Zone Plate is similar to that of a lens, and the focal length of a lens is given by f = \pi/\lambda a. Great job!