# Fresnel diffraction

A Fresnel zone plate is prepared with a drawing program and printed on a laser printer. The printed zone plate has a central zone diameter of 5.00 cm. A photograph of this zone plate is to be reduced on high-resolution film to give a working zone plate with a 1.00 m focal length. If this zone plate is to be used with light of 633 nm, what factor of reduction is required?

Answer

In the computation below I have used the fact that for small angles $\sin(\theta) = \tan(\theta)$ and the maximum of interference needs to be in the center of the image formed on the screen. This is the best explanation I could find for this, that __the image formed needs to be angular separated (free from interference defects).__

Consider the central zone opening of the Fresnel lens as a single slit opening for light diffraction. The condition of maximum on the screen at distance f =1 m is

$(D/2)*sin(\theta) =k*\lambda$

where $sin(\theta) = (D/2)/f$

therefore $(D/2)^2= f*k*\lambda$

for $k=1$ we have

$D=2*\sqrt(f*\lambda) = 2*\sqrt{(1*633*10^{-9})} =2*7.95*10^{-4} =0.00159 =0.0016 m$

the reduction factor is $(5 cm)/(0.0016) m = 0.05/0.0016 =31.42$ times smaller