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Q. For what distance is ray optics a good approximation when the aperture is $4 \,mm$ wide and the wavelength is $500\, nm$?

Wave Optics

Solution:

Fresnel distance, $z_F = \frac{a^2}{\lambda} $
$ = \frac{(4 \times 10^{-3})^2}{500\times 10^{-9}}$
$\therefore z_F = 32\,m$