Question

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

• A. $22m$
• B. $32m$
• C. $42m$
• D. $52m$

Answer 1

Answer: (B)

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