Q.Calculate the distance upto which ray optics is a good approximation for light of wavelength 500 nm falls on an aperture of width 0.5 mm.

Setting the full diffraction spread over distance z ($2z\lambda/a$) equal to the aperture width a gives the Fresnel distance $z=a^2/(2\lambda)=25$ cm for the given values.

Working

A beam passing through an aperture of width $a$ spreads by diffraction; the (half-)angular spread at the first minimum is $\theta\approx\lambda/a$. Ray optics remains a good approximation only up to the distance $z$ at which this diffractive spreading becomes comparable to the aperture size itself. Taking the total (both-sided) spread over distance $z$ as $2z\theta$, and setting it equal to $a$:

$2z\dfrac{\lambda}{a} = a \ \Rightarrow\ z = \dfrac{a^2}{2\lambda}$

Given: $a=0.5\ \text{mm}=5\times10^{-4}$ m, $\lambda=500\ \text{nm}=5\times10^{-7}$ m.

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