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Q. For what distance ray optics is a good approximation when the aperture is $4 \, mm$ and the wavelength of light is $400 \, nm$ ?

NTA AbhyasNTA Abhyas 2022

Solution:

Rays optics is good approximation up to a distance equal to Fresnel's distance $\left(I_{F}\right)$ .
From the formula of Fresnel's distance,
$D_{f}=\frac{a^{2}}{\lambda }$
Here, $a=4\,mm=4\times 10^{- 3}\,m$
$\lambda =400\,nm=400\times 10^{- 9}m$
$D_{f}=\frac{\left(4 \times 1 0^{- 3}\right)^{2}}{4 \times 1 0^{- 7}}$
$=\frac{4 \times 4 \times 1 0^{- 3} \times 1 0^{- 3}}{4 \times 1 0^{- 7}}=40\,m$