Class 12PhysicsWave OpticsFresnel distance, Diffraction through circular slit, concept of Airy Disc, Rayleigh Criterion of Resolution , Fraunhofer diffraction
EasyEstimate the distance for which ray optics is a good approximation for an aperture of $4 \; mm$ and wavelength $400 \; nm$.
(40 M) Fresnel's distance $(Z_{F})$ is the distance for which ray optics is a good approximation.
It is given by the relation:
$Z_{F} = \frac{a^{2}}{\lambda}$
Where:
Aperture width,$a = 4 \; mm = 4 \times 10^{-3} \; m$
Wavelength of light,$\lambda = 400 \; nm = 400 \times 10^{-9} \; m$
Substituting the values:
$Z_{F} = \frac{(4 \times 10^{-3})^{2}}{400 \times 10^{-9}}$
$Z_{F} = \frac{16 \times 10^{-6}}{400 \times 10^{-9}}$
$Z_{F} = \frac{16}{400} \times 10^{3} = 0.04 \times 1000 = 40 \; m$
Therefore,the distance for which ray optics is a good approximation is $40 \; m$.
It is given by the relation:
$Z_{F} = \frac{a^{2}}{\lambda}$
Where:
Aperture width,$a = 4 \; mm = 4 \times 10^{-3} \; m$
Wavelength of light,$\lambda = 400 \; nm = 400 \times 10^{-9} \; m$
Substituting the values:
$Z_{F} = \frac{(4 \times 10^{-3})^{2}}{400 \times 10^{-9}}$
$Z_{F} = \frac{16 \times 10^{-6}}{400 \times 10^{-9}}$
$Z_{F} = \frac{16}{400} \times 10^{3} = 0.04 \times 1000 = 40 \; m$
Therefore,the distance for which ray optics is a good approximation is $40 \; m$.
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MediumKCET 2016
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