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Q. Light of wavelength $600 \,nm$ is incident on an aperture of size $2\, mm$. The distance upto which light can travel such that its spread is less than the size of the aperture is

Wave Optics

Solution:

The distance upto which light can travel is
Fresnel distance, $z_F = \frac{d^2}{\lambda} $
$=\frac{(2\times 10^{-3})^2}{600 \times 10^{-9}} = 6.67\,m$