The direction of the first secondary maximum in the Fraunhofer diffraction pattern at a single slit is given by ($a$ is the width of the slit):

The distance for which ray optics becomes a good approximation for an aperture of $0.3 \ cm$ and a light of wavelength $6000 \ Å$ is (in $m$)

For what distance is ray optics a good approximation when the aperture is $4 \, mm$ wide and the wavelength is $500 \, nm$ (in $, m$)?

Fresnel diffraction is produced due to light rays falling on a small obstacle. The intensity of light at a point on a screen beyond an obstacle depends on

Explain that for a greater distance,the spreading due to diffraction dominates over that due to ray optics. $OR$ Explain the importance of Fresnel distance.

In Fraunhofer diffraction pattern, slit width is $0.2 \, mm$ and screen is at $2 \, m$ away from the lens. If wavelength of light used is $5000 \, \mathring{A}$, then the distance between the first minimum on either side of the central maximum is (assume $\theta$ is small and measured in radians).