# A light ray of wavelength $\lambda$ is passing through a pin hole of diameter D and the effect is observed on a screen placed at a distance L from the pin hole. The approximations of geometrical optics are applicable if :

$\begin {array} {1 1} (a)\;D \leq \lambda & \quad (b)\;\frac{L \lambda}{D^2}=1 \\ (c)\;\frac{L \lambda}{D^2} < < 1 & \quad (d)\;\frac{L \lambda}{D^2} > > 1 \end {array}$

$(C)\;\frac{L \lambda}{D^2} < < 1$
Hence A is the correct answer.
answered Mar 24, 2014 by