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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 :

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

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

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