For double-slit interference, what is the minimum fraction of slit spacing over wavelength that will allow us to still use the small angle approximation?

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Carly Yang · Accepted Answer

Double slit experiment consists of two slits of width d and a screen to observe the interference at distance D. The light of same or different wavelengths passed through the small width slits. Each slit from different diffraction pattern depending on the width of slit and wavelength of light. they again interfere to form the fringe patterns on the screen.In double slit interference, the fringes are of equal width and and spaced equally.The small angle approximation indicates to consider the very small angle between the diffracted ray and horizontal line  d  sin  ⁡  θ  =  n  λFor small angle approximation,   θ  &amp;lt;&amp;lt;&amp;lt; in radians i.e. less than 0.1 radSo, assuming   sin  ⁡  θ  ≈  θFor any fringe,   n  λ  =  d  sin  ⁡  θwhere nis the order of maxima or minimaWe have,      d    λ    =      n          sin      ⁡      θ              d    λ    =      n    θ  for first maxima, n=1      d    λ    =      1    0.1          d    λ    =  10for first minima,      d    λ    =            n      +              1        2                    sin      ⁡      θ              d    λ    =            n      +              1        2              θ  For n=1 and   θ  =  0.1      d    λ    =            1      +              1        2              0.1          d    λ    =  15

For double-slit interference, what is the minimum fraction of slit spacing over wavelength that will allow us to still use the small angle approximation?

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