In a diffraction experiment with waves of wavelength λ, there will be no intensity minima (that is, no dark fringes) if the slit width is small enough. What is the maximum slit width for which this occurs?

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In a diffraction experiment with waves of wavelength   λ, there will be no intensity minima (that is, no dark fringes) if the slit width is small enough. What is the maximum slit width for which this occurs?

zoranovxp · Accepted Answer

For a single slit diffraction, the diffraction equation is given as  d  sin  ⁡  θ  =  n  λd is the slit width  θ is the diffraction anglen is the order of diffraction  λ is the wavelength of incident lightFor a first order diffraction, n=1  d  sin  ⁡  θ  =  λ    d  =      λ          sin      ⁡      θ      The given condition is that no dark fringes should be formed, which in other words means that there should be no diffraction.Bright and dark fringes are formed when light waves having different path differences interfere with each other.This path difference between the light waves arises due to diffraction of the light, as it passes through the slit.The angle of diffraction   θ is the angle between the slit and the normal to the diffracted light wavesIn be order for the light waves to not be diffracted by the slit, this angle   θ should be equal to   π      /    2So, when   θ  =      π    2  , there will be no diffraction  sin  ⁡      π    2    =  1So,   d  =  λThus, there will be no diffraction of light waves of wavelength   λ when the slit width is equal to this wavelengthAnswer: The maximum slit width for no dark fringes must be equal to the wavelength of incident light.

In a diffraction experiment with waves of wavelength lambda

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