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I did an experiment in which I tried to show that the visibility of the interference fringes is related to the relative slit width in a double slit interferometer. In other words if one slit had a width that was twice the width of the other, would that change the visibility of the interference fringes. I thought it would because I assumed that by changing the width of the slits the intensity of the light passing through the slit would change, but now I am not so sure about that. All of the equations I have seen regarding intensity in a double slit interferometer do not indicate a linear relationship between intensity and slit width.

DanielSank
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Clement Decker
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  • You can make interference with slits of any width, you just have to go far enough away to see the pattern. Consider the aperture of a telescope as a slit. What happens to the point spread function when you scale the telescope up? Does it change its shape? – CuriousOne Mar 10 '16 at 23:56

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The fringe pattern is simply the fourier transform of the slit aperture, in 1 or two dimensions.

The fourier transform of a single slit ~ $sinc(x/a)$ where a is the width of the slit.

The fourier transform of a double (equal) slit ~ $cos(x/a)$ eqn.(1)

The fourier transform of two slits , one twice the width of the other is something like $a*cos(x/a) + b*sin(x/a)$

Which, using well known trigonometric identities is simply eqn. (1) displaced slightly along the x direction

there is no change in intensity

Arif Burhan
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