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(A) The fringe width $\beta$ in a diffraction pattern produced by a wire (which acts like a single slit of width $d$) is given by the formula: $\beta

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Decreases

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(A) The fringe width $\beta$ in a diffraction pattern produced by a wire (which acts like a single slit of width $d$) is given by the formula: $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance of the screen from the wire,and $d$ is the diameter of the wire.From the formula,it is clear that the fringe width $\beta$ is inversely proportional to the diameter of the wire $d$ (i.e.,$\beta \propto \frac{1}{d}$).Therefore,when the diameter $d$ of the wire is increased,the fringe width $\beta$ decreases.

In a diffraction pattern produced by a thin wire,what happens to the fringe width when the diameter of the wire is increased?

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