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Single slit diffraction intensity versus Angle

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The single slit diffraction pattern for a monochromatic wave of wavelength Λ incident normally on a narrow slit of width a is described in the Fraunhofer region by the intensity.

I =

Where θ is the deflection angle perpendicular to the incident wave front.

1. What is the value of I(θ) as θ -> 0?
2. Sketch the form of I versus θ for the particular case Λ = a/2. How does the sketch change as Λ decreases?
3. Show that the intensity peak centered on θ = 0 falls to half its central intensity at
θ = sin-1(0.443Λ/a)

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Solution Summary

This 3-page word document explains in detail the solution for a single slit diffraction problem. Solution also include sketches of the intensity pattern for two values of the wavelength. Additional graph is used to solve for theta for a given value of intensity.

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(i) As θ goes to zero,

Hence, as θ goes to zero I(θ) -> I(0)

I have used the fact that

(ii) Following are the graphs of I(θ) verses θ for Λ = a/2 and Λ = a/4.

When the ...

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