Share
Explore BrainMass

# Single slit diffraction intensity versus Angle

The single slit diffraction pattern for a monochromatic wave of wavelength &#955; incident normally on a narrow slit of width a is described in the Fraunhofer region by the intensity.

I =

Where &#952; is the deflection angle perpendicular to the incident wave front.

1. What is the value of I(&#952;) as &#952; -> 0?
2. Sketch the form of I versus &#952; for the particular case &#955; = a/2. How does the sketch change as &#955; decreases?
3. Show that the intensity peak centered on &#952; = 0 falls to half its central intensity at
&#952; = sin-1(0.443&#955;/a)

#### Solution Preview

Following is the text part of the solution. Please see the attached file for complete solution. Equations, diagrams, graphs and special characters will not appear correctly here. Thank you for using ...

#### 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.

\$2.19