1. Single Slit Diffraction
In the previous chapter we calculated the interference pattern from a pair of slits.
One of the assumptions in the calculation was that the slit width was very small compared with the wavelength of the light.  a     
Now we consider the effect of finite slit width. We start with a single slit.
Each part of the slit acts as a source of light waves, and these different light waves interfere.

2. Destructive interference:
Divide the slit in half.  
Ray  travels farther* than ray  by (a/2)sin. Likewise for rays  and . 
a/2   a 
a/2
If this path difference is exactly half a wavelength (corresponding to a phase difference of 180°) then the two waves will cancel each other and destructive interference results.
Destructive interference:
*All rays from the slit are converging at a point P very far to the right and out of the picture.

3. Destructive interference:  
a/2   a 
a/2
If you divide the slit into 4 equal parts, destructive
interference occurs when
If you divide the slit into 6 equal parts, destructive
interference occurs when

4. In general, destructive interference occurs when  
a/2   a 
a/2
In general, destructive interference occurs when
The above equation gives the positions of the dark fringes. The bright fringes are approximately halfway in between.
Applet.

6. Use this geometry for tomorrow’s single-slit homework problems.a 
If  is small,* then it is valid to use the approximation sin   .
( must be expressed in radians.) O x
*The approximation is quite good for angles of 10 or less, and not bad for even larger angles.

7. Single Slit Diffraction Intensity
Your text gives the intensity distribution for the single slit. The general features of that distribution are shown below.
Most of the intensity is in the central maximum. It is twice the width of the other (secondary) maxima.

8. Starting equations for single-slit intensity:
“Toy”