Fraunhofer intensity pattern

[Qyzren]

Homework Statement

Derive the equation describing the Fraunhofer intensity pattern from a single
slit of width b centered at the origin. Then show that the same intensity pattern is obtained for a slit that is not centred at the origin (i.e assume the slit edges are at x = a - b/2 and x = a + b/2). Is this physically reasonable? Discuss briefly.

The Attempt at a Solution

Suppose the coordinates of P are (X, Y, Z), and the coordinates of a point Q on the aperture are (x, y,z) where z = 0.
http://img133.imageshack.us/my.php?image=picbe2.jpg
If the distance QP is d then.
http://img84.imageshack.us/my.php?image=pic2qv3.jpg
Now chuck that in
http://img224.imageshack.us/my.php?image=pic3fl2.jpg
to obtain
http://img81.imageshack.us/my.php?image=pic4lr2.jpg

now, how would i attempt the second part of the question?
would my d² = (X - (x+a))² + (Y-y)² + (Z-z)²
and L² = (x-a)² + Y² + Z² ??

meaning d² = L²[1 + (2ax + x² + y²)/L² - (2xX+2yY)/L²]
so d ~= L[1 - (2xX + 2yY)/L²] like earlier?

or am i doing it completely wrong?

Thank you all for your input.
Thanks

 

[Jhero]

for your question! To show that the same intensity pattern is obtained for a slit that is not centered at the origin, we can use the same approach as before but with the new coordinates for the slit edges. So instead of using x = 0 for the center of the slit, we can use x = a for one edge and x = a + b for the other edge, as shown in the diagram below:

http://img.photobucket.com/albums/v111/bradleyveldman/slit.png

Using this new coordinate system, we can derive the same equation for the intensity pattern as before. So the final result would be:

I(θ) = [sin(πbθ/λ)]^2/[πbθ/λ]^2

where θ is the angle between the direction of the incident wave and the direction of the diffracted wave.

This result is physically reasonable because the intensity pattern is determined by the diffraction of the incident wave by the edges of the slit, not by the position of the slit itself. As long as the slit edges are parallel, the same diffraction pattern will be observed regardless of the position of the slit. This is because the diffraction pattern is a result of the interference of the waves diffracted from each edge, and the relative positions of the edges do not affect this interference.