How Do You Solve These Diffraction Problems?

[mjsteves]

DIFFRACTION PROBLEMS -- please help

tough problems, any help or either one is greatly appreciated

(1)
Consider the single-slit diffraction pattern in figure V.2, which gives the location of the central maximum, and the zeros of the diffraction. If Z=1m and lambda=6*10^-7m, using the data in figure V.2, what is a?

Figure V.2 is a simple graph, basically the figure is a graph with x values of x and y values of I/I_0. The I/I_0 ranges from 0 to 1 in increments of 0.1. The x values range from -0.05 to 0.05 in increments of 0.01. At x=-0.02, I/I_0=0 and at x=0.02, I/I_0 also equals zero. In between -0.02<x<0.02 there is an upside down parabola with a maximum at x=0 or I/I_0=1.

The equations that are listed in the book are:

(a) I=I_0*(sin(alpha)/alpha)^2 where alpha=pi*(a/lambda)*sin(theta)

approximating for small angles:

(b) tan(theta)=sin(theta)

and (c) tan(theta)=x/Z

and,

(d) aplha=pi*(a/lambda)*(x/Z)

my attempt at a solution:

Basically I used the graph to say that I/I_0=0 when x=0.02. Then i used equation (a) and divided each side by I_0 to get I/I_0=(sin(alpha)/alpha)^2 => 0=(sin(alpha)/alpha)^2, so it seems as if the only way the statement could be true would be if alpha was zero. Using equation (d) for alpha=pi*(a/6*10^-7)*(0.02/1), alpha is zero when a=0. This was my answer, but I really think I did it wrong.

(2)
The Moon is about 4*10^8 m from the Earth and has a diameter of about 3.2*10^6 m. You want a radar antenna which just covers the Moon with the central spot of its diffraction pattern. If the radar wavelength is 10 cm, what is the diameter of the radar antenna?

There are no relevant equations that it gives for this problem other than the ones given above, but it says on one page that a function has zeros when alpha=0 or when alpha=n*pi which implies that n*pi=pi*(a/lambda)*(x_n/Z) or X_n=n*(lambda/a)*Z where n=+/- 1, 2, 3, etc.

my attempt at a solution:

First I tried using a proportion with the lengths/wavelength it gave me and then used that to solve for diameter but would be too easy. Then I tried to use the previous equations but then there would be missing variables unless I am supposed to assume the values from problem 1. I now that it should be single-slit. Please help me out!

 

[Redbelly98]

(1)

0=(sin(alpha)/alpha)^2, so it seems as if the only way the statement could be true would be if alpha was zero

There are other values of alpha which make the expression zero.
When is sin(alpha) = 0 ?

... when alpha=0 or when alpha=n*pi

Yes, that's right. Umm, well not the alpha=0 part. alpha=n*pi is correct.

(2)
Your book should (hopefully) have a discussion, and an equation, for diffraction by a circular aperture.

 

[jimagnus]

For a, I used Xn=n*lamba*Z/a , Xn being a zero. The first zero on my graph was .02, so I plugged in Xn=.02 and n=1.