# Homework Help: Phase difference

1. May 15, 2008

### Math Jeans

1. The problem statement, all variables and given/known data
For single-slit diffraction, calculate the first three values of $$\phi$$ (the total phase difference betweenrays from each edge of the slit) that produce subsidiary maximu by

a) using the phasor model

b) setting $$\frac{dI}{d\phi} = 0$$, where I is given by $$I = I_0 \cdot (\frac{sin(\frac{1}{2} \cdot \phi)}{\frac{1}{2} \cdot \phi})^2$$

2. Relevant equations

given

3. The attempt at a solution

I attempted to do part b first as I already knew the answers from part a (fairly straight forward) and was using them to check my answer.

I realized that even though I had used the exact described method, my answer was completely different as my values for phi ended up as:

1) $$\pi$$
2) The first non-zero solution to $$tan(\frac{\phi}{2}) = \frac{\phi}{2}$$
3) $$2 \cdot \pi$$

2. May 16, 2008

### kamerling

The equation for the intensity with single slit diffraction is

$$\frac { \frac {\pi d} {\lambda} sin(\phi) } { \frac {\pi d} {\lambda} }$$

your equation is produced if $d = \frac {\lambda} { 2 \pi}$. d is obviously too small here to get destructive interference at any angle. For the light to go through the slit and reach the screen you need $\frac {- \pi} {2} < \phi < \frac {\pi} { 2 }$

3. May 16, 2008

### Math Jeans

Err. The equation for the intensity is given in the problem and d is not involved.

4. May 17, 2008

### alphysicist

Hi Math Jeans,

When you take the derivative and set it to zero, you find the minima and maxima. The third values you found (2 pi) is a minima (it makes $I$ to be zero); the maxima are found by using your expression $$tan(\frac{\phi}{2}) = \frac{\phi}{2}$$ and finding the first three non-zero solutions.

I did not see how you got the first answer (pi); I don't believe it makes the derivative zero. If you still get it as an answer would you post your expression for the derivative?