Calculate Phase Difference for Single-Slit Diffraction

[Math Jeans]

Homework Statement

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

Homework Equations

given

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

 

[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 }

 

[Math Jeans]

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

 

[alphysicist]

Homework Statement

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

Homework Equations

given

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

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?