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Phase difference

  • Thread starter Math Jeans
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  • #1
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Homework Statement


For single-slit diffraction, calculate the first three values of [tex]\phi[/tex] (the total phase difference betweenrays from each edge of the slit) that produce subsidiary maximu by

a) using the phasor model

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



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) [tex]\pi[/tex]
2) The first non-zero solution to [tex]tan(\frac{\phi}{2}) = \frac{\phi}{2}[/tex]
3) [tex]2 \cdot \pi[/tex]
 

Answers and Replies

  • #2
454
0
The equation for the intensity with single slit diffraction is

[tex] \frac { \frac {\pi d} {\lambda} sin(\phi) } { \frac {\pi d} {\lambda} } [/tex]

your equation is produced if [itex] d = \frac {\lambda} { 2 \pi} [/itex]. 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 [itex] \frac {- \pi} {2} < \phi < \frac {\pi} { 2 } [/itex]
 
  • #3
335
0
Err. The equation for the intensity is given in the problem and d is not involved.
 
  • #4
alphysicist
Homework Helper
2,238
1

Homework Statement


For single-slit diffraction, calculate the first three values of [tex]\phi[/tex] (the total phase difference betweenrays from each edge of the slit) that produce subsidiary maximu by

a) using the phasor model

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



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) [tex]\pi[/tex]
2) The first non-zero solution to [tex]tan(\frac{\phi}{2}) = \frac{\phi}{2}[/tex]
3) [tex]2 \cdot \pi[/tex]

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 [itex]I[/itex] to be zero); the maxima are found by using your expression [tex]tan(\frac{\phi}{2}) = \frac{\phi}{2}[/tex] 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?
 

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