# Homework Help: Single Slit Diffraction question

1. Apr 27, 2013

### carnot cycle

1. The problem statement, all variables and given/known data
I'm having difficulty understanding why this equation makes sense for the following m values.

2. Relevant equations
The following is stated in my textbook.
For dark fringes,
sinθ = (mλ)/a
where λ is the wavelength of light in nanometers and a is the width of the single slit.
m is ±1, ±2, ±3, ±4, etc..

3. The attempt at a solution

I understand where the equation arises from, however, what I am not understanding is why all values of m are used in this equation and not just the odd ones. Because if I go backwards, here is what happens:

The path length difference is (a/2)sinθ, and for destructive interference, the following must hold:
path length difference = (m/2)λ, where m is an odd integer because the waves have to be out of phase to destroy each other through superposition.

So this is true then:
(a/2)sinθ = (m/2)λ but m has to be odd.

My book cancelled the twos in the denominator and divided by a on both sides, but made it so all values of m could work. I find this very confusing. Why isn't it only odd values of m?

2. Apr 27, 2013

### Simon Bridge

If you understood the derivation, how come you you don't see why the m values are not odd? Isn't that part of the derivation?

How did the book arrange things so that all values integer values of m would work?
Have you looked for other examples of the derivation? i.e.
http://www.math.ubc.ca/~cass/courses/m309-03a/m309-projects/krzak/

3. Apr 27, 2013

### technician

The path length difference is (a/2)sinθ, and for destructive interference, the following must hold:
path length difference = (m/2)λ, where m is an odd integer because the waves have to be out of phase to destroy each other through superposition.

I think you might be confusing 2 source interference with single slit diffraction?
For single slit diffraction MINIMA occur when aSinθ = nλ n = 1,2,3.... (NOT 0 !!!)

4. Apr 27, 2013

### carnot cycle

Thanks for the responses. Yes I might be confusing the two. But I thought for dark fringes in general, the wavelengths have to be out of phase, thus corresponding to a path length difference equal to (n+1/2)λ, where n = 0, 1, 2, 3, etc.

5. Apr 27, 2013

### technician

your general view is correct but a single slit is completely different (mathematically) from 2 slits or a diffraction grating.
Ironically....it turns out that the equation to calculate the position of MINIMA for a single slit 'looks like' the equation to calculate the positions of MAXIMA for 2 slits or a diffraction grating.
Even worse !!!....The positions for MAXIMA with the single slit are not even 1/2 way between the MINIMA !!!!!! the maths is more difficult....have you met it??

6. Apr 27, 2013

### carnot cycle

No I have not. But thanks for telling me that they are very different from one another. I was trying to relate the two and got very confused!