# Counting dark fringes

1. Apr 24, 2014

### negation

Dark spots are measured on a screen at +xm, +ym and +zm where z>y>x from the central axis.

since dark fringes are where (m+0.5)λ, would it be right for me to state that the dark spot z has an order of 3.5?

2. Apr 24, 2014

### Simon Bridge

The "order" is always an integer - so x is 1st order, y is 2nd order, and z is 3rd order "dark fringe" from the center. The order number is usually the same as the m number... but it is a tad ambiguous here because you seem to be using "m" to mean two different things.

i.e. when m=2, are the fringes at +2x, +2y and +2z?

3. Apr 24, 2014

### negation

I might have refined my understanding since I posted this hours ago. I was careless in reusing variables. Matlab should have taught me better not to.

To simplify matters, I shall ignore bright fringe. If z is the 3rd order from the central axis where m =0 at the central axis, then m = 3 for dark fringe located at distance z from central axis. But dark fringes are essentially half-integers multiple of λ.
and so we add a 0.5 to it.

the path difference for a 3rd order dark fringe is dsinθ = (m+0.5)λ = (3.5)λ

4. Apr 24, 2014

### Simon Bridge

There is no zeroth order for the dark fringes.
The dark fringes occur where the path difference is an odd number of half-wavelengths.

If m is to be the order number, then your equation needs to be: $d\sin\theta = (m-\frac{1}{2})\lambda: m=1,2,3\cdots$

Note: if n=1,3,5,7,9... and m=1,2,3,4... then n=2m-1

5. Apr 26, 2014

### negation

When do we use (m+0.5)λ and (m-0.5)λ? I'm quite confused by this

6. Apr 26, 2014

### Simon Bridge

You sound like someone working by memorizing equations instead of reading what they are telling you.
That path leads only to confusion.

I showed you exactly when to use the (m-0.5)λ form when I said: If m is to be the order number
That is: if you want the dark fringe for m=1 to be the first minimum either side of the central max, and m=2 to be the second minimum either side of the central max, and so on.

If you number the fringes by m=1,2,3... then you get two dark fringes with m=1 (etc) either side of the central maxima. This makes sense because you can think of the fringes to the right as positive and the fringes to the left as negative... like you'd normally do with an axis.

But if you number the fringes by m=0,1,2,3... then you have two dark fringes with m=0, one either side of the central maxima. Then the one to the left would be the -0th fringe and the one to the right would be the +0th fringe. Does that make sense mathematically?

7. Apr 27, 2014

### negation

In fact I work worst when memorizing. That is why a derivation demostrating the relationship works wonders in my udnerstanding.
I know what you said but the formula sheet at the back of my book says another thing. I thought it might be good to understand the reason behind the difference in the formulation.

8. Apr 27, 2014

### negation

Yes this makes sense.

The third minima on the +ve y-axis has an order m =3 and so (3-0.5)λ = dsinΘ.
Can you verify?

9. Apr 27, 2014

### Simon Bridge

... it's just a different was of counting.

That would be correct.
The first order would be m=1, which is where the path difference is one-half wavelength ... so you have to subtract 1/2 from m=1 to get that right.
The second order would have a path difference of 2-0.5=1.5 wavelengths and so on.

That is the relationship between the path difference and the order.

10. Apr 27, 2014

### negation

And for destructive interference, (m+0.5)λ applies when counting dark fringes below the central maxima-appears to be consistent with the exception that there would be a negative sign for any maxima or minima below the central axis.

11. Apr 27, 2014

### Simon Bridge

m+0.5 appears when you start counting dark fringes from 0.
m-0.5 appears when you start counting dark fringes from 1.

the second has the advantage that the value of m is the same as the order number.

For bright fringes, there is a central one so it makes sense to start counting with the center as order 0.
For dark fringes you can do the math either way - it just makes more sense to start counting from 1.
That is all there is to it.