# Single Slit Bright Fringes

Gold Member
Hi all. So let us say I shine a laser through a slit of width w, the distance to the screen is L, the distance from the center of the pattern formed to the first next bright fringe (next because the center is a bright fringe) is a. Using Fraunhofer diffraction.

Given variables: a, w and L

My book gives the equation for dark fringes. I need the equation for bright fringes. Here is what I got:

Using geometry I got the path difference between two rays to be (w / 2) * sinx [where x is the angle of the rays from the center horizontal]

Now I assume that for a bright fringe to occur, constructive interference must happen and so (w / 2) * sinx = m * P [where m is 1,2,3... and P is wavelength of laser] => sinx = (2m * P) / w

Also using geometry I was able to achieve that tanx = a / L

For small angles:
tanx = sinx so => a / L = (2mP) / w
=> P = (w * a) / (2L * M)

Now plugging in my values into that equation for the first fringe (m = 1) I get the wrong answer, my Laser's wave length is 645 nm, I got 316nm for an answer. (I am assuming a factor of 2 is the problem). Can anyone please help???

You need to take a special approach to analysing a single slit because there is effectively an infinite number of point sources across the width of the slit and the contribution from all of these point sources must be taken into account.
If you had only 2 point sources separated by a distance w then you would expect a max when the path difference given by wSin∅ = nλ. (traditional double slits)
If you imagine a slit of width w then all the point sources will have an effect in the direction ∅
The big surprise is that all of the sources produce a MINIMUM in the directions where a MAX was produced with only 2 point sources.
The trick is to imagine the point source half way across the slit (w/2 from the top of the slit)
In the direction ∅ waves from the top of the slit and those from this mid-point will have a path difference of λ/2 and will cancel out.
Every point in the top half of the slit can be paired with a point in the lower half of the slit to produce cancellation. Therefore MINIMA are produced for slit widths w of λ, 2λ, 3λ......nλ
Minima occur when wSin∅ = nλ
If w = λ then the position of the first (and only) mimimum is when Sin∅ = 1 ie at 90
If w = 2λ then the first minimum is when Sin∅ = 0.5 ie 30. The second min is at 90 and so on.
The positions of the maxima is more difficult to calculate.... they are not half way between the minima!!!!
Hope this helps.

perplexabot
Gold Member
Interesting. Two questions though:

The big surprise is that all of the sources produce a MINIMUM in the directions where a MAX was produced with only 2 point sources.

For my single slit experiment there was a MAX in the center of the pattern, does that mean if I were to perform the double slit experiment I would achieve a MIN at the center?

Therefore MINIMA are produced for slit widths w of λ, 2λ, 3λ......nλ

Why does the width have to be a multiple of λ? Will you not get a minimum for any width? In other words, will you not get a pattern for any width?

Ah no....I should have said that in the straight through position there is no path difference between waves so there is always a Central Max for a single slit.
And Yes you do get minima for any width but the angles they occur at are given by
Sin∅ = nλ/w

perplexabot
Gold Member
I guess I'll have to measure from the center to the first dark fringe and use n = 1. So λ = (w * a) / (n * L)
where a is from the center to the first dark fringe.

Thanks, you have definitely cleared things up.

Yes... I think your equation is correct to find the wavelength

perplexabot