Slit diffraction

What are our observations of the fringe patterns for single slit, double slit and diffraction grating experiments with monochromatic light? And how does this differ with white light?

(I'm asking for things like "central fringe broader and brighter" in single-slit, as I think my knowledge isn't great for things like that. Explanations of these would also be much appreciated. Mathematical derivations are less necessary, I already know things like x=D*lamba/d.)

Yeah I think we'll leave the derivations until you've done a bit more maths... and you know the formulae.

Basically our single slit pattern (Fraunhofer diffraction) has an intensity distribution governed by the square of a sinc function (sinc(x)=sin(x)/x) and so the central fringe is broader and much more intense than outer fringes whose intensity drops of as governed by the intensity function.

for the double slit experiment, the intensity distribution is a cosine square and so all the fringes are the same width. However, this is still overlayed on top of the single slit pattern and so the brightest fringes are in the middle and then they get dimmer towards the edges but there are points where they get brighter again and then dimmer.

see this: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/dslit.html#c1

diffraction grating: the bright fringes will be very thin and you should know the formula for them

For white light, the red light spreads out more so you will observe chromatic seperation of the fringes with blue on the inside and red on the outside. For a grating, you will get a continuous spectra for a source that emits a contiuous spectra and a discrete spectra if otherwise.

OK, so, for all cases exept the broad first fringe in single slit, will all the bright fringes have the same even width and spacing? And for the single and double slit diffraction, will the bright fringes be the same width as the dark fringes (except for the central single-slit fringe)?

For double-slit diffraction and the diffraction grating, bright fringes (antinodes) are produced when waves m*lamba apart intefere, dark fringes (nodes) where waves (m+(1/2))*lamba interfere. What waves interfere to produce fringes for the single slit?

For the single slit the path differences are the same for constructive and destructice interference, just you have to integrate all of the posible start points for the waves... (hence why you don't do it quantatively at A-level). The fringes get progressively narrower and dimmer the further out you go and the dark fringes are not necessarily the same width. (Look up y=sinc²(x) on Wolfram alpha for the single slit pattern)

For the double slit one, all the fringes are the same width and the darka dn light fringes are the same width too.

Thanks, so for single slit diffraction we can only write d*sin(theta)=λ (where d is the width of the slit, theta is the angle at which light enters, λ is the wavelength of the light). We cannot write d*sin(theta)=nλ for single slit because the fringes are not all of even width apart. For double slit diffraction we can write d*sin(theta)=nλ, where the centre fringe has n=0, one fringe out on either side has n=1, the next fringe out from there on either side has n=2, etc., and the same for the diffraction grating (the same way of numbering n also works for single slit, but there's no point because we can't write d*sin(theta)=nλ for single slit diffraction. For the double slit diffraction and diffraction grating, but not single slit diffraction, we can then write W=Dnλ/d (where d is the distance between the centres of the slits, D is the distance from the slits to the screen onto which the pattern is projected, W is the width between the central fringe and fringe number n, on either side, n defined as before).

For all 3 patterns, bright fringes (maxima) are produced where waves n*λ apart interfere; dark fringes (minima) are produced where waves (n+(1/2))*λ interfere.

For single slit, the bright fringes become less wide and less bright the further out you go (as n increases). The central fringe is far wider and brighter than all the others. For double slit, the fringes are all the same width and the same brightness. For the diffraction grating, the bright fringes are narrower and brighter but are each the same width as each other (dark fringes are each the same width as each other).

For all of these cases, if white light is used, red is diffracted more than blue so will appear on the edge of the fringe further out from the central fringe, whereas blue will appear on the fringe edge closest to the central fringe, and in between them will be a continuous spectrum for each fringe.

Can you confirm or correct my understanding?

Any help on that? (my post above this one)