VCE Physics Question Thread!

[BasicAcid]

For diffraction through a single slit, the angular positions of the intensity minima are given by
where m is a constant (not zero), d is the slit width and theta is the angle the line from the slit to the minima makes with a line parallel to the slit opening.
As you can see, if the wavelength approaches the slit width, the first angle approaches ninety degrees, which means the first intensity minima spreads out completely. If the wavelength exceeds the slit width, then you will not have any minima and the wave will spread completely.


*begins rant about VCE physics

So actually answering Tim...Blah's question, there will still be diffraction if wavelength > slit size but just to a lesser extent?

[SocialRhubarb]

Umm, no, there will be more diffraction if wavelength > slit.

[lolipopper]

mrrghaawd make up your mind guys !!! 

[BasicAcid]

Umm, no, there will be more diffraction if wavelength > slit.

I thought maximum diffraction occurred when wavelength/slit = 1?

[lolipopper]

Could someone verify this?

I'm a bit worried now after reading this


And I don't understand any of nlius explanation

nliu you sexy boy, explain it to us in simple terms.

[ECheong]

well essentially, looking back at that dsin(theta)=lamda equation and applying it. If you have if you have lamda/d = 1 you get a theta of 90 degrees. Anything more than that, and it stays at a diffraction angle of 90 degrees.

 If you imagine an aperture and a wave coming through it, it's not possible for the wave to diffract backwards at an angle >90degrees. This is because if you treat each single point on the edge of a wave as a point source (huygen's principle) there's no way for the point source to send a wave 'around the corner' of the aperture and start sending waves backwards.

Reading over this I've realised it's a pretty difficult concept to explain in words LOL, pictures would be better.

Tl;dr maximum diffraction at lamda/d =1 because you don't get any more diffraction if lamda/d>1

[SocialRhubarb]

I haven't read anything outside of nliu1995's points, but they seem to make physical sense.

Basically, if you have a wavelength/slit ratio greater than 1, you don't get any minima and maxima forming, but instead just have a complete diffraction of the light source, which I'm assuming means an equal distribution of light intensity in all directions.

I should clarify that, obviously you can't have more than 90 degrees of diffraction, which occurs when wavelength approaches the slit width. If wavelength is greater than the slit width, you will get 90 degrees of diffraction as well. So if wavelength is greater than slit width, you will get the maximum possible amount of diffraction occurring.

[lolipopper]

I haven't read anything outside of nliu1995's points, but they seem to make physical sense.

Basically, if you have a wavelength/slit ratio greater than 1, you don't get any minima and maxima forming, but instead just have a complete diffraction of the light source, which I'm assuming means an equal distribution of light intensity in all directions.

I should clarify that, obviously you can't have more than 90 degrees of diffraction, which occurs when wavelength approaches the slit width. If wavelength is greater than the slit width, you will get 90 degrees of diffraction as well. So if wavelength is greater than slit width, you will get the maximum possible amount of diffraction occurring.

so the whole screen will be just bright, correct?

[lzxnl]

The point is, if the wavelength is larger than the slit width, the equation cannot be satisfied by any angle theta. Therefore, there IS no intensity minimum. That means the wave spreads out completely.
The equation I gave is a requirement for the waves in the slit to interfere with each other destructively. Failure to satisfy this equation means no complete destructive interference.
The reason why we say in VCE physics that "incomplete diffraction" occurs is that the width of the middle intensity maximum is small. After the first intensity minimum, the intensity drops off significantly.

Read up on diffraction on wiki for diagrams, but be warned, this subject can get maths-heavy. The equation I gave is a requirement for the waves in the slit to interfere with each other destructively. Failure to satisfy this equation means no complete destructive interference.

There's a neat diagram on wiki too; it'll make my point clearer.

[Alwin]

For diffraction through a single slit, the angular positions of the intensity minima are given by
where m is a constant (not zero), d is the slit width and theta is the angle the line from the slit to the minima makes with a line parallel to the slit opening.
As you can see, if the wavelength approaches the slit width, the first angle approaches ninety degrees, which means the first intensity minima spreads out completely. If the wavelength exceeds the slit width, then you will not have any minima and the wave will spread completely.


*begins rant about VCE physics

nliu, they kinda sorta mildly touch this in Synchrotron detail study, but I agree. Explained muchhhh more in depth in A level (eg singapore) or highschool equivalent in china and other asian countries.

Anyways, for those interested from page 496:

kinda tangential, only read it if you want to

A collimated beam of X-rays incident upon a layer of atoms will be scattered. Bragg’s law states that the beams will interfere constructively, producing maxima when the following relationship is satisfied:
2dsinθ = nλ where d is the distance between layers of atoms (m), θ is the angle the X-ray beam makes with the surface, λ is the wavelength of incident X-ray photons (m) and n is the number of the maxima occurring, 1, 2, 3, etc.

What nliu is trying to say is that as d (which can be considered as the slit width) and λ (wavelength) are approaching the same width, ie d/λ -> 1, the equation becomes:
2λsinθ = nλ -> sinθ= n/2.
n=1 is the centre band, so we look for n=2, the band next to it:
sinθ= n/2 -> sinθ= 2/2 -> θ= 90 degrees.
This implies that the wave has defracted so completely that the next band occurs at right angles to the path, ie parallel to the screen.

So, using a "similar" version of Bragg's Law (sadly barely touched upon in VCE) the max defraction occurs when the ratio of wavelength to slit with approaches one.

EDIT: beaten, but hope you get the gist
This is the page he is referring to: http://en.wikipedia.org/wiki/Diffraction#Single-slit_diffraction

From that page:

Numerical approximation of diffraction pattern from a slit of width equal to wavelength of an incident plane wave in 3D spectrum visualization:	Numerical approximation of diffraction pattern from a slit of width equal to five times the wavelength of an incident plane wave in 3D spectrum visualization	Numerical approximation of diffraction pattern from a slit of width four wavelengths with an incident plane wave. The main central beam, nulls, and phase reversals are apparent.

[lzxnl]

nliu, they kinda sorta mildly touch this in Synchrotron detail study, but I agree. Explained muchhhh more in depth in A level (eg singapore) or highschool equivalent in china and other asian countries.

Anyways, for those interested from page 496:
A collimated beam of X-rays incident upon a layer of atoms will be scattered. Bragg’s law states that the beams will interfere constructively, producing maxima when the following relationship is satisfied:
2dsinθ = nλ where d is the distance between layers of atoms (m), θ is the angle the X-ray beam makes with the surface, λ is the wavelength of incident X-ray photons (m) and n is the number of the maxima occurring, 1, 2, 3, etc.

What nliu is trying to say is that as d (which can be considered as the slit width) and λ (wavelength) are approaching the same width, ie d/λ -> 1, the equation becomes:
2λsinθ = nλ -> sinθ= n/2.
n=1 is the centre band, so we look for n=2, the band next to it:
sinθ= n/2 -> sinθ= 2/2 -> θ= 90 degrees.
This implies that the wave has defracted so completely that the next band occurs at right angles to the path, ie parallel to the screen.

So, using a "similar" version of Bragg's Law (sadly barely touched upon in VCE) the max defraction occurs when the ratio of wavelength to slit with approaches one.

Ah Bragg's law. It's similar, although in that case you've got to remember where the two comes from; the path difference is 2 d sin theta, and that's because there are two layers in the crystal lattice and the 2 comes from a return journey from the first layer to the second layer. Look it up on wiki for a diagram.
Alwin, you'll confuse people xP

In double slit interference, the condition is d sin theta = m lambda for CONSTRUCTIVE interference because d sin theta is the path difference.

For single slit diffraction, the argument works differently. Again, look at some diagrams; they'll be better than what I can draw.
But essentially, if we split up the slit into half, to form two minislits of size d/2, then we have a double slit scenario with path difference d/2 * sin theta. This is equal to half a wavelength for minimum, so d/2*sin theta = w/2. d sin theta = w
If we split the slit into quarters, then quarter 1 can interfere destructively with quarter 2 and then 3 with 4. The slit width is now d/4, so the condition is d/4 * sin theta = w/2. d sin theta = 2w. Etc if we split the slit into 6, 8, 10, 12.
That's sort of where it comes from.

Define "max diffraction" though.

[ECheong]

So essentially for this exam (which is all I care about):

- lamda/width < 1 = not a lot of diffraction
- lamda/width = 1 = max diffraction
- lamda/width > 1 = max diffraction

Correct?

At the core of it, pretty much haha

[lzxnl]

See, in the diagram with w=d, there is a beam that spreads across the entire wall, whereas with the third one, the most intense part of the wave is quite limited in range.

[Alwin]

So essentially for this exam (which is all I care about):

- lamda/width < 1 = not a lot of diffraction
- lamda/width = 1 = max diffraction
- lamda/width > 1 = max diffraction

Correct?

pretty much.

*back to discussion with nliu*

For anyone who's interested in what we're talking about, this is the easiest explanation I can find:

scan of a textbook

See, in the diagram with w=d, there is a beam that spreads across the entire wall, whereas with the third one, the most intense part of the wave is quite limited in range.

You referring to the ones in my post?

[lzxnl]

Oh yeah. That thing. I need to start making mine