# What will happen for single slit diffraction if?

If the width of the single slit is much smaller than the wavelength of a monochromatic light, until there is only one point on the wavefront can pass through it?

Thank you

mfb
Mentor
You get something similar to a spherical wave, expanding in all directions (orthogonal to the slit) behind the slit.

What has been said is quite true.
As a good guide you will find that when the slit width = 1 wavelength the angle of the first minimum in the diffraction pattern is 90 degrees.
This means that a slit width of less than 1 wavelength is essentially a 'point source'

sophiecentaur
Gold Member
2020 Award
What has been said is quite true.
As a good guide you will find that when the slit width = 1 wavelength the angle of the first minimum in the diffraction pattern is 90 degrees.
This means that a slit width of less than 1 wavelength is essentially a 'point source'
Hmm. A bit narrower than that, I think, or you will have a first minimum at +/- 90°, iirc, which is not omnidirectional.

Thank you guys.
To double confirm.
So if I use a slit that is same as the wavelength, then I will get my second maxima at +/-90° after my first?
If smaller than wavalength , then is just a point source?

If you use a slit that is 1 wavelength wide the first MINIMA !!!!! (please note MINIMA) occur at +/-90
There is only the central (0 degrees) maximum.
If the slit is 2 wavelengths wide then the first MINIMUM is at +/-30, the second MINIMUM is at +/-90 so there is a maximum between 30 and 90 (not half way between !!!)

I thought the destructive interference only call minima?
So only the centre is maximum , the rest are called minimum?

Maxima occur with CONSTRUCTIVE interference.
minima occur with DESTRUCTIVE interference.
Are you familiar with 2 source interference....the conditions for the formation of MAX and MIN

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mfb
Mentor
If smaller than wavalength , then is just a point source?
This is not a yes/no question. If you make the slit smaller and smaller, it will get a better and better approximation of a point source.

Quite correct, 1 wavelength width should be seen as an upper limit. With this width some of the wave exists out to +/-90 degrees so wave effects can be detected.
Smaller than 1 wavelength and the intensity decreases, make it small enough to be a better point source and the intensity will be so low that detection of wave effects will be difficult.
Pay your money and make the choice.

Is minima , I though is maxima because I mistook that as double slit .thanks

Smaller than the wavelength, better point source ,lower intensity .

Quite correct, 1 wavelength width should be seen as an upper limit. With this width some of the wave exists out to +/-90 degrees so wave effects can be detected.

Pay your money and make the choice.
What do you mean by exists out to +/-90 degrees? Money?

straight through the slit is 0 degrees.
+/- 90 degrees means 90 degrees on each side of the straight through direction.

"pay your money and make the choice"....sorry....that is a colloquialism, more or less means that you can't have it both ways.
better point source means lower intensity; higher intensity means a less good point source.

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sophiecentaur
Gold Member
2020 Award
Is minima , I though is maxima because I mistook that as double slit .thanks

Smaller than the wavelength, better point source ,lower intensity .

What do you mean by exists out to +/-90 degrees? Money?
Have you looked at the single slit diffraction pattern? The first nulls are at 90 to the straight through direction. Did you not write that you are familiar with Fraunhoffer?

Thank you technician and mfb
Have you looked at the single slit diffraction pattern? The first nulls are at 90 to the straight through direction. Did you not write that you are familiar with Fraunhoffer?
I have. But I don't have slit that is same as the wavelength of the monochromatic light used.
I just studied about Fraunhofer's single and double slit. What is wrong?

sophiecentaur
Gold Member
2020 Award
Thank you technician and mfb

I have. But I don't have slit that is same as the wavelength of the monochromatic light used.
I just studied about Fraunhofer's single and double slit. What is wrong?
If you have studied the theory for Fraunhofer's single slit pattern then you must be aware that it works for any width you want. If not, then you need to start with the simple, ideal two slit interference' situation and progress from that.
Do you appreciate that, for two slits of the same width, the resulting pattern can be calculated by 'multiplying' the single slit (wide) pattern by the double slit (fine) pattern? I googled "two slit diffraction pattern" and found an endless supply of hits which give diagrams and 'the maths', plus loads of diagrams - starting with wiki. I am not really sure what extra "visualisation" you are wanting. I can't think of anything simpler than what I have found.

Andy Resnick
If the width of the single slit is much smaller than the wavelength of a monochromatic light, until there is only one point on the wavefront can pass through it?

Thank you
You will generate an evanescent field. The geometry you describe is used a lot for near-field scanning microscopy, for example. The details were worked out in the 1940's, as I recall.

If you have studied the theory for Fraunhofer's single slit pattern then you must be aware that it works for any width you want.
I think I have studied, but I really cant find out why you are so sure it can work for all width? actually last time I wondered how to block light, I thought light is also particle so we can block them if our slit is smaller then their diameter, but now only realise that they are absorbed by opaque not blocked.

If not, then you need to start with the simple, ideal two slit interference' situation and progress from that.
Do you appreciate that, for two slits of the same width, the resulting pattern can be calculated by 'multiplying' the single slit (wide) pattern by the double slit (fine) pattern?
This one I am not sure too, though there are some math I am not sure.

http://web.mit.edu/viz/EM/visualizations/notes/modules/guide14.pdf
I dont know why we can multiply the intensity of the narrow double with the intensity of the wide single slit to get the intensity of the double slit with wide slit.

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You will generate an evanescent field. The geometry you describe is used a lot for near-field scanning microscopy, for example. The details were worked out in the 1940's, as I recall.
Dont know how the NSMM function, knowing its application makes it more interesting.
Thank you.

jtbell
Mentor
I dont know why we can multiply the intensity of the narrow double with the intensity of the wide single slit to get the intensity of the double slit with wide slit.
One can prove this by starting from scratch and setting up an integral for the total amplitude of the light from the two slits, as in the first equation on this page.

http://www.phy.davidson.edu/StuHome/grpatterson/Diffraction and Spatial Filtering/double_slit1.htm

Each of the two integrals, by itself, gives the total amplitude from one of the slits, which turns out to be the "wide single slit" pattern.

The details of evaluating the integrals are left as an exercise for the student.

I'm surprised that the MIT notes that you link to don't use an integral to find the single-slit intensity pattern. Surely MIT students know how to do integrals!

mfb
Mentor
I think I have studied, but I really cant find out why you are so sure it can work for all width?
That is simple: the derivation of the formula does not need any relevant restriction of the width (well, it cannot be negative...).

sophiecentaur
Gold Member
2020 Award
It's all mathematical, I'm afraid. If you look at what the symbols in that attached equation stand for, you will notice that it includes slit width and separation. Hyperphysics has all the formulae and definitions that you need.

One can prove this by starting from scratch and setting up an integral for the total amplitude of the light from the two slits, as in the first equation on this page.

http://www.phy.davidson.edu/StuHome/grpatterson/Diffraction and Spatial Filtering/double_slit1.htm
Really thanks , for this question I really don't what I should type to Mr . Google, luckily I asked here. Thank you.

That is simple: the derivation of the formula does not need any relevant restriction of the width (well, it cannot be negative...).
O I see, is the formula , we Assume the formula works for all width of range except negative.

It's all mathematical, I'm afraid. If you look at what the symbols in that attached equation stand for, you will notice that it includes slit width and separation. Hyperphysics has all the formulae and definitions that you need.
It did have. But difficult to understand.
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinint.html#c3

Really thank you guys .

sophiecentaur
Gold Member
2020 Award
It did have. But difficult to understand.
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinint.html#c3

Really thank you guys .
There are two ways through this. You can just see the graphs and 'accept' that they come from some (appropriate level of) Maths or really get stuck in with the Maths. The general principle of 'multiplying' patterns does make sense and can help with getting the basic theory.

Claude Bile
Another way to think of the Fraunhofer diffraction pattern from a single slit is as a Fourier transform of the aperture transmission function. The normal rules of FTs apply; the smaller the slit width, the broader the diffraction pattern.

If the aperture is super-narrow then the rules of near-field optics must be taken into account (such as the evanescent fields pointed out by Andy).

Claude.

sophiecentaur
Gold Member
2020 Award
Another way to think of the Fraunhofer diffraction pattern from a single slit is as a Fourier transform of the aperture transmission function. The normal rules of FTs apply; the smaller the slit width, the broader the diffraction pattern.

If the aperture is super-narrow then the rules of near-field optics must be taken into account (such as the evanescent fields pointed out by Andy).

Claude.
I have a feeling that the OP wanted an answer at a less elevated level. More on the lines of how the vectors add together to enhance of cancel.
Iirc, if you write an expression for the result, doing it the long way round, by applying Huygen's principle and arriving at an integral, you get the same mathematical expression as you do when you do the FT on the aperture - and that makes the actual calculation easier because you can use FT instead.