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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
Thank you
Hmm. A bit narrower than that, I think, or you will have a first minimum at +/- 90°, iirc, which is not omnidirectional.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'
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.If smaller than wavalength , then is just a point source?
What do you mean by exists out to +/-90 degrees? Money?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.
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?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?
I have. But I don't have slit that is same as the wavelength of the monochromatic light used.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?
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.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?
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 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
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 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 did try,maybe I am not good at it or can you please guide me your reading material link?If not, then you need to start with the simple, ideal two slit interference' situation and progress from that.
This one I am not sure too, though there are some math I am not sure.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?
Dont know how the NSMM function, knowing its application makes it more interesting.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.
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.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.
That is simple: the derivation of the formula does not need any relevant restriction of the width (well, it cannot be negative...).I think I have studied, but I really cant find out why you are so sure it can work for all width?
Really thanks , for this question I really don't what I should type to Mr . Google, luckily I asked here. Thank you.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
O I see, is the formula , we Assume the formula works for all width of range except negative.That is simple: the derivation of the formula does not need any relevant restriction of the width (well, it cannot be negative...).
It did have. But difficult to understand.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.
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.It did have. But difficult to understand.
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinint.html#c3
Really thank you guys .
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.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.