# 0.4 2d and 3d wavefronts  (Page 4/7)

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## Diffraction through a slit

When a wave strikes a barrier with a hole only part of the wave can move through the hole. If the hole is similar in size to the wavelength of the wave then diffraction occurs. The waves that come through the hole no longer looks like a straight wave front. It bends around the edges of the hole. If the hole is small enough it acts like a point source of circular waves.

Now if we allow the wavefront to impinge on a barrier with a hole in it, then only the points on the wavefront that move into the hole can continue emitting forward moving waves - but because a lot of the wavefront has been removed, the points on the edges of the hole emit waves that bend round the edges.

If you employ Huygens' principle you can see the effect is that the wavefronts are no longer straight lines.

Each point of the slit acts like a point source. If we think about the two point sources on the edges of the slit and call them A and B then we can go back to the diagram we had earlier but with some parts blocked by the wall.

If this diagram were showing sound waves then the sound would be louder (constructive interference) in some places and quieter (destructive interference) in others. You can start to see that there will be a pattern (interference pattern) to the louder and quieter places. If we were studying light waves then the light would be brighter in some places than others depending on the interference.

The intensity (how bright or loud) of the interference pattern for a single narrow slit looks like this:

The picture above shows how the waves add together to form the interference pattern. The peaks correspond to places where the waves are adding most intensely and the minima are places where destructive interference is taking place. When looking at interference patterns from light the spectrum looks like:

There is a formula we can use to determine where the peaks and minima are in the interference spectrum. There will be more than one minimum. There are the same number of minima on either side of the central peak and the distances from the first one on each side are the same to the peak. The distances to the peak from the second minimum on each side is also the same, in fact the two sides are mirror images of each other. We label the first minimum that corresponds to a positive angle from the centre as $m=1$ and the first on the other side (a negative angle from the centre) as $m=-1$ , the second set of minima are labelled $m=2$ and $m=-2$ etc.

The equation for the angle at which the minima occur is given in the definition below:

Interference Minima

The angle at which the minima in the interference spectrum occur is:

$sin\theta =\frac{m\lambda }{a}$

where

$\theta$ is the angle to the minimum

$a$ is the width of the slit

$\lambda$ is the wavelength of the impinging wavefronts

$m$ is the order of the mimimum, $m=±1,±2,±3,...$

A slit with a width of 2511 nm has red light of wavelength 650 nm impinge on it. The diffracted light interferes on a surface. At which angle will the first minimum be?

1. We know that we are dealing with interference patterns from the diffraction of light passing through a slit. The slit has a width of 2511 nm which is $2511×{10}^{-9}\phantom{\rule{4pt}{0ex}}\mathrm{m}$ and we know that the wavelength of the light is 650 nm which is $650×{10}^{-9}\phantom{\rule{4pt}{0ex}}\mathrm{m}$ . We are looking to determine the angle to first minimum so we know that $m=1$ .

2. We know that there is a relationship between the slit width, wavelength and interference minimum angles:

$sin\theta =\frac{m\lambda }{a}$

We can use this relationship to find the angle to the minimum by substituting what we know and solving for the angle.

3. $\begin{array}{ccc}\hfill sin\theta & =& \frac{650×{10}^{-9}\mathrm{m}}{2511×{10}^{-9}\mathrm{m}}\hfill \\ \hfill sin\theta & =& \frac{650}{2511}\hfill \\ \hfill sin\theta & =& 0.258861012\hfill \\ \hfill \theta & =& {sin}^{-1}0.258861012\hfill \\ \hfill \theta & =& {15}^{o}\hfill \end{array}$

The first minimum is at 15 degrees from the centre peak.

what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
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Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
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what king of growth are you checking .?
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Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
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biomolecules are e building blocks of every organics and inorganic materials.
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research.net
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sciencedirect big data base
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Introduction about quantum dots in nanotechnology
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absolutely yes
Daniel
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