Diffraction Of Light
Diffraction
Diffraction of light occurs when a light wave passes by a corner or through an opening or slit that is physically the approximate size of, or even smaller than that light's wavelength. Diffraction is the slight bending of light as it passes around the edge of an object. The amount of bending depends on the relative size of the wavelength of light to the size of the opening. If the opening is much larger than the light's wavelength, the bending will be almost unnoticeable. Nov 13, 2015 Diffraction of light occurs when a light wave passes very close to the edge of an object or through a tiny opening such as a slit or aperture. Diffraction maximum. Light of wavelength 750 nm passes through a slit 1.0 x 10-3 mm wide. How wide is the central maximum (a) in degrees, and (b) in.
Light is a transverse electromagnetic wave. Diffraction, and interference are phenomena observed with all waves.
| Diffraction is the tendency of a wave emitted from a finite source or passing through a finite aperture to spread out as it propagates. Diffraction results from the interference of an infinite number of waves emitted by a continuous distribution of source points. According to Huygens' principle every point on a wave front of light can be considered to be a secondary source of spherical wavelets. These waveletspropagate outward with the characteristic speed of the wave. The wavelets emitted by all points on the wave front interfere with each other to produce the traveling wave. Huygens' principle also holds for electromagnetic waves. When studying the propagation of light, we can replace any wave front by a collection of sources distributed uniformly over the wave front, radiating in phase. |
| When light passes through a small opening, comparable in size to the wavelength λ of the light, in an otherwise opaque obstacle, the wave front on the other side of the opening resembles the wave front shown on the right.. |
The light spreads around the edges of the obstacle. This is the phenomenon of diffraction. Diffraction is a wave phenomenon and is also observed with water waves in a ripple tank.
| Water waves in a ripple tank | |
A single large slit: | A single small slit: |
The single slit
| When light passes through a single slit whose width w is on the order of the wavelength of the light, then we can observe a single slit diffraction pattern on a screen that is a distance L >> w away from the slit. The intensity is a function of angle. Huygens' principle tells us that each part of the slit can be thought of as an emitter of waves. All these waves interfere to produce the diffraction pattern. Where crest meets crest we have constructive interference and where crest meets trough we have destructive interference. |
Very far from a point source the wave fronts are essentially plane waves. This is called the Fraunhofer regime, and the diffraction pattern is called Fraunhofer diffraction. The positions of all maxima and minima in the Fraunhofer diffraction pattern from a single slit can be found from the following simple arguments.
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| If the optical path length of two rays differs by λ/2, the two rays interfere destructively. For ray 1 and ray 7 to be half a wavelength out of phase we need (w/2)sinθ = λ/2 or wsinθ = λ. But from geometry, if these two rays interfere destructively, so do rays 2 and 8, 3 and 8, and 6 and 10, 5 and 11, and 6 and 12. In effect, light from one half of the opening interferes destructively and cancels out light from the other half. Destructive interference produces the dark fringes. Dark fringes in the diffraction pattern of a single slit are found at angles θ for which w sinθ = mλ, where m is an integer, m = 1, 2, 3, ... . For the first dark fringe we have w sinθ = λ. When w is smaller than λ , the equation wsinθ = λ has no solution and no dark fringes are produced. |
| If the interference pattern is viewed on a screen a distance L from the slits, then the wavelength can be found from the spacing of the fringes. We have λ = w sinθ/m and sinθ = z/(L2 + z2)1/2), or λ = zw/(m(L2 + z2)1/2), where z is the distance from the center of the interference pattern to the mth dark line in the pattern. If L >> z then (L2 + z2)1/2 ~ L and we can write λ = zw/(mL). |
Problem:
When a monochromatic light source shines through a 0.2 mm wide slit onto a screen 3.5 m away, the first dark band in the pattern appears 9.1 mm from the center of the bright band. What is the wavelength of the light?
- Solution:
z = 9.1 mm = 9.1*10-3 m
L = 3.5 m
w = 0.2 mm = 2*10-4 m
L >> z, therefore λ = zw/(mL)
λ = (9.1*10-3 m)(2*10-4 m)/(3.5 m)
λ = 5.2*10-7 m = 520 nm
Link: Single slit diffraction II
Diffraction, and interference are phenomena observed with all waves.
- Observe single and double slit diffraction with water waves.
Interference
The double slit
| If light is incident onto an obstacle which contains two very small slits a distance d apart, then the wavelets emanating from each slit will interfere behind the obstacle. Waves passing through each slit are diffracted and spread out. At angles where the single slit diffraction pattern produces nonzero intensity, the waves from the two slits can now constructively or destructively interfere. If we let the light fall onto a screen behind the obstacle, we will observe a pattern of bright and dark stripes on the screen, in the region where with a single slit we only observe a diffraction maximum. This pattern of bright and dark lines is known as an interference fringe pattern. The bright lines indicate constructive interference and the dark lines indicate destructive interference. The bright fringe in the middle of the diagram on the right is caused by constructive interference of the light from the two slits traveling the same distance to the screen. It is known as the zero-order fringe. Crest meets crest and trough meets trough. The dark fringes on either side of the zero-order fringe are caused by destructive interference. Light from one slit travels a distance that is ½ wavelength longer than the distance traveled by light from the other slit. Crests meet troughs at these locations. T he dark fringes are followed by the first-order fringes, one on each side of the zero-order fringe. Light from one slit travels a distance that is one wavelength longer than the distance traveled by light from the other slit to reach these positions. Crest again meets crest. |
The diagram on the right shows the geometry for the fringe pattern. If light with wavelength λ passes through two slits separated by a distance d, we will observe constructively interference at certain angles. These angles are found by applying the condition for constructive interference, which is d sinθ = mλ, m = 0, 1, 2, … . The distances from the two slits to the screen differ by an integer number of wavelengths. Crest meets crest. The angles at which dark fringes occur can be found be applying the condition for destructive interference, which is d sinθ = (m+½)λ, m = 0, 1, 2, … . The distances from the two slits to the screen differ by an integer number of wavelengths + ½ wavelength. Crest meets trough. |
If the interference pattern is viewed on a screen a distance L from the slits, then the wavelength can be found from the spacing of the fringes.
We have sinθ = z/(L2 + z2)1/2 and λ = zd/(m(L2 + z2)1/2), where z is the distance from the center of the interference pattern to the mth bright line in the pattern.
If L >> z then (L2 + z2)1/2 ~ L and we can write
λ = zd/(mL).
Link: Physics 2000: Wave Interference
Diffraction grating
| We have seen that diffraction patterns can be produced by a single slit or by two slits. When light encounters an entire array of identical, equally-spaced slits, called a diffraction grating, the bright fringes, which come from constructive interference of the light waves from different slits, are found at the same angles they are found if there are only two slits. But the pattern is much sharper. The figure on the right shows the interference pattern for various numbers of slits. The width of all slits is 50 micrometers and the spacing between all slits is 150 micrometers. The location of the maxima for two slits is also the location of the maxima for multiple slits. The single slit pattern acts as an envelope for the multiple slit patterns. Diffraction gratings contain a large number of parallel, closely spaced slits or grooves. They produce interference maxima at angles θ given by d sinθ = mλ. Because the spacing between the slits is generally very small, the angles θ are generally quite large. We cannot use the small angle approximation for relating wavelength and the position of the maxima on a screen for gratings, but have to use sinθ = z/(L2 + z2)1/2. |
| Diffraction gratings disperse white light into its component colors because different wavelengths produce bright fringes at different angles. d sinθ = mλ, The spectral pattern is repeated on either side of the main pattern. These repetitions are called 'higher order spectra'. There are often many of them, each one fainter than the previous one. If the distance between slits is d, and the angle to a bright fringe of a particular color is θ, the wavelength of the light can be calculated. |
Problem:
The first order bright line appears 0.25 cm from the center bright line when a double slit grating is used. The distance between the slits is 0.5 mm and the screen is 2.7 m from the grating. Find the wavelength.
- Solution:
z = 0.25 cm = 2.5*10-3 m
L = 2.7 m
d = 0.5 mm = 5*10-4 m
L >> z, therefore λ = zd/(mL)
λ = (2.5*10-3 m)(5*10-4 m)/(2.7 m)
λ = 4.63*10-7 m = 463 nm
Problem:
A diffraction grating has 420 lines per mm. The grating is used to observe normally incident light with a wavelength of 440 nm. The grating is placed 1.3 m from a screen. Where on the screen will the first order bright line appear?
- Solution:
d sinθ = λ
d=(1/420) mm=2.38*10-3 mm=2.38*10-6 m
λ=440 nm=4.40*10-7 m
L=1.3 m = distance to the screen.
sinθ = λ/d = 10.65o.
z= L tanq =24.5 cm = distance from the central maximum.
Diffraction patterns can be observed when light passes through a set of regularly spaced slits. For a diffraction to produce an observable pattern, the spacing of the slits must be comparable to the wavelength of the radiation. Visible light has a wavelength range from ~400 nm to ~700 nm. A typical diffraction grating for visible light with 300 grooves per mm has a slit spacing of (1/300)mm = 3 mm = 3000 nm. This spacing is 4 to 8 times larger than the wavelengths of visible light and produces an easily observable pattern.
The wavelengths of x-rays lie in the 1 nm to 1 pm range. A typical diffraction grating will not produce an observable pattern. But the wavelengths of x-rays are comparable to the spacing of atoms in common crystals, and material with a regularly spaced grid of atoms can diffract x-rays and produce diffraction patterns that can be captured on photographic film.
Take a look at the diagram below:
- When X-rays interact with a single particle, it scatters the incident beam uniformly in all directions.
- When X-rays interact with a solid material the scattered beams can add together in a few directions and reinforce each other to yield diffraction. The regularity of the material is responsible for the diffraction of the beams.
You have likely seen diffraction before. For example if you look at a CD when exposed to white light you can see it diffracted into varies wavelengths of color. The pits (or grooves) in the CD are the regularity of the material that causes the diffraction.
Google images: outreach.atnf.csiro.au
Diffraction can occur when any electromagnetic radiation interacts with a periodic structure. The repeat distance of the periodic structure must be about the same wavelength of the radiation. For example, light can be diffracted by a grating having scribed lines arranged on the order of the wavelength of light.
Exercise (PageIndex{1})
So if all electromagnetic radiation can diffract, why are X-rays used in crystallography?
X-ray Diffraction and Bragg’s Law
X-rays have wavelengths on the order of a few angstroms (1 Angstrom = 0.1 nm). This is the typical inter-atomic distance in crystalline solids, making X-rays the correct order of magnitude for diffraction of atoms of crystalline materials.
How are Diffractions Patterns Made?
When X-rays are scattered from a crystalline solid they can constructively interfere, producing a diffracted beam. What does this mean?
Constructive vs. Destructive Interference
Interference occurs among the waves scattered by the atoms when crystalline solids are exposed to X-rays. There are two types of interference depending on how the waves overlap one another.
Constructive interference occurs when the waves are moving in phase with each other. Destructive interference occurs when the waves are out of phase.
Diffraction Of Lights
This constructive interference results in diffraction patterns.
Bragg's Law and Diffraction
The relationship describing the angle at which a beam of X-rays of a particular wavelength diffracts from a crystalline surface was discovered by Sir William H. Bragg and Sir W. Lawrence Bragg and is known as Bragg’s Law
[mathrm{2dsin θ= nλ}]
Diffraction Of Light Pictures
λ = wavelength of the x-ray
θ = scattering angle
n = integer representing the order of the diffraction peak.
d = inter-plane distance of (i.e atoms, ions, molecules)
.jpg "Diffraction Of Light")
Click on the following image below to get to an Applet where you can explore this relationship of Bragg’s Law
www.eserc.stonybrook.edu/ProjectJava/Bragg/
Guide to how to use Applet: There are 2 rays incident on two atomic layers of a crystal (d). At the beginning the scattered rays are in phase and interfering constructively. Bragg’s Law is satisfied and diffraction is occurring. If you click on the details button you can see the detector, which measures how well the phases of the two rays match. When the meter is green it indicates that Bragg’s law is satisfied. You can change three variables (d, λ, and θ) to see how they effect the diffraction.
Exercise (PageIndex{2}): Unit Cell Size from Diffraction Data
The diffraction pattern of copper metal was measured with X-ray radiation of wavelength of 1.315Å. The first order Bragg diffraction peak was found at an angle 2theta of 50.5 degrees. Calculate the spacing between the diffracting planes in the copper metal.