Wolfram Alpha:
Search by keyword:
Astronomy
Chemistry
Classical Mechanics
Classical Physics
Climate Change
Cosmology
Finance and Accounting
Game Theory
General Relativity
Group Theory
Lagrangian and Hamiltonian Mechanics
Macroeconomics
Mathematics
Microeconomics
Nuclear Physics
Particle Physics
Probability and Statistics
Programming and Computer Science
Quantum Computing
Quantum Field Theory
Quantum Mechanics
Semiconductor Reliability
Solid State Electronics
Special Relativity
Statistical Mechanics
String Theory
Superconductivity
Supersymmetry (SUSY) and Grand Unified Theory (GUT)
The Standard Model
Topology
Units, Constants and Useful Formulas
Double Slit Experiment
----------------------
δ = path difference = dsinθ'
For D>>a, θ' ~ θ therefore δ = dsinθ
For a maximum we need dsinθ = mλ
Now if θ is small, sinθ ~ tanθ = y/D. Therefore,
dy/D = mλ
Intensity:
I = I_{0}cos^{2}(π.d.sinθ/λ)
= I_{0}cos^{2}(π.d.y/λ.D) for small θ
It is worth bearing in mind that these results are an idealization assuming an
infinitesimally small width of the slits. A full treatment of the situation,
accounting for the finite width of the slits requires us to take diffraction
into account.
Single Slit Diffraction
-----------------------
Diffraction manifests itself in the apparent bending of waves around small
obstacles and the spreading out of waves past small openings. There are
two types - Fraunhoher and Fresnel. Fraunhofer diffraction deals with the
limiting cases where the light appoaching the diffracting object is parallel
and monochromatic, and where the image plane is at a distance large compared
to the size of the diffracting object. The more general case where these
restrictions are relaxed is called Fresnel diffraction.
Condition for a maximum:
Using the same geometry and reasoning as the double slit, the condition is,
asinθ = mλ
ay/D = mλ
Intensity: I = I_{0}[sin(π.a.sinθ/λ)/(π.a.sinθ/λ)]^{2}
= I_{0}[sin(π.a.y/λD)/(π.a.y/λ.D)^{2} for small θ
Diffraction and Interference Combined
-------------------------------------
Combined diffraction and interference is obtain by multiplying diffraction
intensity for single slit by interference for double slit. Thus,
I = I_{0}[cos^{2}(π.d.sinθ/λ)][sin(π.a.sinθ/λ)/(π.a.sinθ/λ)]^{2}
The first and the second terms in the above equation are referred to as
the "interference factor" and the "diffraction factor". The former yields
the interference substructure, the latter acts as an envelope which sets
limits on the the number of interference peaks.
Diffraction Grating
-------------------
A diffraction grating is a large number of close, parallel equidistant slits
ruled on glass or metal.
The condition for maximum intensity is the same as that for the double
slit. However, as more parallel equidistant slits are added, the intensity
and sharpness of the principal maxima increase and those of the subsiduary
maxima decrease.
Bright or principal maxima are obtained when,
dsinθ = mλ
d can be calculated from the number of slits per metre.
Spectra
-------
Diffraction gratings provides a valuable way of studying spectra. If white
light is incident normally on a diffraction grating, several colored spectra
are observed on either side of the normal. This occurs because the wavelength
of violet light is 3.8 x 10^{-7} m and the wavelength of red light is 7.0 x 10^{-7} m.
Therefore, θ is less for violet than red.
Now,
dsinθ = 2 x 3.8 x 10^{-7} for violet the 2nd order spectrum
= 7.6 x 10^{-7}
and,
dsinθ = 7.0 x 10^{-7} for red the 1st order spectrum
Thus, the red in the 1st order spectrum does not overlap the violet in the
2nd order spectrum. However, overlapping of colors does occur in higher order
spectra.
Ex: Two slits separated by a distance d = 1cm allow light of wavelength
500nm to pass through and interfere. A screen is situated 10m from the slits,
where a pattern of bright and dark fringes are observed. As measured from
the center 0th order fringe, what is the distance to the first dark fringe
in μm?
y = mDλ/d => 1*10*500 x 10^{-9}/10^{-2} => 5*10^{-4} => 500μm
This is the distance between 2 peaks. The distance from the center is this
value/2.
Bragg's Law
-----------
When x-rays are scattered from a crystal lattice, a diffraction pattern is
observed. A plot of the intensity of the scattered waves as a function of
scattering angle shows peaks in the pattern when the following condition
is satisfied.
mλ = 2dsinθ ... Bragg's Law
The condition for maximum intensity allows us to calculate details about
the crystal structure, or if the crystal structure is known, to determine
the wavelength of the x-rays incident upon the crystal.