diffraction physics 202 professor lee carkner lecture 24
Post on 19-Dec-2015
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PAL #23 Interference Applications
Wavelength of laser D = d = y = 3 mm = 0.003 m (between 0 and 1
maxima) y = mD/d = yd/D = (0.003)(0.00025)/(1.02) =
Is this reasonable? No, laser is red and red light has a
wavelength between ~ 600- 700 nm
PAL #23 Interference Applications
What color does a soap film (n=1.33) appear to be if it is 500 nm thick? We need to find the wavelength of the maxima:
= (2Ln) / (m + ½)
= [(2) (500nm) (1.33)] / (m + ½)
= 2660 nm, 887 nm, 532 nm, 380 nm …
Real soap bubbles change thickness due to turbulence and gravity and so the colors shift
Diffraction When light passes though a small aperture
it spreads out
This flaring produces an image with a bright central maximum and progressively fainter maxima at increasing angles
Geometric optics assume point images, but all real images are blurry
Diffraction and Interference
Young’s experiment is an example of light rays from two different apertures producing interference
This is called single slit diffraction
Instead of two rays from two slits, we have a continuum of rays emerging from one slit
Path Length Difference Minima (dark fringes) should occur at the point where
half of the rays are out of phase with the other half
If we assume that the distance to the screen (D) is much larger than the slit width (a) then the path difference is
where d is the distance between the origin points of the two
rays
We will pair up the rays, and find the path length for which each pair cancels out
Location of the Minima Where is the first minima? Since:
L /d = sin
How far apart can a pair of rays get?
For the first minima L must equal /2:
(a/2) sin = /2a sin =
Diffraction Patterns
a sin = m(min) Where is the location of the minima
corresponding to order m
Note that this relationship is the reverse of that for double slit interference [d sin = (m+½): min]
Since waves from the top and bottom half cancel
Intensity
Intensity of maxima decrease with increasing The intensity is proportional to the value of E2,
which in turn depends on the phase difference
= ½ = (a/) sin I = Im [(sin /]2
where Im is the maximum intensity of the pattern
Intensity Variations The intensity falls off rapidly with
linear distance y
Remember tan = y/D
The narrower the slit the broader the maximum
Remember: m = 1,2,3 … minima m = 1.5, 2.5, 3.5 … maxima
Diffraction and Circular Apertures
The location of the minima depend on the wavelength and the diameter (d) instead of slit width:
sin = 1.22 /d For m = 1
The minima and maxima appear as concentric circles
Rayleigh’s Criterion We will consider two near-by point sources to be
resolvable if the central maximum of one lies on the first minimum of the other
For small angles:R = 1.22 /d
This is called Rayleigh’s criterion Small angle is better
Short and large d give better resolution (smaller R)
Resolution Since virtually all imaging devices have
apertures, virtually all images are blurry
If you view two point sources that are very close together, you may not be able to distinguish them
Next Time
Read: 36.7-36.9 Final, Monday, Feb 13, 9-11 am
About 2/3 covers optics About 1/3 covers fluids, SMH and
waves and thermo Four equation sheets given
If the thickness of the middle layer is ½ wavelength, what kind of interference would you see?
a) Constructiveb) Destructivec) None
n=1
n=1.5
n=1.3
If the thickness of the middle layer is ½ wavelength, what kind of interference would you see?
a) Constructiveb) Destructivec) None
n=1.3
n=1.5
n=1.1