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Interference effects for continuous sources:
i) Light bends around corners.
ii) “Shadows” fill in
iii) “Parallel” beams always spread
iv) Limits of resolution of microscopes and telescopes
Diffraction Ch 38Diffraction Ch 38
Fraunhofer Diffraction: (easy math)Source, screen “ at infinity”, ~parallel rays
eg. Laser & narrow slit.
Fresnel Diffraction: (complicated math)
Source distance, object size and screen distance all comparable.
eg.
Source Steel Ball Screen
ShadowPoisson
Spot
Single Slit, width = Single Slit, width = aa
a
θ 54321
Use Huygens’s idea:Treat the slit as a large number of point sources.
x = a sin θ is path difference between waves at two edges
(Fraunhoffer diffraction)
Divide slit in half and add up rays in pairs:
Ray (1) travels farther than ray (3): when it is ½ cycle behind -> Cancel
Ray (2) travels farther than ray (4): when it is ½ cycle behind -> Cancel
Ray (3) travels farther than ray (5): when it is ½ cycle behind -> Cancel
1) ANY two waves that originate at points a/2 away are out of phase by 180o for some angle θ, and cancel.
2) waves from upper half interfere destructively with waves from the upper half
a θ
54321
sin2
a a
a
sin
2sin
2
By dividing the slit into four parts rather than two, we can
similarly obtain:
a
2sin
By dividing the slit into six parts rather than two, we cansimilarly obtain:
a
3sin
Therefore, the general condition for destructiveinterference is:
...3,2,1,sin ma
m
where m is the order number, for dark fringes
m = 3
m = 2
m = 1
m = -1
m = -2
m = -3
m=1.5
m=-1.5
θ ...3,2,1,sin ma
m
The equation:
Gives us the positions(θ) of the MINIMA !!!
The maxima are approxhalf way between theminima.
Quiz:
What would the pattern look like if the width of the slit was increased?
A) The minima and maxima would be spaced very far apart
B) No change, the pattern would look the same
C) The minima and maxima would be spaced very closely together
Example 1: Light of λ=580nm is incident on a slit of width 0.30 mm. The observing screen is placed 2.0 m from the slit.
a) Find the positions of the first dark fringesb) Find the width of the central bright fringe
Example 2: Light of λ=580nm is incident on a slit of width 0.30 mm. The observing screen is placed 2.0 m from the slit.
a) Determine the width of the first order bright fringe.b) How does it compare to the width of the central max?
= 600 nm; central peak is 6cm wide on a screen 3m away. How wide is the slit?
slit
Example Example 33
L=3.0 mθ1
θ1
3 cm =y1
3 cm
Intensity distribution for a single slit of width a
2
max 2
sin ( / 2)( / 2)2
= sin
I I
and a
Intensity
What are the values of m and ?
a sin θ = m λ =m2
m = ±1, ±2, ±3, …(but not m = 0)
θ
54321
siny
∆y
sin
2
y
sin
2y
The intensity is a resultant of all the incremental electric fields from each part of the slit. These are out of phase by ∆β:
sin
2sin
2ayNN
(Since asinθ=mλ)
Example 4:
Show that:
a) the Minima happen when: a sinθ = m λ, m =±1, ±2…
b) to what angles of β/2 does this correspond?
Intensity
β/2
I1I2 I2
I1
π-π-2π 2π-3π 3π
Example 5: Find the intensities of the first two secondary maxima for the single slit Fraunhofer diffraction pattern:
(Hint: assume that this occurs midway between the minima.)
Notes:
1) Central maximum is twice as wide, and much brighter (~ 75% of light)
2) Secondary maxima get fainter as we move to higher orders m
3) Minima are at
4) Secondary maxima are approximately halfway between the minima. That is:
sin , 02
a m mm
1
2sin( ) ( ) for 1,2....
2
a m m
m