Diffraction Applications
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Diffraction Applications
Physics 202Professor Lee
CarknerLecture 26
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PAL #25
The first side pattern is between the m=1 and m=2 diffraction minima:
a sin = and a sin = 2
sin = /a and sin = 2 /a
sin 1 = 650 X 10-9 / 0.08 X 10-3 = 8.125 X 10-3
sin 2 = (2)(650 X 10-9)/0.08 X 10-3 =1.625 X 10-2
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PAL #25 What interference maxima are between
the two angles?
m1 = d sin 1 / and m2 = d sin 2 /m1 = (0.25 X 10-3)(8.125 X 10-3)/650 X 10-9=
m2 = (0.25 X 10-3)(1.625 X 10-2)/650 X 10-9= We should see 3 bright fringes (m = 4,5,6)
in the first side diffraction envelope
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PAL #25
Middle interference fringe is m = 5 sin = (5)()/(d) =
= (a/ sin = [()(0.08 X 10-3 ) /(650 X 10-
9 )] (0.013) = 5.026 rad = (d/) sin =
I = Imcos2 (sin /)2 = Im (1)(0.0358) = 0.036 Im
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PAL #25 Screen is 2 meters away, what is at point 4.3
cm from the center?
Diffraction pattern y/D = m/a, m = ya/D = (4.3X10-2)(0.08X10-3)/(2)
(650X10-9) = 2.65
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Diffraction Gratings
If light of 2 different wavelengths passes through, each will produce a maxima, but they will tend to blur together
This makes lines from different wavelengths easier to distinguish
A system with large N is called a diffraction grating and is useful for spectroscopy
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Maxima From Grating
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Location of Lines
d sin = m where d is the distance between any two slits (or
rulings) on the grating
For polychromatic light, each maxima is composed of many narrow lines (one for each wavelength the incident light is composed of)
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Elemental Lines
When electrons move between these energy levels, they can produce light at a specific wavelength
The pattern of spectral lines can identify the
element
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Spectroscope
This will produce a series of orders, each order containing lines (maxima) over a range of wavelengths
The wavelength of a line corresponds to its position angle
We measure with a optical scope mounted on a vernier position scale Can also take an image of the pattern
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Using Spectroscopy
We want to be able to resolve lines that are close together
How can we achieve this?
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Line Width The narrower the lines, the easier to
resolve lines that are closely spaced in wavelength
hw = /(Nd cos ) where N is the number of slits and d is
the distance between 2 slits
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Dispersion
D =
D = m / d cos For larger m and smaller d the
resulting spectra takes up more space
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Resolving Power The most important property
of a grating is the resolving power, a measure of how well closely separated lines (in ) can be distinguished
R = av/
For example, a grating with R = 10000 could resolve 2 blue lines (= 450 nm) that were separated by 0.045 nm
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Resolving Power of a Grating
R = Nm
Looking at higher orders helps to resolve lines
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Spectral Type
The types of elements present in a star and the transitions they make depends on the temperature
Examples: Very cool stars (T~3000 K) can be identified by
the presence of titanium oxide which cannot exist at high temperatures
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Next Time
Final exam Monday, 9-11 am, SC304 Bring pencil and calculator 4 equation sheets provided Covers 2/3 optics, 1/3 rest of
course
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In a double slit diffraction pattern, what happens to the number of interference maxima in the first side pattern if you double the width of each slit?
a) Increasesb) Decreasesc) Stays the same
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In a double slit diffraction pattern, what happens to the number of interference maxima in the first side pattern if you double the distance between the slits?
a) Increasesb) Decreasesc) Stays the same
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In a double slit diffraction pattern, what happens to the number of interference maxima in the first side pattern if you double the wavelength?
a) Increasesb) Decreasesc) Stays the same
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In a double slit diffraction pattern, what could to do to maximize the number of fringes in the central pattern?
a) Increase a, increase db) Decrease a, decrease dc) Increase a, decrease dd) Decrease a, increase de) You can’t change the number of
fringes in the central pattern