Chapter 37 - Interference and Diffraction A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University A PowerPoint
Chapter 37 - Interference and Diffraction A PowerPoint
Presentation by Paul E. Tippens, Professor of Physics Southern
Polytechnic State University A PowerPoint Presentation by Paul E.
Tippens, Professor of Physics Southern Polytechnic State University
2007
Slide 3
Objectives: After completing this module, you should be able
to: Define and apply concepts of constructive interference,
destructive interference, diffraction, and resolving power.Define
and apply concepts of constructive interference, destructive
interference, diffraction, and resolving power. Describe Youngs
experiment and be able to predict the location of dark and bright
fringes formed from the interference of light waves.Describe Youngs
experiment and be able to predict the location of dark and bright
fringes formed from the interference of light waves. Discuss the
use of a diffraction grating, derive the grating equation, and
apply it to the solution of optical problems.Discuss the use of a
diffraction grating, derive the grating equation, and apply it to
the solution of optical problems.
Slide 4
Diffraction of Light Diffraction is the ability of light waves
to bend around obstacles placed in their path. OceanBeach Water
waves easily bend around obstacles, but light waves also bend, as
evidenced by the lack of a sharp shadow on the wall. Fuzzy Shadow
Light rays
Slide 5
Water Waves A wave generator sends periodic water waves into a
barrier with a small gap, as shown below. A new set of waves is
observed emerging from the gap to the wall.
Slide 6
Interference of Water Waves An interference pattern is set up
by water waves leaving two slits at the same instant.
Slide 7
Youngs Experiment In Youngs experiment, light from a
monochromatic source falls on two slits, setting up an interference
pattern analogous to that with water waves. Light source S1S1
S2S2
Slide 8
The Superposition Principle The resultant displacement of two
simul- taneous waves (blue and green) is the algebraic sum of the
two displacements.The resultant displacement of two simul- taneous
waves (blue and green) is the algebraic sum of the two
displacements. The superposition of two coherent light waves
results in light and dark fringes on a screen. The composite wave
is shown in yellow.The composite wave is shown in yellow.
Constructive Interference Destructive Interference
Conditions for Bright Fringes Bright fringes occur when the
difference in path p is an integral multiple of one wave length.
p1p1p1p1 p2p2p2p2 p3p3p3p3 p4p4p4p4 Path difference p = 0,, 2, 3,
Bright fringes: p = n, n = 0, 1, 2,...
Slide 11
Conditions for Dark Fringes Dark fringes occur when the
difference in path p is an odd multiple of one-half of a wave
length . p1p1p1p1 p2p2p2p2 p3p3p3p3 p3p3p3p3 n = odd n = 1,3,5 Dark
fringes:
Slide 12
Analytical Methods for Fringes x y d sin s1s1 s2s2 d p1p1 p2p2
Bright fringes: d sin = n, n = 0, 1, 2, 3,... Dark fringes: d sin =
n , n = 1, 3, 5,... p = p 1 p 2 p = d sin Path difference
determines light and dark pattern.
Slide 13
Analytical Methods (Cont.) x y d sin s1s1 s2s2 d p1p1 p2p2 From
geometry, we recall that: Bright fringes:Dark fringes: So
that...
Slide 14
Example 1: Two slits are 0.08 mm apart, and the screen is 2 m
away. How far is the third dark fringe located from the central
maximum if light of wavelength 600 nm is used? x = 2 m; d = 0.08 mm
= 600 nm; y = ? = 600 nm; y = ? The third dark fringe occurs when n
= 5 x y d sin s1s1 s2s2 n = 1, 3, 5 Dark fringes: d sin = 5(
/2)
Slide 15
Example 1 (Cont.): Two slits are 0.08 mm apart, and the screen
is 2 m away. How far is the third dark fringe located from the
central maximum if = 600 nm? x = 2 m; d = 0.08 mm = 600 nm; y = ? =
600 nm; y = ? x y d sin s1s1 s2s2 n = 1, 3, 5 y = 3.75 cm
Slide 16
The Diffraction Grating A diffraction grating consists of
thousands of parallel slits etched on glass so that brighter and
sharper patterns can be observed than with Youngs experiment.
Equation is similar. d sin d d sin n n = 1, 2, 3,
Slide 17
The Grating Equation The grating equation:d = slit width
(spacing) = wavelength of light = angular deviation n = order of
fringe 1 st order 2 nd order
Slide 18
Example 2: Light (600 nm) strikes a grating ruled with 300
lines/mm. What is the angular deviation of the 2 nd order bright
fringe? 300 lines/mm n = 2 To find slit separation, we take
reciprocal of 300 lines/mm: Lines/mm mm/line
Slide 19
Example (Cont.) 2: A grating is ruled with 300 lines/mm. What
is the angular deviation of the 2 nd order bright fringe? 2 = 21.1
0 Angular deviation of second order fringe is: 300 lines/mm n = 2 =
600 nm
Slide 20
A compact disk acts as a diffraction grating. The colors and
intensity of the reflected light depend on the orientation of the
disc relative to the eye.
Slide 21
Interference From Single Slit Pattern Exaggerated When
monochromatic light strikes a single slit, diffraction from the
edges produces an interference pattern as illustrated. Relative
intensity The interference results from the fact that not all paths
of light travel the same distance some arrive out of phase.
Slide 22
Single Slit Interference Pattern a/2 a 1 2 4 3 5 Each point
inside slit acts as a source. For rays 1 and 3 and for 2 and 4:
First dark fringe: For every ray there is another ray that differs
by this path and therefore interferes destructively.
Slide 23
Single Slit Interference Pattern a/2 a 1 2 4 3 5 First dark
fringe: Other dark fringes occur for integral multiples of this
fraction /a.
Slide 24
Example 3: Monochromatic light shines on a single slit of width
0.45 mm. On a screen 1.5 m away, the first dark fringe is displaced
2 mm from the central maximum. What is the wavelength of the light?
x = 1.5 m y a = 0.35 mm = ? = 600 nm
Slide 25
Diffraction for a Circular Opening Circular diffraction D The
diffraction of light passing through a circular opening produces
circular interference fringes that often blur images. For optical
instruments, the problem increases with larger diameters D.
Slide 26
Resolution of Images Consider light through a pinhole. As two
objects get closer the interference fringes overlap, making it
difficult to distinguish separate images. d2d2 Separate images
barely seen d1d1 Clear image of each object
Slide 27
Resolution Limit d2d2 Resolution limit Images are just resolved
when central maximum of one pattern coincides with first dark
fringe of the other pattern. Resolution Limit Separate images
Slide 28
Resolving Power of Instruments The resolving power of an
instrument is a measure of its ability to produce well-defined
separate images. Limiting angle of resolution: For small angles,
sin , and the limiting angle of resolution for a circular opening
is: Limiting angle D
Slide 29
Resolution and Distance Limiting Angle of Resolution: sosososop
D Limiting angle o
Slide 30
Example 4: The tail lights ( = 632 nm) of an auto are 1.2 m
apart and the pupil of the eye is around 2 mm in diameter. How far
away can the tail lights be resolved as separate images?
sosososopEye D Tail lights p = 3.11 km
Slide 31
Summary Bright fringes:Dark fringes: Youngs Experiment:
Monochromatic light falls on two slits, producing interference
fringes on a screen. x y d sin s1s1 s2s2 d p1p1 p2p2
Slide 32
Summary (Cont.) The grating equation: d = slit width (spacing)
= wavelength of light = angular deviation n = order of fringe
Slide 33
Summary (Cont.) Pattern Exaggerated Relative Intensity
Interference from a single slit of width a :
Slide 34
Summary (cont.) Limiting Angle of Resolution: sosososop D
Limiting angle o The resolving power of instruments.
Slide 35
CONCLUSION: Chapter 37 Interference and Diffraction