Download Chapter 15 Section 1 Interference

Chapter 15
Section 1 Interference
Preview
• Objectives
• Combining Light Waves
• Demonstrating Interference
• Sample Problem
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 1 Interference
Objectives
• Describe how light waves interfere with each other to
produce bright and dark fringes.
• Identify the conditions required for interference to
occur.
• Predict the location of interference fringes using the
equation for double-slit interference.
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 1 Interference
Combining Light Waves
• Interference takes place only between waves with the
same wavelength. A light source that has a single
wavelength is called monochromatic.
• In constructive interference, component waves
combine to form a resultant wave with the same
wavelength but with an amplitude that is greater than
the either of the individual component waves.
• In the case of destructive interference, the resultant
amplitude is less than the amplitude of the larger
component wave.
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 1 Interference
Interference Between Transverse Waves
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 1 Interference
Combining Light Waves, continued
• Waves must have a constant phase difference for
interference to be observed.
• Coherence is the correlation between the phases of
two or more waves.
– Sources of light for which the phase difference is
constant are said to be coherent.
– Sources of light for which the phase difference is
not constant are said to be incoherent.
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 1 Interference
Combining Light Waves
Click below to watch the Visual Concept.
Visual Concept
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 1 Interference
Demonstrating Interference
• Interference can be demonstrated by passing light
through two narrow parallel slits.
• If monochromatic light is used, the light from the two
slits produces a series of bright and dark parallel
bands, or fringes, on a viewing screen.
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 1 Interference
Conditions for Interference of Light Waves
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 1 Interference
Demonstrating Interference, continued
• The location of interference
fringes can be predicted.
• The path difference is the
difference in the distance
traveled by two beams when
they are scattered in the same
direction from different points.
• The path difference equals
dsin.
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 1 Interference
Interference Arising from Two Slits
Click below to watch the Visual Concept.
Visual Concept
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 1 Interference
Demonstrating Interference, continued
• The number assigned to interference fringes with
respect to the central bright fringe is called the order
number. The order number is represented by the
symbol m.
• The central bright fringe at q = 0 (m = 0) is called the
zeroth-order maximum, or the central maximum.
• The first maximum on either side of the central
maximum (m = 1) is called the first-order maximum.
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 1 Interference
Demonstrating Interference, continued
• Equation for constructive interference
d sin  = ±m m = 0, 1, 2, 3, …
The path difference between two waves =
an integer multiple of the wavelength
• Equation for destructive interference
d sin  = ±(m + 1/2) m = 0, 1, 2, 3, …
The path difference between two waves =
an odd number of half wavelength
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 1 Interference
Sample Problem
Interference
The distance between the two slits is 0.030 mm. The
second-order bright fringe (m = 2) is measured on a
viewing screen at an angle of 2.15º from the central
maximum. Determine the wavelength of the light.
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 1 Interference
Sample Problem, continued
Interference
1. Define
Given:
d = 3.0  10–5 m
m=2
 = 2.15º
Unknown:
=?
Diagram:
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 1 Interference
Sample Problem, continued
Interference
2. Plan
Choose an equation or situation: Use the equation
for constructive interference.
d sin  = m
Rearrange the equation to isolate the unknown:

d sin
m
© Houghton Mifflin Harcourt Publishing Company
Section 1 Interference
Chapter 15
Sample Problem, continued
Interference
3. Calculate
Substitute the values into the equation and solve:
3.0  10


–5

m  sin2.15º 
2
  5.6  10 –7 m  5.6  102 nm
  5.6  102 nm
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 1 Interference
Sample Problem, continued
Interference
4. Evaluate
This wavelength of light is in the visible spectrum.
The wavelength corresponds to light of a yellowgreen color.
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 2 Diffraction
Preview
• Objectives
• The Bending of Light Waves
• Diffraction Gratings
• Sample Problem
• Diffraction and Instrument Resolution
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 2 Diffraction
Objectives
• Describe how light waves bend around obstacles
and produce bright and dark fringes.
• Calculate the positions of fringes for a diffraction
grating.
• Describe how diffraction determines an optical
instrument’s ability to resolve images.
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 2 Diffraction
The Bending of Light Waves
• Diffraction is a change in the direction of a wave
when the wave encounters an obstacle, an opening,
or an edge.
• Light waves form a diffraction pattern by passing
around an obstacle or bending through a slit and
interfering with each other.
• Wavelets (as in Huygens’ principle) in a wave front
interfere with each other.
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 2 Diffraction
Destructive Interference in Single-Slit Diffraction
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 2 Diffraction
The Bending of Light Waves, continued
• In a diffraction pattern, the central maximum is twice
as wide as the secondary maxima.
• Light diffracted by an obstacle also produces a
pattern.
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 2 Diffraction
Diffraction Gratings
• A diffraction grating uses diffraction and interference
to disperse light into its component colors.
• The position of a maximum depends on the
separation of the slits in the grating, d, the order of
the maximum m,, and the wavelength of the light, .
d sin  = ±m
m = 0, 1, 2, 3, …
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 2 Diffraction
Constructive Interference by a Diffraction Grating
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 2 Diffraction
Sample Problem
Diffraction Gratings
Monochromatic light from a helium-neon laser ( =
632.8 nm) shines at a right angle to the surface of a
diffraction grating that contains 150 500 lines/m. Find
the angles at which one would observe the first-order
and second-order maxima.
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 2 Diffraction
Sample Problem, continued
Diffraction Gratings
1. Define
Given:
 = 632.8 nm = 6.328  10–7 m
m = 1 and 2
1
1
d

m
lines 150 500
150 500
m
Unknown:
1 = ?
© Houghton Mifflin Harcourt Publishing Company
2 = ?
Chapter 15
Section 2 Diffraction
Sample Problem, continued
Diffraction Gratings
1. Define, continued
Diagram:
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 2 Diffraction
Sample Problem, continued
Diffraction Gratings
2. Plan
Choose an equation or situation: Use the
equation for a diffraction grating.
d sin  = ±m
Rearrange the equation to isolate the unknown:
 m 
  sin 

d


–1
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 2 Diffraction
Sample Problem, continued
Diffraction Gratings
3. Calculate
Substitute the values into the equation and solve:
For the first-order maximum, m = 1:


–7


6.328

10
m


–1
–1
1  sin    sin 

1
d
 

m 
 150 500

1  5.465º
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 2 Diffraction
Sample Problem, continued
Diffraction Gratings
3. Calculate, continued
For m = 2:
–1  2 
 2  sin  
 d 

–7
2
6.328

10
m

–1
 2  sin 
1

m
150 500


 2  10.98º
© Houghton Mifflin Harcourt Publishing Company






Chapter 15
Section 2 Diffraction
Sample Problem, continued
Diffraction Gratings
4. Evaluate
The second-order maximum is spread slightly more
than twice as far from the center as the first-order
maximum. This diffraction grating does not have high
dispersion, and it can produce spectral lines up to the
tenth-order maxima (where sin  = 0.9524).
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 2 Diffraction
Function of a Spectrometer
Click below to watch the Visual Concept.
Visual Concept
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 2 Diffraction
Diffraction and Instrument Resolution
• The ability of an optical system to distinguish
between closely spaced objects is limited by the
wave nature of light.
• Resolving power is the ability of an optical
instrument to form separate images of two objects
that are close together.
• Resolution depends on wavelength and aperture
width. For a circular aperture of diameter D:
  1.22
© Houghton Mifflin Harcourt Publishing Company

D
Chapter 15
Section 2 Diffraction
Resolution of Two Light Sources
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 3 Lasers
Preview
• Objectives
• Lasers and Coherence
• Applications of Lasers
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 3 Lasers
Objectives
• Describe the properties of laser light.
• Explain how laser light has particular advantages in
certain applications.
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 3 Lasers
Lasers and Coherence
• A laser is a device that produces coherent light at a
single wavelength.
• The word laser is an acronym of “light amplification
by stimulated emission of radiation.”
• Lasers transform other forms of energy into coherent
light.
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 3 Lasers
Comparing Incoherent and Coherent Light
Click below to watch the Visual Concept.
Visual Concept
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 3 Lasers
Laser
Click below to watch the Visual Concept.
Visual Concept
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 3 Lasers
Applications of Lasers
• Lasers are used to measure distances with great
precision.
• Compact disc and DVD players use lasers to read
digital data on these discs.
• Lasers have many applications in medicine.
– Eye surgery
– Tumor removal
– Scar removal
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 3 Lasers
Components of a
Compact Disc Player
© Houghton Mifflin Harcourt Publishing Company
Chapter 15
Section 3 Lasers
Incoherent and Coherent Light
© Houghton Mifflin Harcourt Publishing Company