© Houghton Mifflin Harcourt Publishing Company Preview Objectives Combining Light Waves Demonstrating Interference Sample Problem Chapter 15 Section 1

© Houghton Mifflin Harcourt Publishing Company

Preview

• Objectives

• Combining Light Waves

• Demonstrating Interference

• Sample Problem

Chapter 15 Section 1 Interference


Section 1 InterferenceChapter 15

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.



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.


Chapter 15

Interference Between Transverse Waves

Section 1 Interference


• 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.


Combining Light Waves, continued


Click below to watch the Visual Concept.

Visual Concept


Combining Light Waves



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.


Chapter 15

Conditions for Interference of Light Waves

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.



Visual Concept


Interference Arising from Two Slits




• 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.




• Equation for constructive interferenced sin = ±m m = 0, 1, 2, 3, …

The path difference between two waves = an integer multiple of the wavelength

• Equation for destructive interferenced sin = ±(m + 1/2) m = 0, 1, 2, 3, …

The path difference between two waves = an odd number of half wavelength



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.



Sample Problem, continued

Interference1. DefineGiven: d = 3.0 10–5 m

m = 2= 2.15º

Unknown: = ?

Diagram:




Interference2. Plan

Choose an equation or situation: Use the equation for constructive interference.

d sin = m

Rearrange the equation to isolate the unknown:

d sinm




Interference3. Calculate

Substitute the values into the equation and solve:

–5

–7 2

2

3.0 10 m sin2.15º

2

5.6 10 m 5.6 10 nm

5.6 10 nm




Interference4. Evaluate

This wavelength of light is in the visible spectrum. The wavelength corresponds to light of a yellow-green color.


Preview

• Objectives

• The Bending of Light Waves

• Diffraction Gratings

• Sample Problem

• Diffraction and Instrument Resolution

Chapter 15 Section 2 Diffraction


Section 2 DiffractionChapter 15

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.



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.


Chapter 15

Destructive Interference in Single-Slit Diffraction

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.



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, …


Chapter 15

Constructive Interference by a Diffraction Grating




Sample Problem


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.





1. Define

Given: = 632.8 nm = 6.328 10–7 m

m = 1 and 2

Unknown: = ? 2 = ?

d 1

150 500lines

m

1

150 500m





1. Define, continued

Diagram:





2. Plan

Choose an equation or situation: Use the equation for a diffraction grating.

d sin = ±mRearrange the equation to isolate the unknown:

–1sinm

d





3. Calculate

Substitute the values into the equation and solve:

For the first-order maximum, m = 1:

–7–1 –1

1

1

6.328 10 msin sin

1m

150 500

5.465º

d





3. Calculate, continued

For m = 2:

–12

–7

–12

2

2sin

2 6.328 10 msin

1m

150 500

10.98º

d





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).



Visual Concept

Chapter 15 Section 2 Diffraction

Function of a Spectrometer



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

D


Chapter 15

Resolution of Two Light Sources



Preview

• Objectives

• Lasers and Coherence

• Applications of Lasers

Chapter 15 Section 3 Lasers


Section 3 LasersChapter 15

Objectives

• Describe the properties of laser light.

• Explain how laser light has particular advantages in certain applications.



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.



Visual Concept


Comparing Incoherent and Coherent Light



Visual Concept


Laser



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


Chapter 15

Components of a Compact Disc Player

Section 3 Lasers


Chapter 15

Incoherent and Coherent Light

Section 3 Lasers

Documents

© Houghton Mifflin Harcourt Publishing Company Preview Objectives Combining Light Waves Demonstrating Interference Sample Problem Chapter 15 Section 1