Chapter 15 Preview Objectives Combining Light Waves Section 1 Interference Preview Objectives Combining Light Waves Demonstrating Interference Sample Problem

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.

Chapter 15 Combining Light Waves 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.

Interference Between Transverse Waves Chapter 15 Section 1 Interference Interference Between Transverse Waves

Combining Light Waves, continued 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.

Chapter 15 Combining Light Waves Section 1 Interference Click below to watch the Visual Concept. Visual Concept

Demonstrating Interference 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.

Conditions for Interference of Light Waves Chapter 15 Section 1 Interference Conditions for Interference of Light Waves

Demonstrating Interference, continued 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.

Interference Arising from Two Slits Chapter 15 Section 1 Interference Interference Arising from Two Slits Click below to watch the Visual Concept. Visual Concept

Demonstrating Interference, continued 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.

Demonstrating Interference, continued 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, … an odd number of half wavelength

Chapter 15 Sample Problem Interference 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.

Sample Problem, continued Chapter 15 Section 1 Interference Sample Problem, continued Interference 1. Define Given: d = 3.0  10–5 m m = 2  = 2.15º Unknown:  = ? Diagram:

Sample Problem, continued 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:

Sample Problem, continued Chapter 15 Section 1 Interference Sample Problem, continued Interference 3. Calculate Substitute the values into the equation and solve:

Sample Problem, continued 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 yellow-green color.

Chapter 15 Preview Objectives The Bending of Light Waves Section 2 Diffraction Preview Objectives The Bending of Light Waves Diffraction Gratings Sample Problem Diffraction and Instrument Resolution

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.

The Bending of Light Waves 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.

Destructive Interference in Single-Slit Diffraction Chapter 15 Section 2 Diffraction Destructive Interference in Single-Slit Diffraction

The Bending of Light Waves, continued 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.

Chapter 15 Diffraction Gratings 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, …

Constructive Interference by a Diffraction Grating Chapter 15 Section 2 Diffraction Constructive Interference by a Diffraction Grating

Chapter 15 Sample Problem Diffraction Gratings 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.

Sample Problem, continued Chapter 15 Section 2 Diffraction Sample Problem, continued Diffraction Gratings Define Given:  = 632.8 nm = 6.328  10–7 m m = 1 and 2 Unknown: 1 = ? 2 = ?

Sample Problem, continued Chapter 15 Section 2 Diffraction Sample Problem, continued Diffraction Gratings Define, continued Diagram:

Sample Problem, continued 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:

Sample Problem, continued 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:

Sample Problem, continued Chapter 15 Section 2 Diffraction Sample Problem, continued Diffraction Gratings 3. Calculate, continued For m = 2:

Sample Problem, continued 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).

Function of a Spectrometer Chapter 15 Section 2 Diffraction Function of a Spectrometer Click below to watch the Visual Concept. Visual Concept

Diffraction and Instrument Resolution 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:

Resolution of Two Light Sources Chapter 15 Section 2 Diffraction Resolution of Two Light Sources

Chapter 15 Preview Objectives Lasers and Coherence Section 3 Lasers Preview Objectives Lasers and Coherence Applications of Lasers

Chapter 15 Objectives Describe the properties of laser light. Section 3 Lasers Objectives Describe the properties of laser light. Explain how laser light has particular advantages in certain applications.

Chapter 15 Lasers and Coherence 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.

Comparing Incoherent and Coherent Light Chapter 15 Section 3 Lasers Comparing Incoherent and Coherent Light Click below to watch the Visual Concept. Visual Concept

Chapter 15 Laser Section 3 Lasers Click below to watch the Visual Concept. Visual Concept

Applications of Lasers 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

Components of a Compact Disc Player Chapter 15 Section 3 Lasers Components of a Compact Disc Player

Incoherent and Coherent Light Chapter 15 Section 3 Lasers Incoherent and Coherent Light