I NTERFERENCE AND D IFFRACTION Chapter 15 Holt. Section 1 Interference: Combining Light Waves I nterference takes place only between waves with the same.

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I NTERFERENCE AND D IFFRACTION Chapter 15 Holt

Section 1 Interference: Combining Light Waves I nterference 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 (intensity) 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. 1 2 Resultant (blue)

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. Section 1 Interference: Combining Light Waves 360 o out of Phase

Superposition of Waves Which two waves are in phase and which two waves are out of phase? In PhaseOut of Phase

http://www.kevinkemp.com/homerecordingtutorial/micing.htm A C B 3 2 1

http://www.admet.com/images/SineWave.gif SourceSource SourceSource monochromatic monochromatic and coherent

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. Section 1 Interference: D EMONSTRATING I NTERFERENCE

http://micro.magnet.fsu.edu/primer/java/interference/doublesli t/ Young’s Double Slit Experiment : 1803 and 1807

Holt video

C ONDITIONS FOR I NTERFERENCE OF L IGHT W AVES Section 1 Interference: D EMONSTRATING I NTERFERENCE

The location of interference fringes can be predicted. Bright fringes result from constructive interference and dark fringes result from complete destructive interference. 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 d sin . Constructive interference occurs when d sin  equals, 2, 3, … or m where m= 0, 1, 2, 3, …. Destructive interference occurs when d sin  equal s 1/2, (1+1/2), (2+1/2), … or (m+1/2) where m= 0,1,2,3, …. Section 1 Interference: D EMONSTRATING I NTERFERENCE

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  = 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 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 Section 1 Interference: D EMONSTRATING I NTERFERENCE

Chapter 15 S AMPLE P ROBLEM 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. Section 1 Interference: D EMONSTRATING I NTERFERENCE Given: d = 3.0  10 –5 m m = 2  = 2.15º Unknown: = ? Equation: d sin  = m

T HE B ENDING OF L IGHT W AVES 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. Section 2 Diffraction

In a diffraction pattern, the central maximum is twice as wide as the secondary maxima. Light diffracted by an obstacle also produces a pattern. Section 2 Diffraction: T HE B ENDING OF L IGHT W AVES http://sdsu-physics.org/physics197/phys197.html Diffraction Pattern around a Penny!

A diffraction grating uses diffraction and interference to disperse light into its component colors. A large number of parallel, closely spaced slits constitutes a diffraction grating. 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, … Section 2 Diffraction: D IFFRACTION G RATINGS A grating ruled with 5000 lines/cm has a slit spacing, d, equal to (1/5000) cm = 2 x 10 -4 cm.

D IFFRACTION G RATINGS Section 2 Diffraction

http://www.itp.uni- hannover.de/~zawischa/ITP/multibeam.html Single Slit Monochromatic Single Slit White Light Double Slit Monochromatic Double Slit White Light Three Slits

http://www.itp.uni- hannover.de/~zawischa/ITP/multibeam.html Seven Slits Fifteen Slits Section 2 Diffraction

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. S ECTION 2 D IFFRACTION Given: = 632.8 nm = 6.328  10 –7 m m = 1 and 2 Unknown:   = ?   = ?

S ECTION 2 D IFFRACTION Choose an equation or situation: Use the equation for a diffraction grating. d sin  = ± m 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).