S-110 A.What does the term Interference mean when applied to waves? B.Describe what you think would happened when light interferes constructively. C.Describe what would happen if light went through total destructive interference.

Objectives I can explain Young’s double slit experiment. I can calculate wavelength using interference patterns.

Interference and Diffraction

9.1 Superposition and Interference

Review – Superposition net displacement caused by a combination of waves – the algebraic sum Phase difference 180 o – out of step by half a wavelength Coherent – maintains a constant phase difference Applet

9.2 Young’s Two-Slit Experiment

1801 – Thomas Young Shot monochromatic, coherent light through two slits The result was bands of color called bright fringes Applet

9.2 Young’s Two-Slit Experiment What happened Hugyen’s principle – every point on a wave front can be treated as the source of a new wave When a wave is sent through a barrier two identical wave fronts are created

9.2 Young’s Two-Slit Experiment So when a wave is sent through double slits The fringes are caused by constructive interference – bright fringes destructive interference – dark fringes When we look at the pathway of two rays Ripple Tank

9.2 Young’s Two-Slit Experiment For constructive interference, the path difference must be a multiple of So m – any whole number dsin 

9.2 Young’s Two-Slit Experiment For dark fringes – the waves must be 180 o out of phase, so To measure , we need to go back to the experiment An look at the triangle One side is L, we will call the other side x x L

9.2 Young’s Two-Slit Experiment We can calculate the angle using the law of tangents x L

S-111 A light is shined through two slits that are 1.2 mm apart. If the third order minima is produced on a screen 1.5, how far from the central maxima will it be. Assume that the light used has a frequency of 7.2 x10 14 hz.

S-112 A light is shined through two slits that are 0.50 mm apart. If the fourth order maxima is produced on a screen 0.80 m from the slits, how far from the central maxima will it be. Assume that the light used has a wavelength of 1.8 x10 -7 m.

9.3 Interference in Reflected Waves

Waves that reflect of objects at different locations can interfere Waves can go through a phase change due to reflection

9.3 Interference in Reflected Waves When waves reflect at a boundary as they go From higher n to lower n – no phase change From lower n to higher n – 180 o phase change

9.3 Interference in Reflected Waves Lets look at what happens in an air wedge There is no phase change at the first boundary There is a phase change at the second boundary The paths are essentially the same length in a real interference pattern (or they wouldn’t hit the same part of eye)

9.3 Interference in Reflected Waves So the length of the path is For constructive interference For destructive interference

9.3 Interference in Reflected Waves Thin Films – soap bubbles or oil slicks First ray (phase change) Second ray (no phase change) Solving in terms of

9.3 Interference in Reflected Waves Combining we get So for constructive interference

9.3 Interference in Reflected Waves And for destructive interference

S-113 Light of wavelength 615 nm strikes the surface of an oil film (n=1.55) that is floating on water (n=1.33). The light strikes the surface of the oil at an angle of 22 o. What is the minimum thickness of the oil that would produce a destructive interference pattern?

9.4 Diffraction

Waves diffract (bend) when they pass through barriers Single Slit Diffraction – monochromatic light sent through a single slit will cause an interference pattern The pattern occurs because of the diffraction of light around the edge of the slit Diffraction Applet

9.4 Diffraction Similar to double slit geometry If a is the width of the slit, the first minimum would occur In general

9.4 Diffraction If the distance D is much greater than the slit distance y, then we can use the Approximation Combining with

9.5 Resolution

Resolution – the ability to visually separate closed spaced objects Depends on the aperture (size of lens) In a slit the first dark fringe would be For a circular aperture of diameter D produces a central bright and a dark fringe at an angle

9.5 Resolution Due to this interference, a point source of light will be viewed as a circular image Rayleigh’s Criterion: If the first dark fringe of one circular diffraction patter passes through the center of a second pattern, the two sources responsible for the patterns will appear to be a single source. Applet

9.5 Resolution Examples: pixels on TV’s and Computer

9.6 Diffraction Gratings

9.6 Diffraction Grating Diffraction Grating – a system with a large number of slits 3D glasses Produced most often by taking pictures of slits and putting them on a slide Produces sharp, widely spaced fringes

9.6 Diffraction Grating Patterns are caused by multiple waves interfering This example shows a maxima produced by 10 slits There would be other areas of constructive interference

9.6 Diffraction Grating Because different colors of light have different wavelengths, and diffract differently We can separate colors Equations are the same as double slits Constructive Destructive