1. Chapter 24: Young’s Experiment Llyod’s Mirror Thin Films

2. Hint: Be able to do the homework (both the problems to turn in AND the recommended ones) you’ll do fine on the exam! Wednesday, February 3, 1999 in class Ch. 22, 23, 24 (Sect. 1-6) You may bring one 3”X5” index card (hand-written on both sides), a pencil or pen, and a scientific calculator with you.

3. Will not be due until MONDAY, February 1.

4. (circa 1800) To demonstrate the interference properties of light waves. The screen is used to produce a coherent, in phase light source for the slits in the second screen. Light arrives at the slits of the second screen in phase. monochromatic light source Viewing Screen

5. monochromatic light source crests troughs bright dark Now we’ve created an experiment for light that is completely analogous to the sound in the field around our outdoor stage

6. Zeroth order maximum 1 st 2 nd 0011 dark fringes So the image on the screen will look like the following:

7. A little geometry will allow us to locate the position of the maxima and minima more precisely: d screen L y r1r1 r2r2  For L >> d, r 1 and r 2 are approximately parallel, so that the difference in path length is . NOT TO SCALE  

8. d screen L y r1r1 r2r2    For small  So,

9. The maxima will occur when  is an integral multiple of the wavelength of the light: The minima will occur when  is an odd integral number of half a wavelength of the incident light: bright fringes dark fringes

11. There is another way we can produce two coherent light sources besides using the double slit... source mirror image

12. But the interference pattern that is observed on the screen is EXACTLY THE OPPOSITE of what we saw in Young’s Experiment!!!! It turns out that light undergoes a phase shift of exactly 180 o (one-half a wavelength) when it bounces off the mirror!

13. In general, any time light reflects off the boundary of a material with a higher index of refraction than that in which the light is incident on the surface, this phase shift of 180 o will occur. mirror, n > 1 incident crest trough reflected Air, n = 1

14. No phase change occurs when light reflects off a boundary leading to a medium with a lower index of refraction. Air, n = 1 Water, n = 1.33 Air, n = 1 Phase shift No phase shift

15. The closer to normal the incident light is on the top surface of the layer of water, the closer the reflected rays from the top and the bottom surface will be. Air, n = 1 Water, n = 1.33 Air, n = 1 Phase shift No phase shift

16. If the reflected rays lie on top of one another (as will be the case if the light is normally incident on the top surface), the reflected rays can interfere with one another! Air, n = 1 Water, n = 1.33 Air, n = 1 t The path length difference is 2 t

17. So, if the path difference ( 2 t ) is an integral multiple of the wavelength of the light, we will get destructive interference above the top surface! Here, the relevant wavelength of the light is its wavelength in the medium!

18. For the case of a thin film of index of refraction n surrounded by air, the maxima and minima occur under the following conditions Constructive interference Destructive interference

19. Air, n = 1 Water, n = 1.33 Glass, n = 1.5 t In this case, the blue and red rays suffer a 180 o phase change, so the conditions for maxima and minima are exactly the opposite of what appears on the last slide!