1. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley PowerPoint ® Lectures for University Physics, Twelfth Edition – Hugh D. Young and Roger A. Freedman Lectures by James Pazun Chapter 33 The Nature and Propagation of Light

2. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Introduction A coating of oil on water or a delicate glass prism can create a rainbow. A rainstorm among open patches of daylight can cast a conventional rainbow. Both effects are beautiful and arise from the wavelength dependence of refraction angles. Eyeglasses or contact lenses both use refraction to correct imperfections in the eyeball’s focus on the retina and allow vision correction.

3. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley “Light is a wave,” “Light is a particle” The wave–particle duality of light was not well understood until Albert Einstein won his Nobel Prize in the early 20th century. It was a tenacious understanding of light that led to quantum mechanics and modern physics. Consider Figures 33.1 and 33.2 below. Black-body and laser radiation look so different but are brother and sister.

4. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Wave fronts and rays Light is actually a nearly uncountable number of electromagnetic wave fronts, but analysis of refraction or reflection are made possible by treating light as a ray.

5. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Reflection and refraction Figure 33.5 illustrates both reflection and refraction at once. The storefront window both shows the passersby their reflections and allows them to see inside.

6. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley When light passes from vacuum (index of refraction n = 1) into water (n = 1.333), Q33.1 A. the wavelength increases and the frequency is unchanged. B. the wavelength decreases and the frequency is unchanged. C. the wavelength is unchanged and the frequency increases. D. the wavelength is unchanged and the frequency decreases. E. both the wavelength and the frequency change.

7. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley When light passes from vacuum (index of refraction n = 1) into water (n = 1.333), A33.1 A. the wavelength increases and the frequency is unchanged. B. the wavelength decreases and the frequency is unchanged. C. the wavelength is unchanged and the frequency increases. D. the wavelength is unchanged and the frequency decreases. E. both the wavelength and the frequency change.

8. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley We will consider specular reflections A real surface will scatter and reflect light. Diffuse reflection is the rule, not the exception. We will use specular reflection as we used the ray approximation, to make a very difficult problem manageable. Consider Figure 33.6.

9. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Laws of reflection and refraction Angle of incidence = angle of reflection. Snell’s Law of Refraction considers the slowing of light in a medium other than vacuum … the index of refraction. Consider Figures 33.7 and 33.8.

10. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Light passes from vacuum (index of refraction n = 1) into water (n = 1.333). If the incident angle  a is in the range 0° <  a < 90°, Q33.2 A. the refracted angle is greater than the incident angle. B. the refracted angle is equal to the incident angle. C. the refracted angle is less than the incident angle. D. the answer depends on the specific value of  a.

11. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Light passes from vacuum (index of refraction n = 1) into water (n = 1.333). If the incident angle  a is in the range 0° <  a < 90°, A33.2 A. the refracted angle is greater than the incident angle. B. the refracted angle is equal to the incident angle. C. the refracted angle is less than the incident angle. D. the answer depends on the specific value of  a.

12. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Light passes from a medium of index of refraction n a into a second medium of index of refraction n b. The angles of incidence and refraction are  a and  b respectively. If n a < n b, Q33.3  a >  b and the light speeds up as it enters the second medium. B.  a >  b and the light slows down as it enters the second medium. C.  a <  b and the light speeds up as it enters the second medium. D.  a <  b and the light slows down as it enters the second medium.

13. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Light passes from a medium of index of refraction n a into a second medium of index of refraction n b. The angles of incidence and refraction are  a and  b respectively. If n a < n b, A33.3  a >  b and the light speeds up as it enters the second medium. B.  a >  b and the light slows down as it enters the second medium. C.  a <  b and the light speeds up as it enters the second medium. D.  a <  b and the light slows down as it enters the second medium.

14. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Why should the ruler appear to be bent? The difference in index of refraction for air and water causes your eye to be deceived. Your brain follows rays back to the origin they would have had if not bent. Consider Figure 33.9.

15. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Why should sunsets be orange and red? The light path at sunset is much longer than at noon when the sun is directly overhead. Consider Figure 33.10.

16. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Tabulated indexes of refraction

17. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Goldfish You are looking at a goldfish in a fish tank from the top. It appears that the fish is 30 degrees below the horizontal. Where is the fish?

18. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Total internal reflection I As the angle of incidence becomes more and more acute, the light ceases to be transmitted, only reflected. Consider Figure 33.13.

19. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Total internal reflection II Using clever arrangements of glass or plastic, the applications are mind boggling.

20. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Q33.4 Light passes from a medium of index of refraction n a into a second medium of index of refraction n b. The critical angle for total internal reflection is  crit. In order for total internal reflection to occur, what must be true about n a, n b, and the incident angle  a ? A. n a > n b and  a >  crit B. n a > n b and  a <  crit C. n a  crit D. n a < n b and  a <  crit

21. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Light passes from a medium of index of refraction n a into a second medium of index of refraction n b. The critical angle for total internal reflection is  crit. In order for total internal reflection to occur, what must be true about n a, n b, and the incident angle  a ? A33.4 A. n a > n b and  a >  crit B. n a > n b and  a <  crit C. n a  crit D. n a < n b and  a <  crit

22. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Goldfish 2 Describe what a goldfish sees if another fish swims directly overhead and moves past the angle of total internal reflection. Where is this point of total internal reflection? Is there a spot where you cannot see the goldfish looking down from the top of the tank? Is there a way where you can see the goldfish, but the fish can’t see you?

23. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Dispersion From the discussion of the prism seen in earlier slides, we recall that light refraction is wavelength dependent. This effect is made more pronounced if the index of refraction is higher. “Making a rainbow” is actually more than just appreciation of beauty; applied to chemical systems, the dispersion of spectral lines can be a powerful identification tool. Refer to Figures 33.18 and 33.19 (not shown).

24. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Special case—dispersion and atmospheric rainbows As a person looks into the sky and sees a rainbow, he or she is actually “receiving light signals” from a physical spread of water droplets over many meters (or hundreds of meters) of altitude in the atmosphere. The reds come from the higher droplets and the blues from the lower (as we have seen in the wavelength dependence of light refraction).

25. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Selecting one orientation of the EM wave—the Polaroid A Polaroid filter is a polymer array that can be thought of like teeth in a comb. Hold the comb at arm’s length with the teeth pointing down. Continue the mental cartoon and imagine waves oscillating straight up and down passing without resistance. Any “side-to-side” component and they would be blocked.

26. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Polarization I Read Problem-Solving Strategy 33.2. Follow Example 33.5. By reflection from a surface? Read pages 1139 and 1140 and then refer to Figure 33.27 below.

27. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Polarization II Consider Figure 33.28. Follow Example 33.6, illustrated by Figure 33.29.

28. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Scattering of light The observed colors in the sky depend on the scattering phenomenon. Deep blue sky comes from preferential scattering of photons of shorter wavelength in the visible spectrum. Clouds are white because they scatter all wavelengths efficiently. Refer to Figures 33.32 and 33.33 (not shown).

29. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Huygens’s Principle From the work of Christian Huygens in 1678, the geometrical analysis reveals that every point of a wave front can be considered to be a source of secondary wavelets that spread with a speed equal to the speed of propagation of the wave. Consider Figures 33.34 and 33.35.

30. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Huygens’s Principle II Huygens’s work can form an explanation of reflection and refraction. Refer to Figures 33.36 and 33.37.