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## PROGRAM OF “PHYSICS”

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** PROGRAM OF “PHYSICS”**Lecturer:Dr. DO Xuan Hoi Room 413 E-mail :dxhoi@hcmiu.edu.vn**PHYSICS 4 (Wave and Modern Physics)**02 credits (30 periods) Chapter 1 Mechanical Wave Chapter 2 Properties of Light Chapter 3 Introduction to Quantum Physics Chapter 4 Atomic Physics Chapter 5 Relativity and Nuclear Physics**References :**Halliday D., Resnick R. and Walker, J. (2005), Fundamentals of Physics, Extended seventh edition. John Willey and Sons, Inc. Alonso M. and Finn E.J. (1992). Physics, Addison-Wesley Publishing Company Hecht, E. (2000). Physics. Calculus, Second Edition. Brooks/Cole. Faughn/Serway (2006), Serway’s College Physics, Brooks/Cole. Roger Muncaster (1994), A-Level Physics, Stanley Thornes.**http://ocw.mit.edu/OcwWeb/Physics/index.htm**http://www.opensourcephysics.org/index.html http://hyperphysics.phy-astr.gsu.edu/hbase/HFrame.html http://www.practicalphysics.org/go/Default.html http://www.msm.cam.ac.uk/ http://www.iop.org/index.html . . .**PHYSICS 4**Chapter 2Properties of Light A. WAVE OPTICS The Nature of Light Interference of Light Waves Diffraction Patterns Polarization B. GEOMETRIC OPTICS Light Rays The Laws of Reflection and Refraction Mirrors Thin Lenses**A. WAVE OPTICS**The Nature of Light Interference of Light Waves Diffraction Patterns Polarization**1 The Nature of Light**1.1 Dual nature of light • In some cases light behaves like a wave(classical E & M – light propagation) • In some cases light behaves like a particle (photoelectric effect) • Einstein formulated theory of light: Planck’s constant**1.2 Huygens’s Principle** Light travels at 3.00 x 108 m/s in vacuum (travels slower in liquids and solids) In order to describe propagation: Huygens method : All points on given wave front taken as point sources for propagation of spherical waves Assume wave moves through medium in straight line in direction of rays**2 Interference of Light Waves**2.1 Conditions for interference • light sources must be coherent (must maintain a constant phase with each other) • sources must have identical wavelength • superposition principle must apply**2.2 Young’s Double-Slit Experiment**• Setup: light shines at the plane with two slits • Result: a series of parallel dark and bright bands called fringes**Opposite phases**In phases**M**d1 d2 S1 S2 Interference of Sinusoidal Waves Amplitude of y depends on the path difference : = d2 - d1 Amplitude of y maximum : Amplitude of y equal to 0 :** The path difference in**Young’s Double-Slit Experiment d1 d2** Positions of fringes on the screen**Bright fringes (constructive interference) : Dark fringes (destructive interference) :** Angular Positions of Fringes**Bright regions (constructive interference) : Dark regions (destructive interference) :**A viewing screen is separated from a double-**lit source by 1.2 m. The distance between the two slits is 0.030 mm. The second-order bright fringe is 4.5 cm from the center line. (a)Determine the wavelength of the light. PROBLEM 1 SOLUTION (a) The position of second-order bright fringe :**A viewing screen is separated from a double-**lit source by 1.2 m. The distance between the two slits is 0.030 mm. The second-order bright fringe is 4.5 cm from the center line. (b) Calculate the distance between adjacent bright fringes. PROBLEM 1 SOLUTION (b)**A light source emits visible light of two**wavelengths: 430 nm and 510 nm. The source is used in a double-slit interference experiment in which L = 1.5 m and d = 0.025 mm. Find the separation distance between the third order bright fringes. PROBLEM 2 SOLUTION**PROBLEM 3**SOLUTION**2.3 Intensity distribution of the double-slit.**Interference pattern • light : electromagnetic wave • suppose that the two slits represent coherent sources of sinusoidal waves : the two waves from the slits have the same angular frequency • assume that the two waves have the same amplitudeE0 of the electric field at point P d1 d2**At point P :**Amplitude of E : d1 d2**At point P :**Amplitude of E : The intensity of a wave is proportional to the square of the electric field magnitude at that point :**PROBLEM 4**SOLUTION**I**PROBLEM 4 SOLUTION**PROBLEM 4**SOLUTION**I**PROBLEM 4 SOLUTION**2.4 INTERFERENCE IN THIN FILMS**• Interference effects are commonly observed in thin films, such as thin layers of oil on water or the thin surface of a soap bubble • The varied colors observed when white light is incident on such films result from the interference of waves reflected from the two surfaces of the film**a. Change of phase due to reflection** Interference pattern with Lloyd’s mirror P’ is equidistant from points S and S : the path difference is zero bright fringe at P’ ? We observe a dark fringe at P’ There a 180° phase change produced by reflection CONCLUSION : an electromagnetic wave undergoes a phase change of 180° upon reflection from a medium that has a higher index of refraction than the one in which the wave is traveling.** Change of phase due to reflection**180° phase change No phase change If is the wavelength of the light in free space and if the medium has the refraction index n : The wavelength of light in this medium :**b. Interference in thin films**• Consider a film of uniform thickness t and index of refraction n. Assume that the light rays traveling in air are nearly normal to the two surfaces of the film. • Reflected ray 1 : phase change of 180° with respect to the incident wave. • Reflected ray 2: no phase change • Ray 1 is 180° out of phase with ray 2 a path difference of n /2. • ray 2 travels an extra distance 2t before the waves recombine in the air**Ray 1 is 180° out of phase with ray 2 a path difference**of n /2. • ray 2 travels an extra distance 2t before the waves recombine in the air Condition for constructive interference : Condition for destructive interference :**Calculate the minimum thickness of a soap-**bubble (n = 1.33) film that results in constructive interference in the reflected light if the film is illuminated with light whose wavelength in free space is = 600 nm. PROBLEM 5 SOLUTION Constructive interference : The minimum film thickness : m = 0**A thin, wedge-shaped film of refractive index**n is illuminated with monochromatic light of wavelength , as illustrated in the figure. Describe the interference pattern observed for this case. PROBLEM 6 SOLUTION Dark fringes : Bright fringes :**A thin, wedge-shaped film of refractive index**n is illuminated with monochromatic light of wavelength , as illustrated in the figure. Describe the interference pattern observed for this case. PROBLEM 6 SOLUTION White light : Dark fringes : Bright fringes :**Suppose the two glass plates in the figure are two**microscope slides 10.0 cm long. At one end they are in contact; at the other end they are separated by a piece of paper 0.0200 mm thick. What is the spacing of the interference fringes seen by reflection? Is the fringe at the line of contact bright or dark? Assume monochromatic light with a wavelength in air of = 500nm. PROBLEM 7 SOLUTION Dark fringes :**Suppose the two glass plates in the figure are two**microscope slides 10.0 cm long. At one end they are in contact; at the other end they are separated by a piece of paper 0.0200 mm thick. What is the spacing of the interference fringes seen by reflection? Is the fringe at the line of contact bright or dark? Assume monochromatic light with a wavelength in air of = 500nm. Suppose the glass plates have n = 1.52 and the space between plates contains water (n = 1.33) instead of air. What happens now? PROBLEM 7 SOLUTION In the film of water (n = 1.33), the wavelength :**Suppose the two glass plates in the figure are two**microscope slides 10.0 cm long. At one end they are in contact; at the other end they are separated by a piece of paper 0.0200 mm thick. What is the spacing of the interference fringes seen by reflection? Is the fringe at the line of contact bright or dark? Assume monochromatic light with a wavelength in air of = 500nm. Suppose the upper of the two plates is a plastic material with n = 1.40, the wedge is filled with a silicone grease having n = 1.50, and the bottom plate is a dense flint glass with n = 1.60. What happens now? PROBLEM 8 SOLUTION**SOLUTION**The wavelength in the silicone grease: The two reflected waves from the line of contact are in phase (they both undergo the same phase shift), so the line of contact is at a bright fringe.**c. Newton’s Rings**The convex surface of a lens in contact with a plane glass plate. A thin film of air is formed between the two surfaces. View the setup with monochromatic light : Circular interference fringes Ray 1 : a phase change of 180° upon reflection; ray 2 : no phase change the conditions for constructive and destructive interference**2.5 THE MICHELSON INTERFEROMETER** Principle : splits a light beam into two parts and then recombines the parts to form an interference pattern Use :To measure wavelengths or other lengths with great precision The observer sees an interference pattern that results from the difference in path lengths for rays 1 and 2.**3. Light Diffraction**3.1 Introduction If a wave encounters a barrier that has an opening of dimensions similar to the wavelength, the part of the wave that passes through the opening will spread out(diffract) into the region beyond the barrier That is another proof of wave nature Diffraction Phenomenon WAVE Light produces also the diffraction: a light (with appropriate wavelength) beam going through a hole gives a number of circular fringes**3.2 Fraunhofer diffraction**All the rays passing through a narrow slit are approximately parallel to one another.** Diffraction from narrow slit**Huygens’s principle :Each portion of the slit acts as a source of light waves Divide the slit into two halves Rays 1 and 3 : ray 1 travels farther than ray 3 by an amount equal to the path difference (a/2) sin Rays 2 and 4 : ray 2 travels farther than ray 4 by an amount equal to the path difference (a/2) sin Waves from the upper half of the slit interfere destructively with waves from the lower half when :** Divide the slit into two halves** Divide the slit into four parts : Dark fringes : Divide the slit into six parts : Dark fringes :**Divide the slit into two halves :**Divide the slit into four parts : Dark fringes : Divide the slit into six parts : Dark fringes : The general condition for destructive interference : Positions of dark fringes :**The width of the central bright fringe :**Light of wavelength 580 nm is incident on a slit having a width of 0.300 mm. The viewing screen is 2.00 m from the slit. Find the positions of the first dark fringes and the width of the central bright fringe. PROBLEM 9 SOLUTION The two dark fringes that flank the central bright fringe : m = 1**You pass 633-nm laser light through a narrow**slit and observe the diffraction pattern on a screen 6.0 m away. You find that the distance on the screen between the centers of the first minima outside the central bright fringe is 32 mm. How wide is the slit? PROBLEM 10 SOLUTION Dark fringes : The first minimum : m = 1** Intensity of Single-Slit Diffraction Patterns**The intensity at each point on the screen : where Imax is the intensity at = 0 (the central maximum). We see that minima occur when : (The general condition for destructive interference) a sin/