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April 18 Physics 54 Lecture Professor Henry Greenside

Key Points from Previous Lecture. Sign conventions for do, di, f, m, hi. Today's Topics. Chapter 35:

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April 18 Physics 54 Lecture Professor Henry Greenside

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    1. April 18 Physics 54 Lecture Professor Henry Greenside

    2. Key Points from Previous Lecture

    3. Today’s Topics

    4. Chapter 35: Interference of Waves

    5. Huygens’ Principle Explains Diffraction

    6. Huygens' Principle

    7. Applying Huygens’ Principle to Two Slits

    8. To Understand Interference, Understand Superposition of Waves

    9. Interference of Coherent Waves In the Double-Slit Experiment

    10. Light Particles or Waves? Young’s Double-Slit Experiment!

    11. Interference Preconditions

    12. Conditions for Constructive and Destructive Interference of Two Coherent Waves

    13. Two-slit Experiment: Light, Electrons Are Neither Waves Nor Particles Two-slit experiment gives extremely strange results when you shine light of low intensity so that, on average, only a single photon arrives on the screen at a time. A detector always detects a full lump of light, never part of a photon. Yet, as more and more photons arrive, one gets classical interference pattern of coherent waves arriving at the slits. One comes to a crazy conclusion: somehow the lumpy photon or particle must pass through both slits at the same time and interfere with itself! One gets the same results for electrons or, in fact, for any particle. Interference effects like these have been observed for molecules as big as buckyballs, C60, which is the molecule you get when you replace every vertex on a soccer ball with a carbon atom (a marvelous molecule that won a Nobel prize in chemistry for its discoverer, Richard Small).Two-slit experiment gives extremely strange results when you shine light of low intensity so that, on average, only a single photon arrives on the screen at a time. A detector always detects a full lump of light, never part of a photon. Yet, as more and more photons arrive, one gets classical interference pattern of coherent waves arriving at the slits. One comes to a crazy conclusion: somehow the lumpy photon or particle must pass through both slits at the same time and interfere with itself! One gets the same results for electrons or, in fact, for any particle. Interference effects like these have been observed for molecules as big as buckyballs, C60, which is the molecule you get when you replace every vertex on a soccer ball with a carbon atom (a marvelous molecule that won a Nobel prize in chemistry for its discoverer, Richard Small).

    14. Double-slit is predicted to be dispersive, the spacing or angle depends on the wavelength so non-monochromatic light will separate out.Double-slit is predicted to be dispersive, the spacing or angle depends on the wavelength so non-monochromatic light will separate out.

    15. Proving that Light is a Transverse Wave: Two-slit Experiment Plus Two Linear Polarizers

    16. PRS Question: Answer: (2) Let’s look at the first equation d sin(theta) = m lambda, which gives the conditions that an angle theta must satisfy for a constructive maximum (bright fringe) to occur. Then for fixed fringe number m (say m=1) and for fixed wavelength of light lambda striking the slits, the right side is constant. If the distance d between slits decreases, in order for the product “d sin(theta)” to remain constant in value, the quantity sin(theta) must increase. But for angles 0 <= theta <= Pi/2 (in radians, or 0 <= theta <= 90 in degrees), the sine function is a monotonically increasing function. So in order for sin(theta) to become larger, the angles theta themselves must become larger for each choice of fringe number m. So all the angles corresponding to bright fringes, and also all the angles corresponding to dark fringes, become larger. The fringes spread further apart as d is decreased.Answer: (2) Let’s look at the first equation d sin(theta) = m lambda, which gives the conditions that an angle theta must satisfy for a constructive maximum (bright fringe) to occur. Then for fixed fringe number m (say m=1) and for fixed wavelength of light lambda striking the slits, the right side is constant. If the distance d between slits decreases, in order for the product “d sin(theta)” to remain constant in value, the quantity sin(theta) must increase. But for angles 0 <= theta <= Pi/2 (in radians, or 0 <= theta <= 90 in degrees), the sine function is a monotonically increasing function. So in order for sin(theta) to become larger, the angles theta themselves must become larger for each choice of fringe number m. So all the angles corresponding to bright fringes, and also all the angles corresponding to dark fringes, become larger. The fringes spread further apart as d is decreased.

    17. Worked Example at Whiteboard Use small angle approximation sin(theta) ~ theta and use approximation that x1 ~ L theta1, x2 ~ L theta2. So condition d sin(theta1) = m1 lambda becomes d theta1 ~ m1 lambda so x1 ~ L theta1 ~ L (m1 lambda / d) = Use small angle approximation sin(theta) ~ theta and use approximation that x1 ~ L theta1, x2 ~ L theta2. So condition d sin(theta1) = m1 lambda becomes d theta1 ~ m1 lambda so x1 ~ L theta1 ~ L (m1 lambda / d) =

    18. PRS Question: Answer: (2) This is similar to previous PRS question. If wavelength lambda increases, the right side of the first equation d sin(theta) = m lambda increases for a given fringe number m. But for a fixed spacing d, the left side can become bigger only if sin(theta) becomes larger, which in turn implies that theta itself, the angle corresponding to the mth fringe, increases. So all bright fringes, also all dark fringes, spread further apart.Answer: (2) This is similar to previous PRS question. If wavelength lambda increases, the right side of the first equation d sin(theta) = m lambda increases for a given fringe number m. But for a fixed spacing d, the left side can become bigger only if sin(theta) becomes larger, which in turn implies that theta itself, the angle corresponding to the mth fringe, increases. So all bright fringes, also all dark fringes, spread further apart.

    19. Wavelengths of Light From Double-Slit Interference http://www.brantacan.co.uk/WLFringes.jpg is source of left image, interference for five different colors.http://www.brantacan.co.uk/WLFringes.jpg is source of left image, interference for five different colors.

    20. Application of Interference: Thin Films

    21. Thin Film Interference Versus Two-Slit Interference

    22. Whiteboard Discussion of Thin-Film Interference

    23. Section 15-7 on Waves: Reflection

    24. Light Reflection From Denser Media: l/2 Shift In Position of Wave

    25. Measure Small Thicknesses with Air Wedge A subtlety: why do we consider interference at the lower surface of the tilted glass plate, but ignore interference at the upper surface of the tilted plate?A subtlety: why do we consider interference at the lower surface of the tilted glass plate, but ignore interference at the upper surface of the tilted plate?

    26. Interference From Thin Film

    27. Worked Example 35-7: Thickness of Soap Bubble Skin

    28. PRS: How Many Phase Shifts In Example 35-8? Answer: (5) First step is to count the total number of half-wavelength shifts. Light striking the air-MgF2 coating undergoes one shift since the index of refraction increases from air (n=1) to MgF2 (n=1.38). The transmitted ray continues onward until it strikes the second interface, from MgF2 to glass, which gives another shift. Since there is no shift for transmission, the total number of shifts is 2, which means that the shifts cancel each other out. So the criterion for destructive interference is that the total difference in length traveled by rays ni the MgF2 layer, a distance of 2t where t is the thickness of the layer, must be a half-integer multiple of the wavelength of light in the MgF2 layer. So the answer is (5).Answer: (5) First step is to count the total number of half-wavelength shifts. Light striking the air-MgF2 coating undergoes one shift since the index of refraction increases from air (n=1) to MgF2 (n=1.38). The transmitted ray continues onward until it strikes the second interface, from MgF2 to glass, which gives another shift. Since there is no shift for transmission, the total number of shifts is 2, which means that the shifts cancel each other out. So the criterion for destructive interference is that the total difference in length traveled by rays ni the MgF2 layer, a distance of 2t where t is the thickness of the layer, must be a half-integer multiple of the wavelength of light in the MgF2 layer. So the answer is (5).

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